PAS MATEMATIKA BLOG

Koordinat Kartesius dan Koordinat Kutub serta Kuadran

C. Koordinat Kartesius dan Koordinat Kutub/Polar

Perhatikan ilustrasi berikut

$\begin{array}{|c|c|}\hline \textrm{Kartesius}\: \rightarrow \: \textrm{Kutub}&\textrm{Kutub}\: \rightarrow \: \textrm{Kartesius}\\\hline P(x,y)\: \rightarrow \: P\left ( r,\alpha ^{0} \right )&P\left ( r,\alpha ^{0} \right )\: \rightarrow \: P(x,y)\\\hline \begin{aligned}&r=\sqrt{x^{2}+y^{2}},\\ &\displaystyle \tan \alpha ^{0}=\frac{y}{x},\quad \displaystyle \alpha ^{0}=\arctan \frac{y}{x} \end{aligned}&\left\{\begin{matrix} x=r.\cos \alpha ^{0}\\ \\ y=r.\sin \alpha ^{0} \end{matrix}\right.\\\hline \end{array}$.

D. Perbandingan Trigonometri di Berbagai Kuadran

D. 1 Untuk Sudut 0 sampai dengan 360 derajat.

$\begin{array}{ccc|cccc} \textrm{Kuadran II}&&&&\textrm{Kuadran I}&\\ \left (180^{\circ}-\alpha \right )&&&&\textrm{Semua nilai trigon}&\color{blue}\textbf{positif}\\ &&&&&\\\hline \textrm{Kuadran III}&&&&\textrm{Kuadran IV}&\\ \left (180^{\circ}+\alpha \right )&&&&\left (360^{\circ}-\alpha \right )& \\ \end{array}$.

$\begin{array}{llll} (1).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 90^{0}-\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 90^{0}-\theta \right )=\cos \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 90^{0}-\theta \right )=\sin \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 90^{0}-\theta \right )=\cot \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 90^{0}-\theta \right )=\tan \theta \end{array}$.

$\begin{array}{llll} (2).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 180^{0}-\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 180^{0}-\theta \right )=\sin \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 180^{0}-\theta \right )=-\cos \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 180^{0}-\theta \right )=-\tan \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 180^{0}-\theta \right )=-\cot \theta \end{array}$.

$\begin{array}{llll} (3).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 180^{0}+\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 180^{0}+\theta \right )=-\sin \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 180^{0}+\theta \right )=-\cos \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 180^{0}+\theta \right )=\tan \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 180^{0}+\theta \right )=\cot \theta \end{array}$.

$\begin{array}{llll} (4).&\textrm{Perbandingan Trigonometri untuk Sudut}\: \left ( 360^{0}-\alpha \right )\\ &\textrm{a}.\quad \sin \alpha \rightarrow \sin \left ( 360^{0}-\theta \right )=-\sin \theta \\ &\textrm{b}.\quad \cos \alpha \rightarrow \cos \left ( 360^{0}-\theta \right )=\cos \theta \\ &\textrm{c}.\quad \tan \alpha \rightarrow \tan \left ( 360^{0}-\theta \right )=-\tan \theta \\ &\textrm{d}.\quad \cot \alpha \rightarrow \cot \left ( 360^{0}-\theta \right )=-\cot \theta \end{array}$.

D. 2 Untuk Sudut yang Lain

$\begin{aligned}\textrm{a}.\quad&\begin{cases} \sin \left ( -A \right ) & =-\sin A \\ \cos \left ( -A \right ) & =\cos A \\ \tan \left ( -A \right ) & = -\tan A \end{cases}\\ \textrm{b}.\quad&\begin{cases} \csc \left ( -A \right ) &=-\csc A \\ \sec \left ( -A \right ) &=\sec A \\ \cot \left ( -A \right ) &=-\cot A \end{cases}\\ \textrm{c}.\quad&\begin{cases} \sin \left ( n.360^{\circ}+A \right ) & =\sin A \\ \cos \left ( n.360^{\circ}+A \right ) & =\cos A \\ \tan \left ( n.360^{\circ}+A \right ) & =\tan A \end{cases},\qquad n\in \mathbb{N} \end{aligned}$.

Dengan catata: $0^{\circ}=360^{\circ} =720^{\circ}=1080^{\circ}=n.360^{\circ}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai}\\ &\textrm{a}.\quad\sin 120^{\circ}\\ &\textrm{b}.\quad\cos 240^{\circ}\\ &\textrm{c}.\quad\tan 315^{\circ}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad\sin 120^{\circ}&=\sin \left ( 180^{\circ}-60^{\circ} \right )=\sin 60^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{3},\qquad \color{red}\textrm{atau}\\ &=\sin \left ( 90^{\circ}+30^{\circ} \right )=\cos 30^{\circ}=\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.\quad\cos 240^{\circ}&=\cos \left ( 180^{\circ}+60^{\circ} \right) =-\cos 60^{\circ}\\ &=-\displaystyle \frac{1}{2},\qquad \color{red}\textrm{atau}\\ &=\cos \left ( 270^{\circ}-30^{\circ} \right )=-\sin 30^{\circ}=-\frac{1}{2}\\ \textrm{c}.\quad\tan 315^{\circ}&=\tan \left ( 360^{\circ}-45^{\circ} \right )=-\tan 45^{\circ}\\ &=-1,\qquad \color{red}\textrm{atau}\\ &=\tan \left ( 270^{\circ}+45^{\circ} \right )=-\cot 45^{\circ}=-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Buktikan bahwa}\\\\ &\textrm{a}.\quad \displaystyle \frac{\cos \left ( 90^{\circ}-B \right )}{\sec B}+\frac{\sin \left ( 90^{\circ}-B \right )}{\csc B}=2\sin B\cos B\\\\ &\textrm{b}.\quad \tan C+\tan \left ( 90^{\circ}-C \right )=\sec C.\sec \left ( 90^{\circ}-C \right )\\\\ &\textbf{Bukti}:\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{\cos \left ( 90^{\circ}-B \right )}{\sec B}+\frac{\sin \left ( 90^{\circ}-B \right )}{\csc B}\\ &=\displaystyle \frac{\sin B}{\sec B}+\frac{\cos B}{\csc B}\\ &=\displaystyle \frac{\sin B}{\displaystyle \frac{1}{\cos B}}+\frac{\cos B}{\displaystyle \frac{1}{\sin B}}\\ &=\sin B\cos B+\sin B\cos B\\ &=2\sin B\cos B\qquad\quad \blacksquare \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad&\tan C+\tan \left ( 90^{\circ}-C \right )\\ &=\tan C+\cot C\\ &=\displaystyle \frac{\sin C}{\cos C}+\frac{\cos C}{\sin C}\\ &=\displaystyle \frac{\sin ^{2}C+\cos ^{2}C}{\sin C\cos C}=\displaystyle \frac{1}{\sin C\cos C}\\ &=\displaystyle \frac{1}{\cos C}.\frac{1}{\sin C}\\ &=\sec C.\csc C\\ &=\sec C.\sec \left ( 90^{\circ}-C \right )\qquad\quad \blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah nilai}\\ &\textrm{a}.\quad \tan \left ( A-90^{\circ} \right )\sin \left ( -A \right )\\ &\textrm{b}.\quad\cos 540^{\circ}+\sin 690^{\circ}\\ &\textrm{c}.\quad \sin 2021^{\circ}+\cos 2021^{\circ}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\tan \left ( A-90^{\circ} \right )\sin \left ( -A \right )\\ &=\tan \left ( -\left (90^{\circ}-A \right ) \right )\left ( -\sin A \right )\\ &=-\tan \left ( 90^{\circ}-A \right )\left ( -\sin A \right )\\ &= \tan \left ( 90^{\circ}-A \right )\left ( \sin A \right )\\ &=\cot A.\sin A\\ &=\displaystyle \frac{\cos A}{\sin A}.\sin A\\ &=\cos A \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad&\cos 540^{\circ}+\sin 690^{\circ}\\ &=\cos \left ( 360^{\circ}+180^{\circ} \right )+\sin \left ( 720^{\circ}-30^{\circ} \right )\\ &=\cos \left ( 0^{\circ}+180^{\circ} \right )+\sin \left ( 0^{\circ}-30^{\circ} \right )\\ &=\cos 180^{\circ}+ \sin \left ( -30^{\circ} \right ) \\ &=\cos 180^{\circ}-\sin 30^{\circ}\\ &=-1-\displaystyle \frac{1}{2}\\ &=-\displaystyle \frac{3}{2} \end{aligned} \\ &\begin{aligned}\textrm{c}.\quad&\sin 2021^{\circ}+\cos 2021^{\circ}\\ &=\sin \left ( 5.360^{\circ}+221^{\circ} \right )+\cos \left ( 5.360^{\circ}+221^{\circ} \right )\\ &=\sin \left (0^{\circ}+221^{\circ} \right )+\cos \left (0^{\circ}+221^{\circ} \right )\\ &=\sin 221^{\circ}+\cos 221^{\circ}\\ &=\sin \left ( 180^{\circ}+41^{\circ} \right )+\cos \left ( 180^{\circ}+41^{\circ} \right )\\ &=-\sin 41^{\circ}-\cos 41^{\circ} \end{aligned} \end{array}$

Perbandingan Trigonometri Sudut dalam Segitiga Siku-Siku

B. Perbandingan Trigonometri Suatu Sudut

B. 1 Perbandingan Trigonometri pada Segitiha Siku-Siku

Perhatikanlah ilustrasi dari segitiga ABC siku-siku di C berikut

$\begin{matrix} \sin \alpha =\displaystyle \frac{BC}{AB}\qquad\Leftrightarrow\quad \csc \alpha =\displaystyle \frac{AB}{BC}=\color{red}\displaystyle \frac{1}{\sin \alpha }\\\\ \cos \alpha =\displaystyle \frac{AC}{AB}\qquad\Leftrightarrow \quad \sec \alpha =\displaystyle \frac{AB}{AC}=\color{red}\displaystyle \frac{1}{\cos \alpha }\\\\ \tan \alpha =\displaystyle \frac{BC}{AC}\qquad\Leftrightarrow \quad \cot \alpha =\displaystyle \frac{AC}{BC}=\color{red}\displaystyle \frac{1}{\tan \alpha } \end{matrix}$.

B. 2 Perbandingan Trigonometri untuk Sudut Istimewa.

$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \alpha ^{0}&0^{0}&30^{0}&45^{0}&60^{0}&90^{0}&180^{0}&270^{0}&360^{0}\\\hline \sin \alpha ^{0}&0&\color{red}\displaystyle \frac{1}{2}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}\sqrt{3}&1&0&-1&0\\\hline \cos \alpha ^{0}&1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\color{red}\displaystyle \frac{1}{2}&0&-1&0&1\\\hline \tan \alpha ^{0}&0&\displaystyle \frac{1}{3}\sqrt{3}&1&\sqrt{3}&TD&0&TD&0\\\hline \end{array}$.

B. 3. Tripel Pythagoras pada Segitiga Siku-Siku

$\begin{array}{|c|c|c|c|}\hline a&b&c=\sqrt{a^{2}+b^{2}}&\color{red}\textrm{Tripel}\\\hline 3&4&5&(3,4,5)\\\hline 5&12&13&(5,12,13)\\\hline 7&24&25&(7,24,25)\\\hline 8&15&17&(8,15,17)\\\hline 9&40&41&(9,40,41)\\\hline 11&60&61&(11,60,61)\\\hline 12&35&37&(12,35,37)\\\hline 15&112&113&(15,112,113)\\\hline 16&63&65&(16,63,65)\\\hline 17&144&145&(17,144,145)\\\hline 19&180&181&(19,180,181)\\\hline 20&21&29&(20,21,29)\\\hline 21&220&221&(21,220,221)\\\hline 29&420&421&(29,420,421)\\\hline 30&224&226&(30,224,226)\\\hline 31&480&481&(31,480,481)\\\hline 33&544&545&(33,544,545)\\\hline 35&612&613&(35,612,613)\\\hline 37&684&685&(37,684,685)\\\hline 39&760&761&(39,760,761)\\\hline 41&840&841&(41,840,841)\\\hline 43&924&925&(43,924,925)\\\hline 45&1012&1013&(45,1012,1013)\\\hline 47&1104&1105&(47,1104,1105)\\\hline 48&55&73&(48,55,73)\\\hline 49&1200&1201&(49,1200,1201)\\\hline 51&1300&1301&(51,1300,1301)\\\hline 60&63&87&(60,63,87)\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai perbandingan}\: \: \color{red}\sin \alpha \\ & \color{red}\cos \alpha , \tan \alpha \: \: \color{black}\textrm{untuk segitiga berikut} \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\textrm{Untuk sisi miringnya adalah}\\ &\begin{aligned}\textrm{Sisi miring}&=\sqrt{3^{2}+4^{2}}\\ &=\sqrt{9+16}\\ &=\sqrt{25}\\ &=5 \end{aligned}\\ &\textrm{Sehingga}\\ &\color{red}\sin \alpha =\displaystyle \frac{3}{5},\: \: \: \cos \alpha =\frac{4}{5}\: \: \color{black}\textrm{dan}\: \color{red}\tan \alpha =\frac{3}{4}\\ &\textrm{Dengan langkah sama}\\ \end{aligned}\\&\begin{aligned}\textrm{b}.\quad&\textrm{Untuk sisi miringnya adalah}\\&\begin{aligned}\textrm{Sisi miring}&=\sqrt{5^{2}+12^{2}}\\&=\sqrt{25+144}\\&=\sqrt{169}\\ &=13 \end{aligned}\\ &\textrm{Sehingga}\\ &\color{red}\sin \alpha =\frac{12}{13}\: \: \cos \alpha =\frac{5}{13}\: \: \color{black}\textrm{dan}\: \color{red}\tan \alpha =\frac{12}{5}. \end{aligned} \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai dari}\\ &\begin{array}{llll} \textrm{a}.&\left ( \tan 30^{0}+\sin 30^{0} \right )\cos 30^{0}\\ \textrm{b}.&\left ( \tan 60^{0} \right )^{2}+4\left ( \sin 60^{0} \right )^{2}\\ \textrm{c}.&\tan 60^{0}-\sin 60^{0}-\tan 30^{0}\\ \textrm{d}.&\displaystyle \frac{1+\sin 30^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 30^{0}}\\ \textrm{e}.&\displaystyle \frac{2\tan 30^{0}}{1+\tan ^{2}30^{0}} \end{array}\\\\ &\textbf{Jawab}:\\ \end{array}$.

$.\: \qquad\begin{array}{|l|l|}\hline \begin{array}{llll} \textrm{a}.&\left ( \tan 30^{0}+\sin 30^{0} \right )\cos 30^{0}\\ &=\displaystyle \left ( \frac{1}{3}\sqrt{3}+\frac{1}{2} \right ).\frac{1}{2}\sqrt{3}\\ &=\displaystyle \frac{1}{2}+\frac{1}{4}\sqrt{3} \end{array}&\begin{array}{llll} \textrm{b}.&\left ( \tan 60^{0} \right )^{2}+4\left ( \sin 60^{0} \right )^{2}\\ &\displaystyle =\left ( \sqrt{3} \right )^{2}+4\left ( \frac{1}{2}\sqrt{3} \right )^{2}\\ &\displaystyle =3+3\\ &=6 \end{array}\\\hline \begin{array}{llll} \textrm{c}.&\tan 60^{0}-\sin 60^{0}-\tan 30^{0}\\ &=\displaystyle \sqrt{3}-\frac{1}{2}\sqrt{3}-\frac{1}{3}\sqrt{3}\\ &=\displaystyle \sqrt{3}-\frac{5}{6}\sqrt{3}\\&=\displaystyle \frac{1}{6}\sqrt{3} \end{array}&\begin{array}{llll} \textrm{d}.&\displaystyle \frac{1+\sin 30^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 30^{0}}\\ &=\displaystyle \frac{1+\frac{1}{2}}{\frac{1}{2}}+\frac{\frac{1}{2}\sqrt{3}}{1+\frac{1}{2}}\\ &=\displaystyle 3+\frac{1}{3}\sqrt{3} \end{array}.\\\hline \begin{array}{llll} \textrm{e}.&\displaystyle \frac{2\tan 30^{0}}{1+\tan ^{2}30^{0}}\\ &=\displaystyle \frac{2\times \frac{1}{3}\sqrt{3}}{1+\left ( \frac{1}{3}\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{\frac{2}{3}\sqrt{3}}{1+\frac{1}{3}}\\ &=\displaystyle \frac{\frac{2}{3}\sqrt{3}}{\frac{4}{3}}\\ &=\displaystyle \frac{1}{2}\sqrt{3} \end{array}.&\\\hline \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah ilustrasi berikut}\\ \end{array}$

$.\: \qquad\begin{aligned}&\textrm{Jika Jarak antara kucing seorang pencatat}\\ &\textrm{dan kucing adalah 100 m, maka jarak}\\ &\textrm{Pencatat tersebut dengan seorang tentara}\\ &\textrm{sebagaimana gambar tersebut di atas adalah}?\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan gambar di atas dengan diberikan}\\ &\textrm{tambahan keterangan sebagai berikut} \end{aligned}$.

$.\: \qquad\begin{aligned}&\textrm{Ditanya berpakah panjang jarak } \left ( x+100 \right ) ?\\ &\begin{aligned}\quad&\textrm{Pilih}\: :\qquad \color{blue}y=y\\ &\Leftrightarrow x.\tan 60^{0}=\left ( x+100 \right ).\tan 30^{0}\\ &\Leftrightarrow x.\sqrt{3}=\left ( x+100 \right )\frac{1}{3}\sqrt{3}\\ &\Leftrightarrow 3x=x+100\\ &\Leftrightarrow 2x=100\\ &\Leftrightarrow x=50 \end{aligned}\\ &\textrm{Jadi}\: \: x+100=50+100=\color{red}150 \: \: \color{black}\textrm{meter}. \end{aligned}$.

$\begin{array}{ll}\\ 4. &\textrm{Tentukanlah perbandingan trigonometri}\\ &\quad \angle XOA\: \: \textrm{jika}\: \: A(3,5)\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

$\: \qquad\begin{aligned}&\textrm{Dengan memandang ilustrasi gambar di atas}\\ &\textrm{kita mendapatkan}\: \: \triangle OAA',\: \: \textrm{dengan menggunakan}\\ &\textrm{teorema pythagoras kita mendapatkan}\\ &\begin{aligned}OA^{2}&=\left ( OA' \right )^{2}+\left ( AA' \right )^{2}\\ &=3^{2}+5^{2}\\ &=9+25\\ &=34\\ OA&=\sqrt{34} \end{aligned}.\\ &\textrm{Sehingga akan didapatkan}\\ &\begin{array}{llll} \textrm{a}.&\displaystyle \sin A'OA=\frac{5}{\sqrt{34}}=\frac{5}{34}\sqrt{34}\\ \textrm{b}.&\displaystyle \cos A'OA=\frac{3}{\sqrt{34}}=\frac{3}{34}\sqrt{34}\\ \textrm{c}.&\displaystyle \tan A'OA=\frac{5}{3} \end{array}. \end{aligned}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah nilai dari}\\ &\begin{array}{llll} \textrm{a}.&\left ( \tan 60^{0}+\sin 45^{0} \right )\cos 0^{0}\\ \textrm{b}.&\left ( \tan 30^{0} \right )^{2}+4\left ( \sin 30^{0} \right )^{2}\\ \textrm{c}.&\tan 60^{0}+\sin 60^{0}+\tan 30^{0}\\ \textrm{d}.&\displaystyle \frac{1+\sin 45^{0}}{\sin 30^{0}}+\frac{\cos 30^{0}}{1+\sin 45^{0}}\\ \textrm{e}.&\displaystyle \frac{2\tan 60^{0}}{1+\tan ^{2}60^{0}} \end{array} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukan nilai perbandingan trigonometrinya}\\ &\textrm{dari gambar segitiga berikut} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: \tan \alpha =0,75.\: \: \textrm{Tentukan nilai}\\ &\textrm{perbandingan trigonometri berikut}\\ &\begin{array}{llll} \textrm{a}.&\sin \alpha \\ \textrm{b}.&\cos \alpha \\ \textrm{c}.&\csc \alpha \\ \textrm{d}.&\sec \alpha \\ \textrm{e}.&\cot \alpha \\ \textrm{f}.&\sin^{2} \alpha+\cos ^{2}\alpha \\ \end{array} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: 1-\cos \beta =\displaystyle \frac{9}{25(1+\cos \beta )}\: \: \textrm{Tentukan}\\ &\textrm{nilai dari}\: \: \: \tan \beta \end{array}$.

DAFTAR PUSTAKA

Kurnianingsih, S., Kuntarti, Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 2 Standar Isi 2006. Jakarta: ESIS.
Marwanta, dkk. 2013. Matematika SMA Kelas X. Jakarta: YUDHISTIRA.
Yuana. R.A., Indriyastuti. 2017. Persepktif Matematika untuk Kelas X SMA dan MA Kelompok Matematika Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI.

Trigonometri

A. Ukuran Sudut

Sudut itu sendiri adalah suatu bangun yang dibatasi oleh dua buah sinar atau garis yang berpotongan di sekitar titik potongnya. Sebagai ilustrasinya perhatikanlah gambar berikut

Ada dua buah sinar yaitu $\overrightarrow{OA}$ dan $\overrightarrow{OB}$ yang bertemu atau berpotongan di titik $O$ den terbentuklah $\angle AOB$.

Sedangkan berkaitan dengan pengukuran sudut nantinya minimal kita mengenal 3 jenis, yaitu: derajat, radian, dan gone atau grade. Berikut hubungan ketiga jenis ikuran sudut yang dimaksud beserta lambang/notasi penulisannya. Untuk anak ditingkat MA/SMA ukuran sudut yang dikenalkan biasanya derajat dan radian, akan tetapi untuk anak SMK ada satu lagi yaitu satuan gone atau grade.

$\begin{array}{|c|}\hline 1\: \: \textrm{keliling}\: \: \bigcirc =360^{0}=2\pi \: \: \textrm{radian}=400^{g}\\\hline \color{red}\textrm{atau}\\\hline \displaystyle \frac{1}{2}\: \: \textrm{keliling}\: \: \bigcirc =180^{0}=\pi \: \textrm{radian}=200^{9}\\\hline \color{red}\textrm{boleh juga}\\\hline \displaystyle \frac{1}{4}\: \: \textrm{keliling}\: \: \bigcirc =90^{0}=\displaystyle \frac{1}{2}\pi \: \textrm{radian}=100^{9}\\\hline \color{red}\textrm{Jika ingin diperkecil lagi}\\\hline \displaystyle \frac{1}{8}\: \: \textrm{keliling}\: \: \bigcirc =45^{0}=\displaystyle \frac{1}{4}\pi \: \textrm{radian}=50^{9}\\\hline \end{array}$.

Perhatikan ilustrasi berikut

Sebagai catatan, terkadang sudut dalam satuan derajat dipresentasikan dalam bentuk seksagesimal, yaitu: diubah dalam satuan menit dan detik.

$\begin{array}{|l|}\hline \bullet \quad \displaystyle 1^{\circ}={60}'={3600}''\\\hline \bullet \quad {1}'={60}''\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Ubahlah sudut berikut ke satuan}\\ & \textrm{yang diminta}\\ &\textrm{a}.\quad 1^{\circ}=....\: \textrm{rad}\\ &\textrm{b}.\quad 1\: \textrm{rad}=....\: ^{0}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}& \begin{aligned}\textrm{a}.\quad\qquad 360^{0}&=2\pi \: \textrm{rad}\\ 360\times 1^{0}&=2\pi \: \textrm{rad}\\ 1^{0}&=\displaystyle \frac{2\pi }{360}\\ &=\displaystyle \frac{\pi }{180}\: \textrm{rad} \end{aligned}\\&\begin{aligned}\textrm{b}.\quad\qquad 2\pi \: \textrm{rad}&=360^{0}\\ 2\pi \times 1\: \textrm{rad}&=360^{0}\\ 1\: \textrm{rad}&=\left ( \displaystyle \frac{360}{2\pi } \right )^{0}\\ &=\left ( \displaystyle \frac{180}{\pi } \right )^{0} \end{aligned} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Ubahlah sudut berikut ke satuan }\\ &\textrm{yang diminta}\\&\textrm{a}.\quad 53,24^{\circ}=....\: \textrm{(dalam sexagesimal)}\\ &\textrm{b}.\quad 23^{\circ}{12}'{24}''=....\: ^{0}\\\\&\textbf{Jawab}:\\&\begin{aligned}\textrm{a}.\quad53,24^{0}&=53^{0}+0,24^{0}\\ &=53^{0}+0,24\times 1^{0}\\ &=53^{0}+0,24\times {60}'\\ &=53^{0}+{14,4}'\\ &=53^{0}+{14}'+{0,4}'\\ &=53^{0}+{14}'+0,4\times {1}'\\ &=53^{0}+{14}'+0,4\times {60}''\\ &=53^{0}+{14}'+{24}''\\ 53,24^{0}&=53^{0}{14}'{24}'' \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad 23^{0}{12}'{24}''&=23^{0}+12\times {1}'+24\times {1}''\\ &=23^{0}+12\times \left ( \frac{1}{60} \right )^{0}+24\times \left ( \frac{1}{3600} \right )^{0}\\&=23^{0}+0,2^{0}+0,00\overline{666}^{0}\\ &=23,20\overline{666}^{0}\end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Ubahlah ke dalam sudut-sudut berikut dalam radian}\\ &\begin{array}{llllllll}\\ \textrm{a}.&3^{0}&\textrm{e}.&90^{0}&\textrm{i}.&210^{0}&\textrm{m}.&300^{0}\\ \textrm{b}.&15^{0}&\textrm{f}.&120^{0}&\textrm{j}.&225^{0}&\textrm{n}.&315^{0}\\ \textrm{c}.&30^{0}&\textrm{g}.&135^{0}&\textrm{k}.&240^{0}&\textrm{0}.&12^{0}{24}'\\ \textrm{d}.&60^{0}&\textrm{h}.&150^{0}&\textrm{l}.&270^{0}&\textrm{p}.&30^{0}{12}'{30}'' \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bahwa}:\quad 180^{0}=180\times 1^{0}=\pi \: rad\Rightarrow 1^{0}=\displaystyle \frac{\pi }{180}\: rad \\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad 3^{0}&=\cdots \: \pi \: rad\\ 3^{0}&=\displaystyle \frac{3}{180}\pi \: rad\\ &=\displaystyle \frac{1}{60}\pi \: rad \end{aligned}&\begin{aligned}\textrm{m}.\quad 300^{0}&=\cdots \: \pi \: rad\\ 300^{0}&=\displaystyle \frac{300}{180}\pi \: rad\\ &=\displaystyle \frac{5}{3}\pi \: rad\\ \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad 30^{0}&=\cdots \: \pi \: rad\\ 30^{0}&=\displaystyle \frac{30}{180}\pi \: rad\\ &=\displaystyle \frac{1}{6}\pi \: rad\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{o}.\quad 12^{0}{24}'&=\cdots \: \pi \: rad\\ 12^{0}{24}'&=\displaystyle \frac{12+\left ( \displaystyle \frac{24}{60} \right )}{180}\pi \: rad\\ &=\displaystyle \frac{12+0,4}{180}\pi \: rad\\ &=\displaystyle \frac{12,4}{180}\pi \: rad\\ &=\displaystyle \frac{31}{450}\pi \: rad \end{aligned}\\\hline \begin{aligned}\textrm{f}.\quad 120^{0}&=\cdots \: \pi \: rad\\ 120^{0}&=\displaystyle \frac{120}{180}\pi \: rad\\ &=\displaystyle \frac{2}{3}\pi \: rad \end{aligned}&\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Ubahlah ke dalam sudut-sudut berikut dalam derajat}\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\pi \: rad&\textrm{e}.&\displaystyle \frac{4}{3}\pi \: rad\\\\ \textrm{b}.&\displaystyle \frac{3}{5}\pi \: rad&\textrm{f}.&2 \: rad\\\\ \textrm{c}.&\displaystyle \frac{2}{9}\pi \: rad&\textrm{g}.&12\: rad\\\\ \textrm{d}.&\displaystyle \frac{7}{12}\pi \: rad&\textrm{h}.&15\pi \: rad \end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Ingat bahwa}:\quad \pi \: rad=180^{0}\Rightarrow 1\: rad=\displaystyle \frac{180^{0}}{\pi } \\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \displaystyle \frac{1}{2}\pi \: rad&=\displaystyle \frac{1}{2}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=90^{0} \end{aligned}&\begin{aligned}\textrm{b}.\quad \displaystyle \frac{3}{5}\pi \: rad&=\displaystyle \frac{3}{5}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=108^{0} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad \displaystyle \frac{7}{12}\pi \: rad&=\displaystyle \frac{7}{12}\pi \times \displaystyle \frac{180^{0}}{\pi }\\ &=105^{0}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad 2\: rad&=2\times \displaystyle \frac{180^{0}}{\pi }\\ &=2\times \displaystyle \frac{180^{0}}{\frac{22}{7}}\\ &=\displaystyle \frac{1260^{0}}{11}\\ &=114,\overline{54}\, ^{0} \end{aligned}\\\hline \end{array} \end{array}$.

$\LARGE\colorbox{yellow}{LATIHAN SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Ubahlah sudut berikut ke satuan }\\ &\textrm{yang diminta}\\ &\textrm{a}.\quad27^{0}=.....\: \textrm{rad}\\ &\textrm{b}.\quad 28\: \textrm{rad}=....^{0}\\ &\textrm{c}.\quad 31,35^{\circ}=....\: \textrm{(dalam sexagesimal)}\\ &\textrm{d}.\quad 30^{\circ}{24}'{12}''=....\: ^{0} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Ubahlah sudut berikut ke satuan }\\ &\textrm{yang diminta}\\ &\textrm{a}.\quad 135^{0}=.....\: \textrm{rad}\\ &\textrm{b}.\quad 6\: \textrm{rad}=....^{0}\\ &\textrm{c}.\quad 23,45^{\circ}=....\: \textrm{(dalam sexagesimal)}\\ &\textrm{d}.\quad 45^{\circ}{50}'{36}''=....\: ^{0} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Silahkan dicoba sendiri soal yang belum}\\ &\textrm{dibahas} \end{array}$.

Vektor pada Ruang

$\color{blue}\textrm{A. Vektor Di Ruang}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|c|c|}\hline \textrm{Nama}&\textbf{R}^{3}\\\hline \textrm{Vektor Satuan}&\textrm{Ruang (Bidang XYZ)}\\\hline \hat{e}_{\bar{a}}=\displaystyle \frac{\bar{a}}{\left | \bar{a} \right |}&\begin{cases} i= &\textrm{vektor satuan} \\ &\textrm{yang searah sumbu X}\\ j= &\textrm{Vektor satuan}\\ &\textrm{yang searah sumbu Y}\\ k=&\textrm{Vektor satuan}\\ &\textrm{searah sumbu Z} \end{cases} \\\hline \textrm{Vektor nol}&\overrightarrow{O}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\\\hline \textrm{Vektor posisi}&\overrightarrow{OP}=\vec{p}=\begin{pmatrix} p_{1}\\ p_{2}\\ p_{3} \end{pmatrix}=p_{1}\bar{i}+p_{2}\bar{j}+p_{3}\bar{j}\\\hline \textrm{Besar Vektor}&\overrightarrow{OP}=\sqrt{p_{1}^{2}+p_{2}^{2}+p_{3}^{2}}\\\hline \end{array}$

$\color{blue}\textrm{B. Operasi Vektor}$

$\color{blue}\textrm{1. Sifat-Sifat Aljabar Vektor}$

$\begin{array}{|l|l|l|}\hline 1.&\textrm{Komutatif penjumlahan}&\vec{a}+\vec{b}=\vec{b}+\vec{a}\\\hline 2.&\textrm{Asosiatif penjumlahan}&\left ( \vec{a}+\vec{b} \right )+\vec{c}=\vec{a}+\left ( \vec{b}+\vec{c} \right )\\\hline 3.&\textrm{Elemen Identitas}&\vec{a}+\vec{0}=\vec{0}+\vec{a}=\vec{a}\\\hline 4.&\textrm{Invers Penjumlahan}&\vec{a}+\left ( -\vec{a} \right )=\left ( -\vec{a} \right )+\vec{a}=\vec{0}\\\hline 5.&\textrm{Perkalian dengan skalar}&k\left ( l\vec{a} \right )=\left ( kl \right )\vec{a}\\ &&k\left ( \vec{a}+ \vec{b} \right )=k\vec{a}+k\vec{b}\\ &&k\left ( \vec{a}- \vec{b} \right )=k\vec{a}-k\vec{b}\\\hline 6.&\begin{aligned}&\textrm{Jika A, B, dan C segaris }\\ &\color{blue}\textrm{(Kolinear)} \end{aligned}&\begin{cases} \overrightarrow{AB}=k\overrightarrow{BC} \\ \overrightarrow{AC}=k\overrightarrow{AB} \\ dll \end{cases}\\\hline \end{array}$.

$\begin{array}{|c|c|}\hline \color{blue}\textrm{Vektor}&\color{blue}\textrm{Contoh}\\\hline \vec{z}=a\vec{i}+b\vec{j}+c\vec{k}&\begin{aligned}&\textrm{diketahui}\: \: \vec{p}=\vec{i}-2\vec{j}+2\vec{k}\\ &\textrm{maka pangjang vektor}\: \: \vec{p}\: \: \textrm{adalah}\\ &\left | \vec{p} \right |=\sqrt{1^{2}+(-2)^{2}+2^{2}}\\ &\quad\: \: =\sqrt{1+4+4}=\sqrt{9}=3 \end{aligned}\\\hline &\begin{aligned}&\textrm{Vektor satuan dari}\: \: \vec{p}\: \: \textrm{adalah}\\ &\vec{e}_{\vec{p}}=\frac{\vec{p}}{\left | \vec{p} \right |}=\displaystyle \frac{\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}}{3}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}=\displaystyle \begin{pmatrix} \frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3} \end{pmatrix} \end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\overrightarrow{b}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix},\: \: \textrm{dan}\: \: \overrightarrow{c}=\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix},\\ & \textrm{tentukanlah hasil dari}\\ &\textrm{a}.\quad \overrightarrow{a}+\overrightarrow{b}\\ &\textrm{b}.\quad 6\overrightarrow{a}+2\overrightarrow{b}\\ &\textrm{c}.\quad 2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad&\overrightarrow{a}+\overrightarrow{b}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 2+(-3)\\ 1+(-5)\\ (-4)+2 \end{pmatrix}\\ &=\begin{pmatrix} 2-3\\ 1-5\\ -4+2 \end{pmatrix}=\color{red}\begin{pmatrix} -1\\ -4\\ -2 \end{pmatrix}\\ \textrm{b}.\quad&6\overrightarrow{a}+2\overrightarrow{b}=6\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+2\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ 6-10\\ -24+4 \end{pmatrix}=\color{red}\begin{pmatrix} 12\\ -4\\ -20 \end{pmatrix}\\ \textrm{c}.\quad&2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &2\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}-\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}+\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} 4+3+7\\ 2+5+0\\ -8-2+4 \end{pmatrix}=\color{red}\begin{pmatrix} 17\\ 7\\ -6 \end{pmatrix}\\ \textrm{d}.\quad&\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}=\cdots \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\textrm{tentukanlah}\: \: \left | \overrightarrow{a} \right |\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\left | \overrightarrow{a} \right |&=\sqrt{2^{2}+1^{2}+(-4)^{2}}\\ &=\sqrt{4+1+16}\\ &=\color{red}\sqrt{21} \end{aligned} \end{array}$

$\color{blue}\textrm{2. Perkalian Skalar Dua Vektor}$

Konsep perkalian skalar dua buah vektor di ruang sama persis dengan konsep di bidang, yaitu:

$\color{red}\overrightarrow{a}\cdot \overrightarrow{b}=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$.

Misalkan diketahui

$\color{red}\begin{aligned}&\overrightarrow{a}=\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}, \overrightarrow{b}=\begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix},\: \: \color{black}\textrm{maka}\\ & \overrightarrow{a} \cdot \overrightarrow{b} =\color{black}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}\cdot \begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix}\\ &\qquad\quad =\color{black}a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: t\: \: \textrm{jika}\\ & \overrightarrow{p}=3\bar{i}+t\bar{j}+\bar{k}\: \: \textrm{dan}\: \: \overrightarrow{p}\cdot \overrightarrow{p}=13\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\overrightarrow{p}\cdot \overrightarrow{p}=13\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |\left | \overrightarrow{p} \right |\cos 0^{\circ}=13,\\ &\qquad\qquad \color{blue}\textrm{ingat bahwa}\: \: \angle \left ( \overrightarrow{p},\overrightarrow{p} \right )=0^{\circ}\\ &\qquad\qquad \color{blue}\textrm{dan nilai}\: \: \cos 0^{\circ}=1,\\ & \color{black}\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |^{2}.1=13\Leftrightarrow \left | \overrightarrow{p} \right |^{2}=13\\ &\Leftrightarrow \left (\sqrt{3^{2}+t^{2}+1^{2}} \right )^{2}=13\\ &\Leftrightarrow 3^{2}+t^{2}+1^{2}=13\\ &\Leftrightarrow 9+t^{2}+1=13\\ &\Leftrightarrow t^{2}=13-9-1=10\\ &\Leftrightarrow t^{2}=3\\ &\Leftrightarrow t=\color{red}\pm \sqrt{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: \overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{Jika}\: \: \overrightarrow{p}\: \: \textrm{tegak lurus}\: \: \overrightarrow{q}\: \: \textrm{maka}\\ &\textrm{tentukanlah nilai}\: \: t\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{dengan}\: \: \overrightarrow{p}\: \: \textrm{dan}\: \: \overrightarrow{q}\: \: \textrm{tegak lurus}\\ &\textrm{artinya}\: \: \color{blue}\angle \left ( \overrightarrow{p},\overrightarrow{q} \right )=90^{\circ}.\: \color{black}\textrm{Sehingga}\\ &\textrm{nilai}\: \: \color{blue}\cos 90^{\circ}=0\\ &\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cos \theta \\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cdot 0=0\\ &\Leftrightarrow \: \begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\cdot \begin{pmatrix} 4\\ -1\\ t \end{pmatrix}=0\\ &\Leftrightarrow \: (-2)(4)+(1)(-1)+(3)(t)=0\\ &\Leftrightarrow \: -8-1+3t=0\\ &\Leftrightarrow \: 3t=9\\ &\Leftrightarrow \: t=\color{red}3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: \left | \overline{u} \right |=6\: ,\: \left | \overline{v} \right |=4\sqrt{3},\: \: \textrm{dan}\: \: \left | \overline{u}-\overline{v} \right |=8\\ &\textrm{tentukanlah nilai dari}\\ &\textrm{a}.\quad \overline{u}\bullet \overline{v}\\ &\textrm{b}.\quad \left | \overline{u}+\overline{v} \right |\\ &\textrm{c}.\quad \textbf{cosinus}\: \: \textrm{sudut antara}\: \: \overline{u}\: \: \textrm{dan}\: \: \overline{v}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\: \cdots \\ &\: 2.\overline{u}\bullet \overline{v}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}-\left | \overline{u}-\overline{v} \right |^{2}\\ &\: 2.\overline{u}\bullet \overline{v}=6^{2}+(4\sqrt{3})^{2}-8^{2}\\ &\: 2. \overline{u}\bullet \overline{v}=36+48-64=84-64=20\\ &\quad \overline{u}\bullet \overline{v}=\displaystyle \frac{20}{2}=\color{red}10\\ \textrm{b}.\quad &\left | \overline{u}+\overline{v} \right |^{2}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}+2.\overline{u}\bullet \overline{v}\\ &\left | \overline{u}+\overline{v} \right |^{2}=6^{2}+(4\sqrt{3})^{2}+20\\ &\: \: \quad\qquad =84+20=104\\ &\left | \overline{u}+\overline{v} \right |=\color{red}\sqrt{104}\\ \textrm{c}.\quad &\cos \angle (\overline{u},\, \overline{v})=\displaystyle \frac{\overline{u}\bullet \overline{v}}{\left |\overline{u} \right |.\left | \overline{v} \right |}=\frac{10}{6.(4\sqrt{3})}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &\quad\qquad\qquad =\displaystyle \frac{10\sqrt{3}}{72}=\color{red}\frac{5}{36}\sqrt{3}\\ &\color{blue}\textbf{Berikut ilustrasi gambarnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Diketahui}\: \: \left | \vec{a} \right |=\sqrt{3}\: ,\left | \vec{b} \right |=1,\: \: \textrm{dan}\: \: \left | \vec{a}-\vec{b} \right |=1\\ &\textrm{maka panjang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&2\sqrt{2}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\color{red}\sqrt{7}&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \, \textrm{sebagaimana pada soal}\\ \left | \vec{a}-\vec{b} \right |^{2}&=\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}-2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ 1^{2}&=\left ( \sqrt{3} \right )^{2}+1^{2}-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta &=3\\ \textrm{maka pan}&\textrm{jang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\\ \left | \vec{a}+\vec{b} \right |&=\sqrt{\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta}\\ &=\sqrt{\left ( \sqrt{3} \right )^{2}+1^{2}+3}\\ &=\sqrt{3+1+3}\\ &=\color{red}\sqrt{7} \end{aligned} \end{array}$

RANGKUMAN

Klik di sini

DAFTAR PUSTAKA

Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
Miyanto, Aksin, N., Suparno. 2021. Buku Interaktif Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas X Semester 2. Yogyakarta: INTAN PARIWARA.
Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Contoh Soal 8 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 36.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{0}^{n^{2}}\left [ \sqrt{x} \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \frac{(n+1)n(n+1)}{6}\\ \textrm{b}.\quad \displaystyle \frac{n(n-1)(n-2)}{6}\\ \textrm{c}.\quad \displaystyle \frac{n(n+1)(2n+1)}{6}\\ \textrm{d}.\quad \displaystyle \frac{n(n-1)(3n+1)}{6}\\ \textrm{e}.\quad \color{red}\displaystyle \frac{n(n-1)(4n+1)}{6}\end{array}\\\\ &\textbf{Jawab}:\\ \end{array}$.

$.\: \qquad\begin{aligned}&\displaystyle \int_{0}^{n^{2}}\left [ \sqrt{x} \right ]\: dx\\ &=\displaystyle \int_{0}^{1}\left [\sqrt{0} \right ]\: dx+\int_{1}^{2^{2}}\left [\sqrt{1} \right ]\: dx+\int_{2^{2}}^{3^{2}}\left [\sqrt{2^{2}} \right ]\: dx \\ &\quad +\displaystyle \int_{3^{2}}^{4^{2}}\left [ \sqrt{3^{2}} \right ]\: dx+\cdots +\int_{(n-1)^{2}}^{n^{2}}\left [ \sqrt{(n-1)^{2}} \right ]\: dx\\ &=\displaystyle \int_{0}^{1}0\: dx+\int_{1}^{4}1\: dx+\int_{4}^{9}2\: dx+\int_{9}^{16}3\: dx\\ &\quad +\cdots +\int_{(n-1)^{2}}^{n^{2}}(n-1)\: dx\\ &=0+x|_{1}^{4}+2x|_{4}^{9}+3x|_{9}^{16}+\cdots +(n-1)x|_{(n-1)^{2}}^{n^{2}}\\ &=0+(4-1)+2(9-4)+3(16-9)+\cdots +(n-1)\left ( 2n-1 \right )\\ &=3+2.5+3.7+4.9+5.11+\cdots +(n-1)(2n-1)\\ &=3+10+21+36+55+\cdots +(n-1)(2n-1)\\ &\quad (\textrm{Deretnya berupa barisan aritmetika tingkat 2})\\ &=\displaystyle \frac{n(n-1)(4n+1)}{6} \end{aligned}$.

DAFTAR PUSTAKA

Hutahaean. 1980. Kalkulus Diferensial dan Integral I. Jakarta: GRAMEDIA.
Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar ISI 2006. Jakarta: ESIS.
Nugroho, P.A., Gunarto, D. 2013. Big Bank Soal + Bahas Matematika SMA/MA Kelas 1,2,3. Jakarta: WAHYUMEDIA.
Sukino. 2015. Matematika Kelompok Peminatan Matematika dan Ilmu Alam. Jakarta: ERLANGGA.
Suparmin, S., Malau, A. 2014. Seri Kinomatika: Mainstream Matematika Dasar & Matematika IPA untuk Siswa SMA/MA Kelompok MIA. Bandung: YRAMA WIDYA.

Youtube:

Calculus | Find the Integral of the Floor Function of x

https://www.youtube.com/watch?v=CMme8XgfEJI

Contoh Soal 7 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 31.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{2} \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 3\\ \textrm{b}.\quad \displaystyle 3\frac{1}{2}\\ \textrm{c}.\quad \color{red}\displaystyle 4\\ \textrm{d}.\quad \displaystyle 4\frac{1}{2}\\ \textrm{e}.\quad \displaystyle 5\end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.

$.\: \qquad\begin{aligned}&\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{2} \right ]\: dx\\&=\displaystyle \int_{-1}^{-\frac{1}{2}}-1\: dx+\int_{-\frac{1}{2}}^{\frac{1}{2}}0\: dx+\int_{\frac{1}{2}}^{\frac{3}{2}}1\: dx+\int_{\frac{3}{2}}^{\frac{5}{2}}2\: dx +\int_{\frac{5}{2}}^{3}3\: dx\\ &=-1|_{-1}^{-\frac{1}{2}}+0|_{-\frac{1}{2}}^{\frac{1}{2}}+x|_{\frac{1}{2}}^{\frac{3}{2}}+2x|_{\frac{3}{2}}^{\frac{5}{2}}+3x|_{\frac{5}{2}}^{3}\\ &=-\left ( -\displaystyle \frac{1}{2}+1 \right )+0+\left ( \displaystyle \frac{3}{2}-\frac{1}{2} \right )+2\left ( \frac{5}{2}-\frac{3}{2} \right ) +3\left ( 3-\displaystyle \frac{5}{2} \right )\\&=-\displaystyle \frac{1}{2}+0+1+2+\frac{3}{2}\\ &=\color{red}\displaystyle 4\\ &\textrm{Perhatikan tabel berikut} \end{aligned}$.

$\begin{array}{ll}\\ 32.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{3} \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 3\\ \textrm{b}.\quad \color{red}\displaystyle 3\frac{1}{3}\\ \textrm{c}.\quad \displaystyle 4\frac{2}{3}\\ \textrm{d}.\quad \displaystyle 5\frac{1}{3}\\ \textrm{e}.\quad \displaystyle 6\frac{1}{3}\end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.

$.\: \qquad\begin{aligned}&\displaystyle \int_{-1}^{3}\left [ x+\displaystyle \frac{1}{3} \right ]\: dx\\&=\displaystyle \int_{-1}^{-\frac{1}{3}}-1\: dx+\int_{-\frac{1}{3}}^{\frac{2}{3}}0\: dx+\int_{\frac{2}{3}}^{\frac{5}{3}}1\: dx+\int_{\frac{5}{3}}^{\frac{8}{3}}2\: dx +\int_{\frac{8}{3}}^{3}3\: dx\\ &=-1|_{-1}^{-\frac{1}{3}}+0|_{-\frac{1}{3}}^{\frac{2}{3}}+x|_{\frac{2}{3}}^{\frac{5}{3}}+2x|_{\frac{5}{3}}^{\frac{8}{3}}+3x|_{\frac{8}{3}}^{3}\\ &=-\left ( -\displaystyle \frac{1}{3}+1 \right )+0+\left ( \displaystyle \frac{5}{3}-\frac{2}{3} \right )+2\left ( \frac{8}{3}-\frac{5}{3} \right ) +3\left ( 3-\displaystyle \frac{8}{3} \right )\\&=-\displaystyle \frac{2}{3}+0+1+2+\frac{3}{3}\\ &=\color{red}\displaystyle 3\frac{1}{3} \end{aligned}$.

$\begin{array}{ll}\\ 33.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{0}^{n}\left [ x \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle n^{2}\\ \textrm{b}.\quad \color{red}\displaystyle \frac{n(n-1)}{2}\\ \textrm{c}.\quad \displaystyle \frac{n(n+1)}{2}\\ \textrm{d}.\quad \displaystyle \frac{(n+1)(n+2)}{2}\\ \textrm{e}.\quad \displaystyle \frac{n^{2}+1}{2}\end{array}\\\\ &\textbf{Jawab}:\\ & \end{array}$.

$.\: \qquad\begin{aligned}&\displaystyle \int_{0}^{n}f(x)\: dx=\displaystyle \int_{0}^{n}\left [ x \right ]\: dx\\&=\displaystyle \int_{0}^{1}0\: dx+\int_{1}^{2}1\: dx+\int_{2}^{3}2\: dx+\cdots +\int_{n-1}^{n}(n-1)\: dx\\ &=0+x|_{1}^{2}+2x|_{2}^{3}+\cdots +(n-1)x|_{n-1}^{n}\\ &=0+(2-1)+2(3-2)+\cdots +(n-1)(n-(n-1))\\&=0+1+2+\cdots (n-1)\\ &=\displaystyle \frac{(n-1)((n-1)+1)}{2}\\ &=\color{red}\displaystyle \frac{n(n-1)}{2}\\ &\textrm{Perhatikan tabel berikut} \end{aligned}$.

$\begin{array}{ll}\\ 34.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-2022}^{2022}\left [ x \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int_{-2022}^{2022}\left [ x \right ]\: dx\\ &=\displaystyle \int_{-2022}^{-2021}\left [ x \right ]\: dx+\displaystyle \int_{-2021}^{-2020}\left [ x \right ]\: dx+\cdots +\int_{2021}^{2022}\left [ x \right ]\: dx\\ &=-2022+(-2021)+\cdots +2020+2021\\ &=\color{red}-2022 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 35.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-2022}^{2022}(x-\left [ x \right ])\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}\left [ x \right ]dx\\ \textrm{e}.\quad \color{red}\displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int_{-2022}^{2022}(x-\left [ x \right ])\: dx\\ &=\displaystyle \int_{-2022}^{2022}x\: dx-\left (\displaystyle \int_{-2022}^{-2021}\left [ x \right ]\: dx+\cdots +\int_{2021}^{2022}\left [ x \right ]\: dx \right )\\ &\textrm{Sebelumnya perlu diingat bahwa}\\ &\quad \displaystyle \int_{-2022}^{2022}x\: dx\: \: \textrm{adalah}\: \textrm{fungsi ganjil, maka}=0\\ &\quad \textrm{Sehingga nilai akhirnya adalah}\\ &=0-(-2022)\qquad \textrm{lihat pembahasan sebelumnya}\\ &=\color{red}2022 \end{aligned} \end{array}$.

Contoh Soal 6 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 26.&(\textbf{SBMPTN Mat IPA 2013})\\&\textrm{Diketahui sebuah lingkaran berjari-jari}\: \: 2\\ &\textrm{sebagaimana ilustrasi dibawah. Jika tali}\\ &\textrm{busur pada gambar berjarak 1 dari garis}\\ &\textrm{tengah, maka luas daerah di atas tali busur}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 2\int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{b}.\quad \color{red}\displaystyle 2\int_{0}^{\sqrt{3}}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{1}\sqrt{4-x^{2}}\: dx\\ \textrm{e}.\quad \displaystyle 2\int_{0}^{\sqrt{3}}\sqrt{4-x^{2}}\: dx\end{array} \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Diketahui bahwa sebuah lingkaran}\\ &\textrm{dengan}\: \: r=2,\: \textrm{artinya}\: \: x^{2}+y^{2}=r^{2}=2^{2}=4\\ &\textrm{Diketahui pula persamaan garis}\: \: y=1,\: \textrm{maka}\\&x^{2}+1^{2}=4\Leftrightarrow x^{2}=4-1=3\Leftrightarrow x=\pm \sqrt{3} \end{aligned}$

$.\: \qquad\begin{aligned}&\textrm{Sehingga luas daerah di atas tali busur}\\ &\textrm{yaitu berupa}\: \: \textbf{temberang}\: \textrm{adalah}\\ &=\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}\left ( f_{2}(x)-f_{1}(x) \right )dx\\ &=\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ &\textrm{ingat fungsi genap pada pembahasan}\\ &\textrm{sebelumnya, karena}\: f(x)=\sqrt{4-x^{2}}=f(-x)\\ &=\color{red}2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx \end{aligned}$.

$\begin{array}{ll}\\ 27.&(\textbf{SBMPTN Mat IPA 2013})\\&\textrm{Diketahui sebuah lingkaran berjari-jari}\: \: 2\\ &\textrm{sebagaimana ilustrasi dibawah. Jika tali}\\ &\textrm{busur pada gambar berjarak 1 dari garis}\\ &\textrm{tengah, maka luas daerah di bawah tali busur}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 2\pi -2\int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{b}.\quad \displaystyle 2\pi - 2\int_{0}^{\sqrt{3}}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{c}.\quad \displaystyle 4\pi -\int_{0}^{1}\left ( \sqrt{4-x^{2}}-1 \right )dx\\ \textrm{d}.\quad \color{red}\displaystyle 4\pi -2\int_{0}^{\sqrt{3}}\left (\sqrt{4-x^{2}}-1 \right )\: dx\\ \textrm{e}.\quad \displaystyle 4\pi -2\int_{0}^{\sqrt{3}}\sqrt{4-x^{2}}\: dx\end{array}\\\\&\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dengan ilustrasi sama dengan no.24 di atas}\\ &\textrm{Kita dapat tentukan luas wilayah di bawah}\\ &\textbf{tembereng},\: \textrm{yaitu}:\\&=\textrm{Luas lingkaran penuh itu}-\textbf{tembereng}\\ &=\pi .r^{2}-2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx\\ &=\pi .(2)^{2}-2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx\\&=\color{red}4\pi -2\displaystyle \int_{0}^{\sqrt{3}}\left ( \displaystyle \sqrt{4-x^{2}}-1 \right )\: dx \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 28.&(\textbf{SBMPTN Mat IPA 2013})\\&\textrm{Luas daerah yang dibatasi oleh kurva}\\ &y=2-x^{2}\: \: \textrm{dan}\: \: y=\left | x \right |\: \: \textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle 2\int_{-1}^{0}\left (-x^{2}+x+2 \right )dx\\ \textrm{b}.\quad \displaystyle 2\int_{-1}^{0}\left ( -x^{2}-x+2 \right )dx\\ \textrm{c}.\quad \displaystyle 2\int_{-1}^{0}\left ( -x^{2}+x-2 \right )dx\\ \textrm{d}.\quad \displaystyle \int_{-1}^{1}\left (-x^{2}-x+2 \right )\: dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{1}\left ( -x^{2}+x+2 \right )\: dx\end{array} \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Kita tentukan poin batas-batasnya, yaitu}:\\ &\textrm{Titik potong kurva dengan sumbu-X},\: y=0\\ &y=2-x^{2}=0\Leftrightarrow (\sqrt{2}+x)(\sqrt{2}-x)=0\\ &\Leftrightarrow x=-\sqrt{2}\: \: \textrm{atau}\: \: x=\sqrt{2}\\ &\textrm{Titik potong kurva dengan garis, yaitu}:\\ &2-x^{2}=\left | x \right |\Leftrightarrow x^{2}+\left | x \right |-2=0\\ &\bullet \: \: x\geq 0,\: \left | x \right |=x\\ &\: \quad \Leftrightarrow x^{2}+x-2=(x+2)(x-1)=0\\ &\: \quad \Leftrightarrow x=-2\: (\textbf{TM})\: \: \textrm{atau}\: \: x=1\\ &\bullet \: \: x< 0,\: \left | x \right |=-x\\ &\: \quad \Leftrightarrow x^{2}-x-2=(x-2)(x+1)=0\\ &\: \quad \Leftrightarrow x=2\: (\textbf{TM})\: \: \textrm{atau}\: \: x=-1\\ &\textrm{Perhatikanlah ilustrasi berikut} \end{aligned}$.

$.\: \qquad\begin{aligned}&\textrm{Misalkan}\: \: f_{1}(x)=2-x^{2}\: \: \textrm{dan}\: \: f_{2}(x)=\left | x \right |\\ &\textrm{maka luas arsirannya adalah}:\\ &\textbf{Luas arsiran}=\: 2\displaystyle \int_{-1}^{0}\left (f_{1}(x)-f_{2}(x) \right )dx\\ &=2\displaystyle \int_{-1}^{0}\left ( 2-x^{2}-\left | x \right | \right )dx\\ &=2\displaystyle \int_{-1}^{0}\left (2-x^{2}-(-x) \right )dx\\ &=2\displaystyle \int_{-1}^{0}\left ( -x^{2}+x+2 \right )dx \end{aligned}$.

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{-1}^{3}\left [ x \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle 2\\ \textrm{b}.\quad \displaystyle 4\\ \textrm{c}.\quad \displaystyle 6\\ \textrm{d}.\quad \displaystyle 8\\ \textrm{e}.\quad \displaystyle 10\end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi berikut} \end{array}$.

Jika diperjelas lagi

$.\: \qquad\begin{aligned}&\displaystyle \int_{-1}^{3}f(x)\: dx=\displaystyle \int_{-1}^{3}\left [ x \right ]\: dx\\&=\displaystyle \int_{-1}^{0}-1\: dx+\int_{0}^{1}0\: dx+\int_{1}^{2}1\: dx+\int_{2}^{3}2\: dx\\ &=(-x)|_{-1}^{0}+0+x|_{1}^{2}+2x|_{2}^{3}\\ &=\left ( 0-(-(-1)) \right )+0+(2-1)+2(3-2)\\ &=-1+0+1+2\\ &=\color{red}2 \end{aligned}$.

$\begin{array}{ll}\\ 30.&\textrm{Jika}\: \: \left [ x \right ]\: \: \textrm{menyatakan bilangan bulat}\\ &\textrm{terbesar yang}\leq x,\: \textrm{maka nilai dari}\\ &\displaystyle \int_{a}^{b}\left [ x \right ]\: dx+\int_{a}^{b}\left [ -x \right ]\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle a+b\\ \textrm{b}.\quad \displaystyle 0\\ \color{red}\textrm{c}.\quad \displaystyle a-b\\ \textrm{d}.\quad \displaystyle 2a\\ \textrm{e}.\quad \displaystyle 2b\end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan tabel berikut}\\ &\textrm{Misalkan}\: \: f(x)=\left [ x \right ],\: g(x)=\left [ -x \right ],\: \textrm{dan}\\ &h(x)=f(x)+g(x)=\left [ x \right ]+\left [ -x \right ] \end{array}$..

$.\: \qquad\begin{array}{|c|c|c|c|}\hline \textrm{Interval}\: \: & c\leq f(x)< c+1&c\leq g(x)< c+1&\left [ h(x) \right ]=c\\\hline \vdots &\vdots &\vdots &\vdots \\\hline -3\leq x< -2&-3\leq x< -2&2< x\leq 3&-3+2=-1\\\hline -2\leq x< -1&-2\leq x< -1&1\leq x< 2&-2+1=-1\\\hline -1\leq x< 0&-1\leq x< 0&0\leq x< 1&-1+0=-1\\\hline 0\leq x< 1&0\leq x< 1&-1\leq x< 0&0+-1=-1\\\hline 1\leq x< 2&1\leq x< 2&-2\leq x< -1&1+-2=-1\\\hline 2\leq x< 3&2\leq x< 3&-3\leq x< -2&2+-3=-1\\\hline \vdots &\vdots &\vdots &\vdots \\\hline a\leq x< b &a\leq x< b &-b\leq x< -a &\color{red}a-b \\\hline \end{array}$.

Contoh Soal 5 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 21.&\textrm{Pada gambar berikut, inetgral yang}\\ &\textrm{menyatakan luas daerah yang diarsir adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \int_{0}^{3}(2x^{2}-8x+6)dx\\ \textrm{b}.\quad \displaystyle \int_{0}^{1}(x^{2}-4x+3)dx-\int_{1}^{3}(x^{2}-4x+3)dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}(x^{2}+4x+3)dx+\int_{1}^{3}(x^{2}+4x+3)dx\\ \textrm{d}.\quad \color{red}\displaystyle \int_{0}^{1}(2x^{2}-8x+6)dx-\int_{1}^{3}(2x^{2}-8x+6)dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{1}(2x^{2}+8x+6)dx-\int_{1}^{3}(2x^{2}+8x+6)dx\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Kita tentukan poin batas-batasnya, yaitu}:\\&y=ax^{2}+bx+c=d(x-1)(x-3)\\ &\textrm{karena kurva melalui}\: \: (0,6),\: \textrm{maka}\\ &6=d(0-1)(0-3)\Rightarrow d=2\\&\textrm{Sehingga fungsi kurvanya adalah}:\\ &y=f(x)=2(x-1)(x-3)\\ &\Leftrightarrow y=2(x^{2}-4x+3)\\ &\Leftrightarrow y=2x^{2}-8x+6\\ &\textrm{Selanjutnya penentuan daerah arsir}\\ &\textrm{tinggal menyesuaikan dengan gambar} \end{aligned}$.

$\begin{array}{ll}\\ 22.&\textrm{Jika}\: \: f\: \: \textrm{fungsi ganjil, maka nilai}\\ &\displaystyle \int_{-2022}^{2022}f(x)\: dx\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}f(x)dx\\ \textrm{c}.\quad \color{red}\displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}f(x)dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perlu diketahui bahwa}\\ &f(x)=\begin{cases} \textrm{fungsi genap} &\Rightarrow f(-x)=f(x) \\ \textrm{fungsi ganjil} &\Rightarrow f(-x)=-f(x) \end{cases}\\ &\textrm{Untuk}\\ &\begin{array}{|c|c|}\hline f\: \textrm{fungsi genap}&\displaystyle \int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\\\hline f\: \textrm{fungsi ganjil}&\displaystyle \int_{-a}^{a}f(x)dx=0\qquad\qquad\: \\\hline \end{array}\\ &\textrm{Karena}\: \: f(x)\: \: \textrm{fungsi ganjil, maka}\\ &\color{red}\displaystyle \int_{-2022}^{2022}f(x)\: dx=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 23.&\textrm{Jika}\: \: f\: \: \textrm{fungsi genap, maka nilai}\\ &\displaystyle \int_{-2022}^{2022}f(x)\: dx\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}f(x)dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \color{red}\displaystyle 2\int_{0}^{2022}f(x)dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perlu diketahui bahwa}\\ &f(x)=\begin{cases} \textrm{fungsi genap} &\Rightarrow f(-x)=f(x) \\ \textrm{fungsi ganjil} &\Rightarrow f(-x)=-f(x) \end{cases}\\ &\textrm{Untuk}\\ &\begin{array}{|c|c|}\hline f\: \textrm{fungsi genap}&\displaystyle \int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\\\hline f\: \textrm{fungsi ganjil}&\displaystyle \int_{-a}^{a}f(x)dx=0\qquad\qquad\: \\\hline \end{array}\\ &\textrm{Karena}\: \: f(x)\: \: \textrm{fungsi genap, maka}\\ &\color{red}\displaystyle \int_{-2022}^{2022}f(x)\: dx=2\displaystyle \int_{0}^{2022}f(x)dx \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 24.&\textrm{Nilai dari}\: \: \displaystyle \int_{-2022}^{2022}\left | x \right |\: dx\: =....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle -2022\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}\left | x \right |dx\\ \textrm{c}.\quad \displaystyle 0\\ \textrm{d}.\quad \color{red}\displaystyle 2\int_{0}^{2022}\left | x \right |dx\\ \textrm{e}.\quad \displaystyle 2022\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Karena}\: \: f(x)=\left | x \right |,\: \: \textrm{adalah fungsi genap},\\ & \textrm{maka}\\ &\displaystyle \int_{-2022}^{2022}\left | x \right |\: dx\: =\color{red}\displaystyle 2\int_{0}^{2022}\left | x \right |dx=2022^{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 25.&\textrm{Nilai dari}\: \: \displaystyle \int_{-2022}^{2022}x\left | x \right |\: dx\: =....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle\frac{ -2022^{3}}{3}\\ \textrm{b}.\quad \displaystyle -2\int_{0}^{2022}x\left | x \right |dx\\ \textrm{c}.\quad \color{red}\displaystyle 0\\ \textrm{d}.\quad \displaystyle 2\int_{0}^{2022}x\left | x \right |dx\\ \textrm{e}.\quad \displaystyle \frac{2022^{3}}{3}\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Karena}\: \: f(x)=x\left | x \right |,\: \: \textrm{adalah fungsi ganjil},\\ & \textrm{maka}\\ &\displaystyle \int_{-2022}^{2022}x\left | x \right |\: dx\: =\color{red}0 \end{aligned} \end{array}$

Contoh Soal 4 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 16.&\textrm{Luas daerah yang batasi oleh parabola}\\ &y=4x-x^{2},\: y=-2x+8,\: \: \textrm{dan sumbu-Y}\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 4\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{b}.\quad \color{red}\displaystyle 6\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{c}.\quad \displaystyle 12\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{d}.\quad \displaystyle 20\frac{2}{3}\: \: \textrm{satuan luas}\\ \textrm{e}.\quad \displaystyle 30\frac{2}{3}\: \: \textrm{satuan luas}\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Luas arsiran}=\textrm{Luas trapesium}-\textrm{luas II}\\&\color{blue}\textrm{Proses penyelesaian 1}\\ &\textrm{Kurva}\: \: y=4x-x^{2},\: \textrm{maka koordinat titik}\\ &\textrm{puncaknya adalah}:(p,q).\: \textrm{Pilih}\: y=(x)=0\\ &\Leftrightarrow y=f(x)=4x-x^{2}=x(4-x)=0\\ &\Leftrightarrow x_{1}=0\: \: \textrm{atau}\: \: x_{2}=4\\ &\textrm{dengan}\: \: p=\displaystyle \frac{x_{1}+x_{2}}{2}=\frac{0+4}{2}=2,\: \: \textrm{dan}\\ &q=y=f(2)=4.2-2^{2}=8-4=4\\ &\textrm{Sehingga}\: \: (p,q)=(2,4)\\ &\color{blue}\textrm{Proses penyelesaian 2}\\ &\textrm{Untuk garis}\: \: y=-2x+8,\: \: \textrm{maka koordinat}\\ &\textrm{titik potongnya terhadap sumbu-X dan Y}:\\ &\begin{array}{|c|c|c|}\hline x&0&4\\\hline y&8&0\\\hline (x,y)&(0,8)&(4,0)\\\hline \end{array}\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{aligned}$.

$.\: \qquad\begin{aligned}&\textrm{Luas arsiran}=\textrm{Luas trapesium}-\textrm{luas II}\\ &=\displaystyle \frac{(8+4)\times 2}{2}-\displaystyle \int_{0}^{2}4x-x^{2}\: dx\\ &=12-\left ( \left (2x^{2}-\displaystyle \frac{1}{3}x^{3} \right )\: \: |_{0}^{2} \right )\\ &=12-\left ( \left (8-\displaystyle \frac{8}{3} \right )-0 \right )\\ &=12+\displaystyle \frac{8}{3}-8=\color{red}6\frac{2}{3}\: \: \color{black}\textrm{satuan luas} \end{aligned}$.

$\begin{array}{ll}\\ 17.&\textrm{Luas daerah yang batasi oleh kurva}\: \: y=4-x^{2}\\ &y=-x+2,\: \: \textrm{dan}\: \: 0\leq x\leq 2\: \: \textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \frac{8}{3}\: \: \textrm{satuan luas}\\ \textrm{b}.\quad \color{red}\displaystyle \frac{10}{3}\: \: \textrm{satuan luas}\\ \textrm{c}.\quad \displaystyle \frac{14}{3}\: \: \textrm{satuan luas}\\ \textrm{d}.\quad \displaystyle \frac{16}{3}\: \: \textrm{satuan luas}\\ \textrm{e}.\quad \displaystyle \frac{26}{3}\: \: \textrm{satuan luas}\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\color{blue}\textrm{Proses penyelesaian 1}\\&\textrm{Kurva}\: \: y=4-x^{2},\: \textrm{maka koordinat titik}\\ &\textrm{puncaknya adalah}:(p,q).\: \textrm{Pilih}\: y=(x)=0\\ &\Leftrightarrow y=f(x)=4-x^{2}=(2+x)(2-x)=0\\ &\Leftrightarrow x_{1}=-2\: \: \textrm{atau}\: \: x_{2}=2\\ &\textrm{dengan}\: \: p=\displaystyle \frac{x_{1}+x_{2}}{2}=\frac{-2+2}{2}=0,\: \: \textrm{dan}\\ &q=y=f(0)=4-0^{2}=4\\ &\textrm{Sehingga}\: \: (p,q)=(0,4)\\ &\color{blue}\textrm{Proses penyelesaian 2}\\ &\textrm{Untuk garis}\: \: y=-x+2,\: \: \textrm{maka koordinat}\\ &\textrm{titik potongnya terhadap sumbu-X dan Y}:\\ &\begin{array}{|c|c|c|}\hline x&0&2\\\hline y&2&0\\\hline (x,y)&(0,2)&(2,0)\\\hline \end{array}\\&\textrm{Perhatikan ilustrasi gambar berikut} \end{aligned}$.

$.\: \qquad\begin{aligned}&\textrm{Luas arsiran}=\textrm{Dibawah kurva}-\textrm{segitiga}\\&=\displaystyle \int_{0}^{2}4-x^{2}\: dx-\displaystyle \frac{1}{2}\times \textrm{alas}\times \textrm{tinggi}\\ &=\left (4x-\displaystyle \frac{1}{3}x^{3} \right )|_{0}^{2}-\displaystyle \frac{2\times 2}{2}\\ &=8-\displaystyle \frac{8}{3}-2=6-2\displaystyle \frac{2}{3}=3\frac{1}{3}\\ &=\color{red}\frac{10}{3}\: \: \color{black}\textrm{satuan luas} \end{aligned}$.

$\begin{array}{ll}\\ 18.&\textrm{Luas daerah di bawah kurva}\: \: y=-x^{2}+8x\\ &\textrm{di atas}\: \: y=6x-24\: \: \textrm{dan terletak di }\\ &\textrm{kuadran I adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(x^{2}-2x-24)\: dx\\ \textrm{b}.\quad \color{red}\displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(-x^{2}+2x+24)\: dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{6}(-x^{2}+8x)\: dx+\int_{6}^{8}(-x^{2}+2x+24)\: dx\\ \textrm{d}.\quad \displaystyle \int_{4}^{6}(6x-24)\: dx+\int_{6}^{8}(-x^{2}+8x)\: dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{4}(6x-24)\: dx+\int_{0}^{6}(-x^{2}+8x)\: dx\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Kita tentukan poin batas-batasnya dulu}\\&\textrm{Titik potong kurva dengan sumbu-X, yaitu}:\\ &y=-x^{2}+8x=x(-x+8)=0\\ &\Leftrightarrow x=0\: \: \textrm{atau}\: \: x=8\\ &\textrm{Titik potong garis dengan sumbu-X, yaitu}:\\ &y=6x-24=0\Leftrightarrow x=6\\ &\textrm{Dan titik potong kurva dengan garis adalah}:\\&y=y\Leftrightarrow -x^{2}+8x=6x-24\\ &\Leftrightarrow -x^{2}+2x+24=0\Leftrightarrow (-x-4)(x-6)=0\\ &\Leftrightarrow x_{1}=-4\: \: \textrm{atau}\: \: x_{2}=6\\ &\textrm{Sehingga luas arsiran}=\textrm{Luas I}+\textrm{Luas II}\\ &=\displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(-x^{2}+8x)-(6x-24)dx\\ &=\color{red}\displaystyle \int_{0}^{4}(-x^{2}+8x)\: dx+\int_{4}^{6}(-x^{2}+2x+24)\: dx\\ &\textrm{Perhatikan ulustrasi berikut} \end{aligned}$

$\begin{array}{ll}\\ 19.&\textrm{Luas daerah yang diarsir pada gambar}\\ &\textrm{di bawah dapat dirumuskan dengan}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx+\int_{b}^{d}g(x)dx-\int_{b}^{c}f(x)dx\\ \textrm{b}.\quad \displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx+\int_{b}^{d}\left ( g(x)-f(x) \right )dx\\ \textrm{c}.\quad \displaystyle \int_{a}^{d}\left ( f(x)-g(x) \right )dx\\ \textrm{d}.\quad \displaystyle \int_{a}^{d}\left ( f(x)-g(x) \right )dx-\int_{c}^{d}g(x)\: dx\\ \textrm{e}.\quad \displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx+\int_{c}^{d}\left ( g(x)-f(x) \right )dx \end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textbf{Jawab}:\\ &\textrm{Cukup jelas bahwa}\\ &\textrm{pada gambar di atas pada selang}\\ &\bullet\quad a\leq x\leq b:\: \displaystyle \int_{a}^{b}\left ( f(x)-g(x) \right )dx\\ &\bullet\quad b\leq x\leq d:\: \displaystyle \int_{b}^{d}\left (g(x)-f(x) \right )dx\\ &\qquad\qquad\qquad \textrm{dikurangi luasan yang}\\&\qquad\qquad\qquad \textrm{berselang}\: \: b\leq x\leq c \end{aligned}$.

$\begin{array}{ll}\\ 20.&\textrm{Daerah}\: \: \textrm{R}\: \: \textrm{di kuadran II, dibatasi oleh}\\ &\textrm{grafik}\: \: y=x^{2},\: y=x+2\: \: \textrm{dan}\: \: y=0\\ &\textrm{Integral yang sesuai kondisi di atas adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle \int_{-2}^{-1}\left ( x+2 \right )dx+\int_{-1}^{0}x^{2}dx\\ \textrm{b}.\quad \displaystyle \int_{-1}^{-2}\left ( x+2 \right )dx+\int_{-1}^{0}x^{2}dx\\ \textrm{c}.\quad \displaystyle \int_{-1}^{-2}x^{2}dx+\int_{-1}^{0}(x+2)dx\\ \textrm{d}.\quad \displaystyle \int_{0}^{2}\left (x^{2}+x+2 \right )dx\\ \textrm{e}.\quad \displaystyle \int_{0}^{2}\left ( -x^{2}+x+2 \right )dx\end{array}\\\\ &\textbf{Jawab}:\\ &\textrm{Perhatikan ilustrasi gambar berikut} \end{array}$.

$.\: \qquad\begin{aligned}&\textrm{Dari gukup jelas bahwa}\\ &\textrm{daerah arsiran akan terformulasikan}\\ &\displaystyle \int_{-2}^{-1}\left ( x+2 \right )dx+\int_{-1}^{0}x^{2}dx\\ &\textrm{Berikut sedikit uraian prosesnya}\\ &\textbf{Sebagai langkah awal},\: \textrm{tentukan dulu}\\ &\textrm{titik potong kura dengan garis, yaitu}:\\ &y=y\Leftrightarrow x^{2}=x+2\Leftrightarrow x^{2}-x-2=0\\ &\Leftrightarrow (x+1)(x-2)=0\\ &\Leftrightarrow x=-1\: \: \textrm{atau}\: \: x=2\\ &\textbf{Langkah berikutnya}\\ &\textrm{Kita tentukan daerah arsiran, yaitu}\\ &\textrm{seperti terformulasikan di atas} \end{aligned}$.

Contoh Soal 3 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 11.&\textrm{Diketahui}\: \: \displaystyle \int f(x)\: dx=ax^{2}+bx+c\\ &\textrm{dengan}\: \: a\neq 0.\: \textrm{Jika}\: a,\: f(a),\: 2b\: \: \textrm{membentuk}\\ &\textrm{barisan arimetika dan}\: \: f(b)=6,\: \: \textrm{maka}\\ &\textrm{nilai}\: \: \displaystyle \int_{0}^{1}f(x)\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle \frac{17}{4}\\ \textrm{b}.\quad \displaystyle \frac{21}{4}\\ \textrm{c}.\quad \displaystyle \frac{25}{4}\\ \textrm{d}.\quad \displaystyle \frac{13}{4}\\ \textrm{e}.\quad \displaystyle \frac{11}{4}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int f(x)\: dx=ax^{2}+bx+c\\ &\Rightarrow f(x)=2ax+b\\ &\Rightarrow f(a)=2a^{2}+b\\ &\Rightarrow f(b)=2ab+b=6\\ &\textrm{Karena}\: \: a,f(a),2b\: \: \textrm{membentuk barisan}\\ &\textrm{aritmetika, maka}\: \: f(a)=\displaystyle \frac{a+2b}{2}\\ &\textrm{Sehingga}\\ &f(a)=2a^{2}+b=\displaystyle \frac{a+2b}{2}\\ &\Leftrightarrow 4a^{2}-a=0\Leftrightarrow a(4a-1)=0\\ &\Leftrightarrow a=0\: \: \textrm{atau}\: \: a=\displaystyle \frac{1}{4}\\ &\textrm{Dan karena}\: \: a\neq 0,\: \: \textrm{maka}\: \: a=\displaystyle \frac{1}{4}\Rightarrow b=4\\ &\textrm{Selanjutnya}\\ &f(x)=2ax+b=\displaystyle \frac{1}{2}x+4\\ &\textrm{Dari fakta di atas, maka}\\ &\displaystyle \int_{0}^{1} f(x)\: dx=\int_{0}^{1} \displaystyle \frac{1}{2}x+4\: dx\\ &=\displaystyle \frac{1}{4}x^{2}+4x|_{0}^{1}\\ &=\left ( \displaystyle \frac{1}{4}+4 \right )-0=\color{red}\displaystyle \frac{17}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Diketahui}\: \: f(x)=a+bx\: \: \textrm{dan}\: \: F(x)\\ &\textrm{adalah anti turunan}\: \: f(x).\: \textrm{Jika}\\ &F(1)-F(0)=3\: ,\: \textrm{maka}\: \: 2a+b\\ &\textrm{adalah}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 10\\ \textrm{b}.\quad \color{red}\displaystyle 6\\ \textrm{c}.\quad \displaystyle 5\\ \textrm{d}.\quad \displaystyle 4\\ \textrm{e}.\quad \displaystyle 3\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Diketahui bahwa}\\ &f(x)=a+bx,\: \: \textrm{sedangkan}\\ &F(x)=\displaystyle \int f(x)\: dx=\int (a+bx)\: dx\\ &\qquad =ax+\displaystyle \frac{1}{2}bx^{2}+C\\ &\textrm{Dan diketahui pula}\\ &\begin{array}{ll} F(0)=a(1)+\displaystyle \frac{1}{2}b(1)^{2}+C&\\ F(1)=a(0)+\displaystyle \frac{1}{2}b(0)^{2}+C&-\\\hline \: \: \quad 3=a+\displaystyle \frac{1}{2}b \end{array}\\ &\textrm{Sehingga}\\ &2a+b=2\left ( a+\displaystyle \frac{1}{2}b \right )=2.3=\color{red}6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Jika}\: \: y=\displaystyle \frac{1}{3}\left ( x^{3}+\frac{3}{x} \right ),\: \textrm{maka}\\ &\displaystyle \int_{1}^{2}\sqrt{4+\left (\displaystyle \frac{dy}{dx} \right )^{2}}\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \frac{13}{14}\\ \textrm{b}.\quad \displaystyle \frac{13}{6}\\ \textrm{c}.\quad \color{red}\displaystyle \frac{15}{6}\\ \textrm{d}.\quad \displaystyle \frac{16}{6}\\ \textrm{e}.\quad \displaystyle \frac{17}{6}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&y=\displaystyle \frac{1}{3}\left ( x^{3}+\frac{3}{x} \right )\\ &\Leftrightarrow dy=\left (x^{2}-\displaystyle \frac{1}{x^{2}} \right )\: dx\\ &\Leftrightarrow \displaystyle \frac{dy}{dx}=\left (x^{2}-\displaystyle \frac{1}{x^{2}} \right )\\ &\Leftrightarrow \left ( \displaystyle \frac{dy}{dx} \right )^{2}=\left (x^{2}-\displaystyle \frac{1}{x^{2}} \right )^{2}=\displaystyle x^{4}-2+\displaystyle \frac{1}{x^{4}}\\ &\textrm{maka}\\ &\displaystyle \int_{1}^{2}\sqrt{4+\left (\displaystyle \frac{dy}{dx} \right )^{2}}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{4+\left ( \displaystyle x^{4}-2+\displaystyle \frac{1}{x^{4}} \right )}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{\left ( \displaystyle x^{4}+2+\displaystyle \frac{1}{x^{4}} \right )}\: dx\\ &=\displaystyle \int_{1}^{2}\sqrt{\left ( \displaystyle x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}}\: dx\\ &=\displaystyle \int_{1}^{2}x^{2}+\displaystyle \frac{1}{x^{2}}\: dx\\&=\displaystyle \frac{1}{3}x^{3}-\displaystyle \frac{1}{x}|_{1}^{2}\\ &=\left ( \displaystyle \frac{8}{3}-\frac{1}{2} \right )-\left ( \displaystyle \frac{1}{3}-1 \right )\\ &=\displaystyle \frac{16-3-2+6}{6}\\&=\color{red}\displaystyle \frac{17}{6} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Luas daerah yang diarsir pada}\\ &\textrm{gambar berikut adalah 12}\\ &\textrm{satuan luas, maka nilai}\: \: a=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 4\\ \textrm{b}.\quad \displaystyle 5\\ \textrm{c}.\quad \color{red}\displaystyle 6\\ \textrm{d}.\quad \displaystyle 7\\ \textrm{e}.\quad \displaystyle 8\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textrm{Luas arsiran}=12\\ &=4.a-\displaystyle \int_{1}^{4}\displaystyle \frac{a}{(4-1)(4-7)}(x-1)(x-7)\: dx\\ &=4a-\displaystyle \int_{1}^{4}\displaystyle -\frac{a}{9}(x^{2}-8x+7)\: dx\\ &=4a- \displaystyle \int_{1}^{4}\displaystyle \frac{a}{9}(-x^{2}+8x-7)\: dx\\ &=4a-\left (-\displaystyle \frac{a}{27}x^{3}+\frac{4a}{9}x^{2}-\frac{7a}{9}x|_{1}^{4} \right )\\ &=4a+\displaystyle \frac{64a}{27}-\frac{64a}{9}+\frac{28a}{9}-\frac{a}{27}+\frac{4a}{9}-\frac{7a}{9}\\ &=4a-\displaystyle \frac{18a}{9}=2a=12\Leftrightarrow a=\color{red}6\\ &\textrm{Berikut ilustrasi gambar lengkapnya} \end{aligned}$.

$\begin{array}{ll}\\ 15.&\textrm{Luas daerah yang diarsir pada}\\ &\textrm{gambar berikut dapat dinyatakan}\\&\textrm{dengan}\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \int_{0}^{2}(3x-x^{2})\: dx\\ \textrm{b}.\quad \displaystyle \int_{0}^{2}(x+3)\: dx-\int_{0}^{2}x^{2}\: dx\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}(x+3)\: dx-\int_{0}^{2}x^{2}\: dx\\ \textrm{d}.\quad \displaystyle \int_{0}^{1}(x+3-x^{2})\: dx-\int_{1}^{2}x^{2}\: dx\\ \textrm{e}.\quad \color{red}\displaystyle \int_{0}^{1}(x+3-x^{2})\: dx-\int_{1}^{2}\left (4-x^{2} \right )\: dx\end{array}\\\\ \end{array}$.

$.\: \qquad\begin{aligned}&\textrm{Daerah yang diarsir di atas dibatasi}\\ &\textrm{oleh 4 buah batas, yaitu}\\ &\bullet \quad \textrm{Garis}\: \: y=4\\ &\bullet \quad \textrm{Kurva}\: \: y=x^{2}\\ &\bullet \quad \textrm{Garis}\: \: x=0,\: \: \textrm{dan}\\ &\bullet \quad \textrm{Garis yang melalui titik}\: (0,3)\: \textrm{dan}\: (1,4)\\ &\begin{array}{|c|c|}\hline \textrm{Titik}&\textrm{Garis}\\\hline \begin{aligned}&{0,3}\\ &\textrm{dan}\\ &(1,4)\\ & \end{aligned}&\begin{aligned}\displaystyle y&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})+y_{1}\\ y&=\displaystyle \frac{4-3}{1-0}(x-0)+3\\ y&=x+3 \end{aligned}\\\hline \end{array}\\ &\textrm{Sehingga luas arisrannya adalah}:\\ &=\color{red}\displaystyle \int_{0}^{1}(x+3)-x^{2}\: dx+\int_{1}^{2}4-x^{2}\: dx \end{aligned}$.

Contoh Soal 2 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{ll}\\ 6.&\textrm{Hasil dari}\: \: \displaystyle \int_{2}^{4}\left ( -x^{2}+6x-8 \right )dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \frac{38}{3}\\ \textrm{b}.\quad \displaystyle \frac{26}{3}\\ \textrm{c}.\quad \displaystyle \frac{20}{3}\\ \textrm{d}.\quad \displaystyle \frac{16}{3}\\ \textrm{e}.\quad \color{red}\displaystyle \frac{4}{3} \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int_{2}^{4}\left ( -x^{2}+6x-8 \right )dx\\ &=-\displaystyle \frac{1}{3}x^{3}+3x^{2}-8x|_{2}^{4}\\ &=\left ( -\displaystyle \frac{1}{3}(4)^{3}+3(4)^{2}-8(4) \right )-\left ( -\displaystyle \frac{1}{3}(2)^{3}+3(2)^{2}-8(2) \right )\\ &=\left (-\displaystyle \frac{64}{3}+48-32 \right )-\left ( -\displaystyle \frac{8}{3}+12-16 \right )\\ &=-\displaystyle \frac{56}{3}+20=-\displaystyle \frac{56+60}{3}=\color{red}\displaystyle \frac{4}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Hasil dari}\: \: \displaystyle \int_{0}^{2}x^{2}\left ( x+2 \right )dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 6\\ \textrm{b}.\quad 6\displaystyle \frac{1}{3}\\ \textrm{c}.\quad 6\displaystyle \frac{2}{3}\\ \textrm{d}.\quad \color{red}9\displaystyle \frac{1}{3}\\ \textrm{e}.\quad \displaystyle 20 \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \displaystyle \int_{0}^{2}x^{2}\left ( x+2 \right )dx=\displaystyle \int_{0}^{2}x^{3}+2x^{2}\: \: dx\\ &=\displaystyle \frac{1}{4}x^{4}+\frac{2}{3}x^{3}|_{0}^{2}\\ &=\left ( -\displaystyle \frac{1}{4}(2)^{4}+\frac{2}{3}(2)^{3} \right )-\left ( 0 \right )\\ &=4+\displaystyle \frac{16}{3}=\color{red}\displaystyle \frac{4}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Nilai dari}\: \: \displaystyle \int_{0}^{2}18x\sqrt{3x^{2}+4}\: \: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle 4\\ \textrm{b}.\quad 16\\ \textrm{c}.\quad \color{red}112\\ \textrm{d}.\quad 128\\ \textrm{e}.\quad \displaystyle 168 \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Misalkan}\\ &u=3x^{2}+4\Rightarrow du=6x\: dx\\ &\textrm{Selanjutnya}\\ &\displaystyle \int_{0}^{2}18x\sqrt{3x^{2}+4}\: \: dx=\displaystyle \int_{0}^{2}3\sqrt{u}\: \: dx\\ &=\displaystyle \int_{0}^{2}3u^{.^{\frac{1}{2}}}\: \: dx=2u^{.^{\frac{3}{2}}}\: |_{0}^{2}\\ &=2\left ( 3x^{2}+4 \right )^{.^{\frac{3}{2}}}|_{0}^{2}\\ &=2\left ( 3.2^{2}+4 \right )^{.^{\frac{3}{2}}}-(2(0+4)^{.^{\frac{3}{2}}})\\ &=2.(16)^{.^{\frac{3}{2}}}=2.4^{3}-2.2^{3}=128-16\\ &=\color{red}112 \end{aligned} \end{array}$ .

$\begin{array}{ll}\\ 9.&\textrm{Hasil substitusi dari}\: \: u=x+1\: \: \textrm{pada}\\ &\displaystyle \int_{0}^{1}x^{2}\sqrt{x+1}\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \color{red}\displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{b}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{c}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{d}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du\\ \textrm{e}.\quad \displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: du \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Misalkan}\\ &u=x+1\Rightarrow x=u-1\Rightarrow dx=du\\ &\textrm{Selanjutnya}\\ &\displaystyle \int_{0}^{1}(u-1)^{2}\sqrt{u}\: \: du \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Nilai dari}\: \: \displaystyle \int_{0}^{1}5x(1-x)^{6}\: dx=\: ....\\ &\begin{array}{ll} \textrm{a}.\quad \displaystyle \frac{75}{56}\\ \textrm{b}.\quad \displaystyle \frac{10}{56}\\ \textrm{c}.\quad \color{red}\displaystyle \frac{5}{56}\\ \textrm{d}.\quad -\displaystyle \frac{7}{56}\\ \textrm{e}.\quad -\displaystyle \frac{10}{56}\end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\textrm{Diketahui soal bentuk}\: \textbf{integral parsial}\\ &\textrm{dengan metode}\: \textbf{Tanzalin}\: \textrm{diperoleh}\\ &\begin{array}{|c|c|}\hline \textrm{Diturunkan}&\textrm{Diintegralkan}\\\hline \color{red}5x&(1-x)^{6}\\\hline \color{blue}5&\color{red}-\displaystyle \frac{1}{7}(1-x)^{7}\\\hline 0&\color{blue}\displaystyle \frac{1}{56}(1-x)^{8}\\\hline \end{array}\\ &\textrm{Sehingga}\\ &\displaystyle \int_{0}^{1}5x(1-x)^{6}\: dx\\ &=(5x)\left ( -\displaystyle \frac{1}{7}(1-x)^{7} \right )-(5)\left ( \displaystyle \frac{1}{56}(1-x)^{8} \right )\: |_{0}^{1}\\ &=(0-0)-\left ( 0-\left ( \displaystyle \frac{5}{56} \right ) \right )\\ &=\color{red}\displaystyle \frac{5}{56} \end{aligned} \end{array}$