Contoh Soal 8 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 36.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_0^{n^2}\left[\sqrt{x}\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{(n+1)n(n+1)}{6}}\\ \text{b}.{\displaystyle \frac{n(n-1)(n-2)}{6}}\\ \text{c}.{\displaystyle \frac{n(n+1)(2n+1)}{6}}\\ \text{d}.{\displaystyle \frac{n(n-1)(3n+1)}{6}}\\ \text{e}.{\displaystyle \frac{n(n-1)(4n+1)}{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_0^{n^2}\left[\sqrt{x}\right]dx}\\ & ={\displaystyle {\int }_0^1\left[\sqrt{0}\right]dx+{\int }_1^{2^2}\left[\sqrt{1}\right]dx+{\int }_{2^2}^{3^2}\left[\sqrt{2^2}\right]dx}\\ & +{\displaystyle {\int }_{3^2}^{4^2}\left[\sqrt{3^2}\right]dx+\cdots +{\int }_{(n-1{)}^2}^{n^2}\left[\sqrt{(n-1{)}^2}\right]dx}\\ & ={\displaystyle {\int }_0^{1}0dx+{\int }_1^{4}1dx+{\int }_4^{9}2dx+{\int }_9^{16}3dx}\\ & +\cdots +{\int }_{(n-1{)}^2}^{n^2}(n-1)dx\\ & =0+x{|}_1^4+2x{|}_4^9+3x{|}_9^{16}+\cdots +(n-1)x{|}_{(n-1{)}^2}^{n^2}\\ & =0+(4-1)+2(9-4)+3(16-9)+\cdots +(n-1)(2n-1)\\ & =3+2.5+3.7+4.9+5.11+\cdots +(n-1)(2n-1)\\ & =3+10+21+36+55+\cdots +(n-1)(2n-1)\\ & (\text{Deretnya berupa barisan aritmetika tingkat 2})\\ & ={\displaystyle \frac{n(n-1)(4n+1)}{6}}\end{array}$.

DAFTAR PUSTAKA

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

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Contoh Soal 7 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 31.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{2}}]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 3}\\ \text{b}.{\displaystyle 3\frac{1}{2}}\\ \text{c}.{\displaystyle 4}\\ \text{d}.{\displaystyle 4\frac{1}{2}}\\ \text{e}.{\displaystyle 5}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{2}}]dx}\\ & ={\displaystyle {\int }_{-1}^{-\frac{1}{2}}-1dx+{\int }_{-\frac{1}{2}}^{\frac{1}{2}}0dx+{\int }_{\frac{1}{2}}^{\frac{3}{2}}1dx+{\int }_{\frac{3}{2}}^{\frac{5}{2}}2dx+{\int }_{\frac{5}{2}}^{3}3dx}\\ & =-1{|}_{-1}^{-\frac{1}{2}}+0{|}_{-\frac{1}{2}}^{\frac{1}{2}}+x{|}_{\frac{1}{2}}^{\frac{3}{2}}+2x{|}_{\frac{3}{2}}^{\frac{5}{2}}+3x{|}_{\frac{5}{2}}^3\\ & =-(-{\displaystyle \frac{1}{2}+1})+0+\left({\displaystyle \frac{3}{2}-\frac{1}{2}}\right)+2(\frac{5}{2}-\frac{3}{2})+3(3-{\displaystyle \frac{5}{2}})\\ & =-{\displaystyle \frac{1}{2}+0+1+2+\frac{3}{2}}\\ & ={\displaystyle 4}\\ & \text{Perhatikan tabel berikut}\end{array}$.

$\begin{array}{l}\\ 32.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{3}}]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 3}\\ \text{b}.{\displaystyle 3\frac{1}{3}}\\ \text{c}.{\displaystyle 4\frac{2}{3}}\\ \text{d}.{\displaystyle 5\frac{1}{3}}\\ \text{e}.{\displaystyle 6\frac{1}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^3[x+{\displaystyle \frac{1}{3}}]dx}\\ & ={\displaystyle {\int }_{-1}^{-\frac{1}{3}}-1dx+{\int }_{-\frac{1}{3}}^{\frac{2}{3}}0dx+{\int }_{\frac{2}{3}}^{\frac{5}{3}}1dx+{\int }_{\frac{5}{3}}^{\frac{8}{3}}2dx+{\int }_{\frac{8}{3}}^{3}3dx}\\ & =-1{|}_{-1}^{-\frac{1}{3}}+0{|}_{-\frac{1}{3}}^{\frac{2}{3}}+x{|}_{\frac{2}{3}}^{\frac{5}{3}}+2x{|}_{\frac{5}{3}}^{\frac{8}{3}}+3x{|}_{\frac{8}{3}}^3\\ & =-(-{\displaystyle \frac{1}{3}+1})+0+\left({\displaystyle \frac{5}{3}-\frac{2}{3}}\right)+2(\frac{8}{3}-\frac{5}{3})+3(3-{\displaystyle \frac{8}{3}})\\ & =-{\displaystyle \frac{2}{3}+0+1+2+\frac{3}{3}}\\ & ={\displaystyle 3\frac{1}{3}}\end{array}$.

$\begin{array}{l}\\ 33.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_0^n\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle n^2}\\ \text{b}.{\displaystyle \frac{n(n-1)}{2}}\\ \text{c}.{\displaystyle \frac{n(n+1)}{2}}\\ \text{d}.{\displaystyle \frac{(n+1)(n+2)}{2}}\\ \text{e}.{\displaystyle \frac{n^2+1}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& {\displaystyle {\int }_0^{n}f(x)dx={\displaystyle {\int }_0^n\left[x\right]dx}}\\ & ={\displaystyle {\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx+\cdots +{\int }_{n-1}^n(n-1)dx}\\ & =0+x{|}_1^2+2x{|}_2^3+\cdots +(n-1)x{|}_{n-1}^n\\ & =0+(2-1)+2(3-2)+\cdots +(n-1)(n-(n-1))\\ & =0+1+2+\cdots (n-1)\\ & ={\displaystyle \frac{(n-1)((n-1)+1)}{2}}\\ & ={\displaystyle \frac{n(n-1)}{2}}\\ & \text{Perhatikan tabel berikut}\end{array}$.

$\begin{array}{l}\\ 34.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-2022}^{2022}\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left[x\right]dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left[x\right]dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_{-2022}^{2022}\left[x\right]dx}\\ & ={\displaystyle {\int }_{-2022}^{-2021}\left[x\right]dx+{\displaystyle {\int }_{-2021}^{-2020}\left[x\right]dx+\cdots +{\int }_{2021}^{2022}\left[x\right]dx}}\\ & =-2022+(-2021)+\cdots +2020+2021\\ & =-2022\end{array}\end{array}$.

$\begin{array}{l}\\ 35.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-2022}^{2022}(x-\left[x\right])dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left[x\right]dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left[x\right]dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_{-2022}^{2022}(x-\left[x\right])dx}\\ & ={\displaystyle {\int }_{-2022}^{2022}xdx-\left({\displaystyle {\int }_{-2022}^{-2021}\left[x\right]dx+\cdots +{\int }_{2021}^{2022}\left[x\right]dx}\right)}\\ & \text{Sebelumnya perlu diingat bahwa}\\ & {\displaystyle {\int }_{-2022}^{2022}xdx\text{adalah}\text{fungsi ganjil, maka}=0}\\ & \text{Sehingga nilai akhirnya adalah}\\ & =0-(-2022)\text{lihat pembahasan sebelumnya}\\ & =2022\end{array}\end{array}$.

Contoh Soal 6 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 26.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Diketahui sebuah lingkaran berjari-jari}2\\ & \text{sebagaimana ilustrasi dibawah. Jika tali}\\ & \text{busur pada gambar berjarak 1 dari garis}\\ & \text{tengah, maka luas daerah di atas tali busur}\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{b}.{\displaystyle 2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{c}.{\displaystyle {\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{d}.{\displaystyle 2{\int }_0^1\sqrt{4-x^2}dx}\\ \text{e}.{\displaystyle 2{\int }_0^{\sqrt{3}}\sqrt{4-x^2}dx}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Diketahui bahwa sebuah lingkaran}\\ & \text{dengan}r=2,\text{artinya}{x}^2+y^2=r^2=2^2=4\\ & \text{Diketahui pula persamaan garis}y=1,\text{maka}\\ & x^2+1^2=4\iff x^2=4-1=3\iff x=\pm \sqrt{3}\end{array}$

$.\begin{array}{rl}& \text{Sehingga luas daerah di atas tali busur}\\ & \text{yaitu berupa}\mathbf{\text{temberang}}\text{adalah}\\ & ={\displaystyle {\int }_{-\sqrt{3}}^{\sqrt{3}}(f_2(x)-f_1(x))dx}\\ & ={\displaystyle {\int }_{-\sqrt{3}}^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ & \text{ingat fungsi genap pada pembahasan}\\ & \text{sebelumnya, karena}f(x)=\sqrt{4-x^2}=f(-x)\\ & =2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\end{array}$.

$\begin{array}{l}\\ 27.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Diketahui sebuah lingkaran berjari-jari}2\\ & \text{sebagaimana ilustrasi dibawah. Jika tali}\\ & \text{busur pada gambar berjarak 1 dari garis}\\ & \text{tengah, maka luas daerah di bawah tali busur}\\ & \text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2\pi -2{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{b}.{\displaystyle 2\pi -2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{c}.{\displaystyle 4\pi -{\int }_0^1(\sqrt{4-x^2}-1)dx}\\ \text{d}.{\displaystyle 4\pi -2{\int }_0^{\sqrt{3}}(\sqrt{4-x^2}-1)dx}\\ \text{e}.{\displaystyle 4\pi -2{\int }_0^{\sqrt{3}}\sqrt{4-x^2}dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dengan ilustrasi sama dengan no.24 di atas}\\ & \text{Kita dapat tentukan luas wilayah di bawah}\\ & \mathbf{\text{tembereng}},\text{yaitu}:\\ & =\text{Luas lingkaran penuh itu}-\mathbf{\text{tembereng}}\\ & =\pi .r^2-2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\\ & =\pi .(2{)}^2-2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\\ & =4\pi -2{\displaystyle {\int }_0^{\sqrt{3}}\left({\displaystyle \sqrt{4-x^2}-1}\right)dx}\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& (\mathbf{\text{SBMPTN Mat IPA 2013}})\\ & \text{Luas daerah yang dibatasi oleh kurva}\\ & y=2-x^2\text{dan}y=\left|x\right|\text{adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle 2{\int }_{-1}^0(-x^2+x+2)dx}\\ \text{b}.{\displaystyle 2{\int }_{-1}^0(-x^2-x+2)dx}\\ \text{c}.{\displaystyle 2{\int }_{-1}^0(-x^2+x-2)dx}\\ \text{d}.{\displaystyle {\int }_{-1}^1(-x^2-x+2)dx}\\ \text{e}.{\displaystyle {\int }_0^1(-x^2+x+2)dx}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Kita tentukan poin batas-batasnya, yaitu}:\\ & \text{Titik potong kurva dengan sumbu-X},y=0\\ & y=2-x^2=0\iff (\sqrt{2}+x)(\sqrt{2}-x)=0\\ & \iff x=-\sqrt{2}\text{atau}x=\sqrt{2}\\ & \text{Titik potong kurva dengan garis, yaitu}:\\ & 2-x^2=\left|x\right|\iff x^2+\left|x\right|-2=0\\ & \bullet x\ge 0,\left|x\right|=x\\ & \iff x^2+x-2=(x+2)(x-1)=0\\ & \iff x=-2(\mathbf{\text{TM}})\text{atau}x=1\\ & \bullet x<0,\left|x\right|=-x\\ & \iff x^2-x-2=(x-2)(x+1)=0\\ & \iff x=2(\mathbf{\text{TM}})\text{atau}x=-1\\ & \text{Perhatikanlah ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Misalkan}{f}_1(x)=2-x^2\text{dan}{f}_2(x)=\left|x\right|\\ & \text{maka luas arsirannya adalah}:\\ & \mathbf{\text{Luas arsiran}}=2{\displaystyle {\int }_{-1}^0(f_1(x)-f_2(x))dx}\\ & =2{\displaystyle {\int }_{-1}^0(2-x^2-\left|x\right|)dx}\\ & =2{\displaystyle {\int }_{-1}^0(2-x^2-(-x))dx}\\ & =2{\displaystyle {\int }_{-1}^0(-x^2+x+2)dx}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_{-1}^3\left[x\right]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 2}\\ \text{b}.{\displaystyle 4}\\ \text{c}.{\displaystyle 6}\\ \text{d}.{\displaystyle 8}\\ \text{e}.{\displaystyle 10}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

Jika diperjelas lagi

$.\begin{array}{rl}& {\displaystyle {\int }_{-1}^{3}f(x)dx={\displaystyle {\int }_{-1}^3\left[x\right]dx}}\\ & ={\displaystyle {\int }_{-1}^0-1dx+{\int }_0^{1}0dx+{\int }_1^{2}1dx+{\int }_2^{3}2dx}\\ & =(-x){|}_{-1}^0+0+x{|}_1^2+2x{|}_2^3\\ & =(0-(-(-1)))+0+(2-1)+2(3-2)\\ & =-1+0+1+2\\ & =2\end{array}$.

$\begin{array}{l}\\ 30.& \text{Jika}\left[x\right]\text{menyatakan bilangan bulat}\\ & \text{terbesar yang}\le x,\text{maka nilai dari}\\ & {\displaystyle {\int }_a^b\left[x\right]dx+{\int }_a^b[-x]dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle a+b}\\ \text{b}.{\displaystyle 0}\\ \text{c}.{\displaystyle a-b}\\ \text{d}.{\displaystyle 2a}\\ \text{e}.{\displaystyle 2b}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan tabel berikut}\\ & \text{Misalkan}f(x)=\left[x\right],g(x)=[-x],\text{dan}\\ & h(x)=f(x)+g(x)=\left[x\right]+[-x]\end{array}$..

$.\begin{array}{|cccc|}\hline \text{Interval}& c\le f(x)<c+1& c\le g(x)<c+1& [h(x)]=c\\ ⋮& ⋮& ⋮& ⋮\\ -3\le x<-2& -3\le x<-2& 2<x\le 3& -3+2=-1\\ -2\le x<-1& -2\le x<-1& 1\le x<2& -2+1=-1\\ -1\le x<0& -1\le x<0& 0\le x<1& -1+0=-1\\ 0\le x<1& 0\le x<1& -1\le x<0& 0+-1=-1\\ 1\le x<2& 1\le x<2& -2\le x<-1& 1+-2=-1\\ 2\le x<3& 2\le x<3& -3\le x<-2& 2+-3=-1\\ ⋮& ⋮& ⋮& ⋮\\ a\le x<b& a\le x<b& -b\le x<-a& a-b\\ \hline\end{array}$.

Contoh Soal 5 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 21.& \text{Pada gambar berikut, inetgral yang}\\ & \text{menyatakan luas daerah yang diarsir adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^3(2x^2-8x+6)dx}\\ \text{b}.{\displaystyle {\int }_0^1(x^2-4x+3)dx-{\int }_1^3(x^2-4x+3)dx}\\ \text{c}.{\displaystyle {\int }_0^1(x^2+4x+3)dx+{\int }_1^3(x^2+4x+3)dx}\\ \text{d}.{\displaystyle {\int }_0^1(2x^2-8x+6)dx-{\int }_1^3(2x^2-8x+6)dx}\\ \text{e}.{\displaystyle {\int }_0^1(2x^2+8x+6)dx-{\int }_1^3(2x^2+8x+6)dx}\end{array}\\ \end{array}$.

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

$\begin{array}{l}\\ 22.& \text{Jika}f\text{fungsi ganjil, maka nilai}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}f(x)dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}f(x)dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perlu diketahui bahwa}\\ & f(x)=\{\begin{array}{ll}\text{fungsi genap}& \Rightarrow f(-x)=f(x)\\ \text{fungsi ganjil}& \Rightarrow f(-x)=-f(x)\end{array}\\ & \text{Untuk}\\ & \begin{array}{|cc|}\hline f\text{fungsi genap}& {\displaystyle {\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx}\\ f\text{fungsi ganjil}& {\displaystyle {\int }_{-a}^{a}f(x)dx=0}\\ \hline\end{array}\\ & \text{Karena}f(x)\text{fungsi ganjil, maka}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx=0}\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Jika}f\text{fungsi genap, maka nilai}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}f(x)dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}f(x)dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Perlu diketahui bahwa}\\ & f(x)=\{\begin{array}{ll}\text{fungsi genap}& \Rightarrow f(-x)=f(x)\\ \text{fungsi ganjil}& \Rightarrow f(-x)=-f(x)\end{array}\\ & \text{Untuk}\\ & \begin{array}{|cc|}\hline f\text{fungsi genap}& {\displaystyle {\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx}\\ f\text{fungsi ganjil}& {\displaystyle {\int }_{-a}^{a}f(x)dx=0}\\ \hline\end{array}\\ & \text{Karena}f(x)\text{fungsi genap, maka}\\ & {\displaystyle {\int }_{-2022}^{2022}f(x)dx=2{\displaystyle {\int }_0^{2022}f(x)dx}}\end{array}\end{array}$.

 $\begin{array}{l}\\ 24.& \text{Nilai dari}{\displaystyle {\int }_{-2022}^{2022}\left|x\right|dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle -2022}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}\left|x\right|dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}\left|x\right|dx}\\ \text{e}.{\displaystyle 2022}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Karena}f(x)=\left|x\right|,\text{adalah fungsi genap},\\ & \text{maka}\\ & {\displaystyle {\int }_{-2022}^{2022}\left|x\right|dx={\displaystyle 2{\int }_0^{2022}\left|x\right|dx={2022}^2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Nilai dari}{\displaystyle {\int }_{-2022}^{2022}x\left|x\right|dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{-{2022}^3}{3}}\\ \text{b}.{\displaystyle -2{\int }_0^{2022}x\left|x\right|dx}\\ \text{c}.{\displaystyle 0}\\ \text{d}.{\displaystyle 2{\int }_0^{2022}x\left|x\right|dx}\\ \text{e}.{\displaystyle \frac{{2022}^3}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Karena}f(x)=x\left|x\right|,\text{adalah fungsi ganjil},\\ & \text{maka}\\ & {\displaystyle {\int }_{-2022}^{2022}x\left|x\right|dx=0}\end{array}\end{array}$

Contoh Soal 4 Materi Integral Tentu Fungsi Aljabar

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

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Luas arsiran}=\text{Luas trapesium}-\text{luas II}\\ & \text{Proses penyelesaian 1}\\ & \text{Kurva}y=4x-x^2,\text{maka koordinat titik}\\ & \text{puncaknya adalah}:(p,q).\text{Pilih}y=(x)=0\\ & \iff y=f(x)=4x-x^2=x(4-x)=0\\ & \iff x_1=0\text{atau}{x}_2=4\\ & \text{dengan}p={\displaystyle \frac{x_1+x_2}{2}=\frac{0+4}{2}=2,\text{dan}}\\ & q=y=f(2)=4.2-2^2=8-4=4\\ & \text{Sehingga}(p,q)=(2,4)\\ & \text{Proses penyelesaian 2}\\ & \text{Untuk garis}y=-2x+8,\text{maka koordinat}\\ & \text{titik potongnya terhadap sumbu-X dan Y}:\\ & \begin{array}{|ccc|}\hline x& 0& 4\\ y& 8& 0\\ (x,y)& (0,8)& (4,0)\\ \hline\end{array}\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Luas arsiran}=\text{Luas trapesium}-\text{luas II}\\ & ={\displaystyle \frac{(8+4)\times 2}{2}-{\displaystyle {\int }_0^{2}4x-x^{2}dx}}\\ & =12-\left((2x^2-{\displaystyle \frac{1}{3}{x}^3}){|}_0^2\right)\\ & =12-((8-{\displaystyle \frac{8}{3}})-0)\\ & =12+{\displaystyle \frac{8}{3}-8=6\frac{2}{3}\text{satuan luas}}\end{array}$.

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

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Proses penyelesaian 1}\\ & \text{Kurva}y=4-x^2,\text{maka koordinat titik}\\ & \text{puncaknya adalah}:(p,q).\text{Pilih}y=(x)=0\\ & \iff y=f(x)=4-x^2=(2+x)(2-x)=0\\ & \iff x_1=-2\text{atau}{x}_2=2\\ & \text{dengan}p={\displaystyle \frac{x_1+x_2}{2}=\frac{-2+2}{2}=0,\text{dan}}\\ & q=y=f(0)=4-0^2=4\\ & \text{Sehingga}(p,q)=(0,4)\\ & \text{Proses penyelesaian 2}\\ & \text{Untuk garis}y=-x+2,\text{maka koordinat}\\ & \text{titik potongnya terhadap sumbu-X dan Y}:\\ & \begin{array}{|ccc|}\hline x& 0& 2\\ y& 2& 0\\ (x,y)& (0,2)& (2,0)\\ \hline\end{array}\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Luas arsiran}=\text{Dibawah kurva}-\text{segitiga}\\ & ={\displaystyle {\int }_0^{2}4-x^{2}dx-{\displaystyle \frac{1}{2}\times \text{alas}\times \text{tinggi}}}\\ & =(4x-{\displaystyle \frac{1}{3}{x}^3}){|}_0^2-{\displaystyle \frac{2\times 2}{2}}\\ & =8-{\displaystyle \frac{8}{3}-2=6-2{\displaystyle \frac{2}{3}=3\frac{1}{3}}}\\ & =\frac{10}{3}\text{satuan luas}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Luas daerah di bawah kurva}y=-x^2+8x\\ & \text{di atas}y=6x-24\text{dan terletak di }\\ & \text{kuadran I adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(x^2-2x-24)dx}\\ \text{b}.{\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+2x+24)dx}\\ \text{c}.{\displaystyle {\int }_0^6(-x^2+8x)dx+{\int }_6^8(-x^2+2x+24)dx}\\ \text{d}.{\displaystyle {\int }_4^6(6x-24)dx+{\int }_6^8(-x^2+8x)dx}\\ \text{e}.{\displaystyle {\int }_0^4(6x-24)dx+{\int }_0^6(-x^2+8x)dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Kita tentukan poin batas-batasnya dulu}\\ & \text{Titik potong kurva dengan sumbu-X, yaitu}:\\ & y=-x^2+8x=x(-x+8)=0\\ & \iff x=0\text{atau}x=8\\ & \text{Titik potong garis dengan sumbu-X, yaitu}:\\ & y=6x-24=0\iff x=6\\ & \text{Dan titik potong kurva dengan garis adalah}:\\ & y=y\iff -x^2+8x=6x-24\\ & \iff -x^2+2x+24=0\iff (-x-4)(x-6)=0\\ & \iff x_1=-4\text{atau}{x}_2=6\\ & \text{Sehingga luas arsiran}=\text{Luas I}+\text{Luas II}\\ & ={\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+8x)-(6x-24)dx}\\ & ={\displaystyle {\int }_0^4(-x^2+8x)dx+{\int }_4^6(-x^2+2x+24)dx}\\ & \text{Perhatikan ulustrasi berikut}\end{array}$ 

$\begin{array}{l}\\ 19.& \text{Luas daerah yang diarsir pada gambar}\\ & \text{di bawah dapat dirumuskan dengan}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_b^{d}g(x)dx-{\int }_b^{c}f(x)dx}\\ \text{b}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_b^d(g(x)-f(x))dx}\\ \text{c}.{\displaystyle {\int }_a^d(f(x)-g(x))dx}\\ \text{d}.{\displaystyle {\int }_a^d(f(x)-g(x))dx-{\int }_c^{d}g(x)dx}\\ \text{e}.{\displaystyle {\int }_a^b(f(x)-g(x))dx+{\int }_c^d(g(x)-f(x))dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Cukup jelas bahwa}\\ & \text{pada gambar di atas pada selang}\\ & \bullet a\le x\le b:{\displaystyle {\int }_a^b(f(x)-g(x))dx}\\ & \bullet b\le x\le d:{\displaystyle {\int }_b^d(g(x)-f(x))dx}\\ & \text{dikurangi luasan yang}\\ & \text{berselang}b\le x\le c\end{array}$.

$\begin{array}{l}\\ 20.& \text{Daerah}\text{R}\text{di kuadran II, dibatasi oleh}\\ & \text{grafik}y=x^2,y=x+2\text{dan}y=0\\ & \text{Integral yang sesuai kondisi di atas adalah}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_{-2}^{-1}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ \text{b}.{\displaystyle {\int }_{-1}^{-2}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ \text{c}.{\displaystyle {\int }_{-1}^{-2}{x}^{2}dx+{\int }_{-1}^0(x+2)dx}\\ \text{d}.{\displaystyle {\int }_0^2(x^2+x+2)dx}\\ \text{e}.{\displaystyle {\int }_0^2(-x^2+x+2)dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Dari gukup jelas bahwa}\\ & \text{daerah arsiran akan terformulasikan}\\ & {\displaystyle {\int }_{-2}^{-1}(x+2)dx+{\int }_{-1}^0{x}^{2}dx}\\ & \text{Berikut sedikit uraian prosesnya}\\ & \mathbf{\text{Sebagai langkah awal}},\text{tentukan dulu}\\ & \text{titik potong kura dengan garis, yaitu}:\\ & y=y\iff x^2=x+2\iff x^2-x-2=0\\ & \iff (x+1)(x-2)=0\\ & \iff x=-1\text{atau}x=2\\ & \mathbf{\text{Langkah berikutnya}}\\ & \text{Kita tentukan daerah arsiran, yaitu}\\ & \text{seperti terformulasikan di atas}\end{array}$.

Contoh Soal 3 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 11.& \text{Diketahui}{\displaystyle \int f(x)dx=a{x}^2+bx+c}\\ & \text{dengan}a\ne 0.\text{Jika}a,f(a),2b\text{membentuk}\\ & \text{barisan arimetika dan}f(b)=6,\text{maka}\\ & \text{nilai}{\displaystyle {\int }_0^{1}f(x)dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{17}{4}}\\ \text{b}.{\displaystyle \frac{21}{4}}\\ \text{c}.{\displaystyle \frac{25}{4}}\\ \text{d}.{\displaystyle \frac{13}{4}}\\ \text{e}.{\displaystyle \frac{11}{4}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int f(x)dx=a{x}^2+bx+c}\\ & \Rightarrow f(x)=2ax+b\\ & \Rightarrow f(a)=2a^2+b\\ & \Rightarrow f(b)=2ab+b=6\\ & \text{Karena}a,f(a),2b\text{membentuk barisan}\\ & \text{aritmetika, maka}f(a)={\displaystyle \frac{a+2b}{2}}\\ & \text{Sehingga}\\ & f(a)=2a^2+b={\displaystyle \frac{a+2b}{2}}\\ & \iff 4a^2-a=0\iff a(4a-1)=0\\ & \iff a=0\text{atau}a={\displaystyle \frac{1}{4}}\\ & \text{Dan karena}a\ne 0,\text{maka}a={\displaystyle \frac{1}{4}\Rightarrow b=4}\\ & \text{Selanjutnya}\\ & f(x)=2ax+b={\displaystyle \frac{1}{2}x+4}\\ & \text{Dari fakta di atas, maka}\\ & {\displaystyle {\int }_0^{1}f(x)dx={\int }_0^1{\displaystyle \frac{1}{2}x+4dx}}\\ & ={\displaystyle \frac{1}{4}{x}^2+4x{|}_0^1}\\ & =\left({\displaystyle \frac{1}{4}+4}\right)-0={\displaystyle \frac{17}{4}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 13.& \text{Jika}y={\displaystyle \frac{1}{3}(x^3+\frac{3}{x}),\text{maka}}\\ & {\displaystyle {\int }_1^2\sqrt{4+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{13}{14}}\\ \text{b}.{\displaystyle \frac{13}{6}}\\ \text{c}.{\displaystyle \frac{15}{6}}\\ \text{d}.{\displaystyle \frac{16}{6}}\\ \text{e}.{\displaystyle \frac{17}{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& y={\displaystyle \frac{1}{3}(x^3+\frac{3}{x})}\\ & \iff dy=(x^2-{\displaystyle \frac{1}{x^2}})dx\\ & \iff {\displaystyle \frac{dy}{dx}=(x^2-{\displaystyle \frac{1}{x^2}})}\\ & \iff {\left({\displaystyle \frac{dy}{dx}}\right)}^2={(x^2-{\displaystyle \frac{1}{x^2}})}^2={\displaystyle x^4-2+{\displaystyle \frac{1}{x^4}}}\\ & \text{maka}\\ & {\displaystyle {\int }_1^2\sqrt{4+{\left({\displaystyle \frac{dy}{dx}}\right)}^2}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{4+\left({\displaystyle x^4-2+{\displaystyle \frac{1}{x^4}}}\right)}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{\left({\displaystyle x^4+2+{\displaystyle \frac{1}{x^4}}}\right)}dx}\\ & ={\displaystyle {\int }_1^2\sqrt{{\left({\displaystyle x^2+{\displaystyle \frac{1}{x^2}}}\right)}^2}dx}\\ & ={\displaystyle {\int }_1^2{x}^2+{\displaystyle \frac{1}{x^2}dx}}\\ & ={\displaystyle \frac{1}{3}{x}^3-{\displaystyle \frac{1}{x}{|}_1^2}}\\ & =\left({\displaystyle \frac{8}{3}-\frac{1}{2}}\right)-\left({\displaystyle \frac{1}{3}-1}\right)\\ & ={\displaystyle \frac{16-3-2+6}{6}}\\ & ={\displaystyle \frac{17}{6}}\end{array}\end{array}$.

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

$.\begin{array}{rl}& \text{Luas arsiran}=12\\ & =4.a-{\displaystyle {\int }_1^4{\displaystyle \frac{a}{(4-1)(4-7)}(x-1)(x-7)dx}}\\ & =4a-{\displaystyle {\int }_1^4{\displaystyle -\frac{a}{9}(x^2-8x+7)dx}}\\ & =4a-{\displaystyle {\int }_1^4{\displaystyle \frac{a}{9}(-x^2+8x-7)dx}}\\ & =4a-(-{\displaystyle \frac{a}{27}{x}^3+\frac{4a}{9}{x}^2-\frac{7a}{9}x{|}_1^4})\\ & =4a+{\displaystyle \frac{64a}{27}-\frac{64a}{9}+\frac{28a}{9}-\frac{a}{27}+\frac{4a}{9}-\frac{7a}{9}}\\ & =4a-{\displaystyle \frac{18a}{9}=2a=12\iff a=6}\\ & \text{Berikut ilustrasi gambar lengkapnya}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Luas daerah yang diarsir pada}\\ & \text{gambar berikut dapat dinyatakan}\\ & \text{dengan}....\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^2(3x-x^2)dx}\\ \text{b}.{\displaystyle {\int }_0^2(x+3)dx-{\int }_0^2{x}^{2}dx}\\ \text{c}.{\displaystyle {\int }_0^1(x+3)dx-{\int }_0^2{x}^{2}dx}\\ \text{d}.{\displaystyle {\int }_0^1(x+3-x^2)dx-{\int }_1^2{x}^{2}dx}\\ \text{e}.{\displaystyle {\int }_0^1(x+3-x^2)dx-{\int }_1^2(4-x^2)dx}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \text{Daerah yang diarsir di atas dibatasi}\\ & \text{oleh 4 buah batas, yaitu}\\ & \bullet \text{Garis}y=4\\ & \bullet \text{Kurva}y=x^2\\ & \bullet \text{Garis}x=0,\text{dan}\\ & \bullet \text{Garis yang melalui titik}(0,3)\text{dan}(1,4)\\ & \begin{array}{|cc|}\hline \text{Titik}& \text{Garis}\\ \begin{array}{rl}& 0,3\\ & \text{dan}\\ & (1,4)\\ & \end{array}& \begin{array}{rl}{\displaystyle y}& =\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1\\ y& ={\displaystyle \frac{4-3}{1-0}(x-0)+3}\\ y& =x+3\end{array}\\ \hline\end{array}\\ & \text{Sehingga luas arisrannya adalah}:\\ & ={\displaystyle {\int }_0^1(x+3)-x^{2}dx+{\int }_1^{2}4-x^{2}dx}\end{array}$.

Contoh Soal 2 Materi Integral Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 6.& \text{Hasil dari}{\displaystyle {\int }_2^4(-x^2+6x-8)dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{38}{3}}\\ \text{b}.{\displaystyle \frac{26}{3}}\\ \text{c}.{\displaystyle \frac{20}{3}}\\ \text{d}.{\displaystyle \frac{16}{3}}\\ \text{e}.{\displaystyle \frac{4}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_2^4(-x^2+6x-8)dx}\\ & =-{\displaystyle \frac{1}{3}{x}^3+3x^2-8x{|}_2^4}\\ & =(-{\displaystyle \frac{1}{3}(4{)}^3+3(4{)}^2-8(4)})-(-{\displaystyle \frac{1}{3}(2{)}^3+3(2{)}^2-8(2)})\\ & =(-{\displaystyle \frac{64}{3}+48-32})-(-{\displaystyle \frac{8}{3}+12-16})\\ & =-{\displaystyle \frac{56}{3}+20=-{\displaystyle \frac{56+60}{3}={\displaystyle \frac{4}{3}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Hasil dari}{\displaystyle {\int }_0^2{x}^2(x+2)dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 6}\\ \text{b}.6{\displaystyle \frac{1}{3}}\\ \text{c}.6{\displaystyle \frac{2}{3}}\\ \text{d}.9{\displaystyle \frac{1}{3}}\\ \text{e}.{\displaystyle 20}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\displaystyle {\int }_0^2{x}^2(x+2)dx={\displaystyle {\int }_0^2{x}^3+2x^{2}dx}}}\\ & ={\displaystyle \frac{1}{4}{x}^4+\frac{2}{3}{x}^3{|}_0^2}\\ & =(-{\displaystyle \frac{1}{4}(2{)}^4+\frac{2}{3}(2{)}^3})-\left(0\right)\\ & =4+{\displaystyle \frac{16}{3}={\displaystyle \frac{4}{3}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai dari}{\displaystyle {\int }_0^{2}18x\sqrt{3x^2+4}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle 4}\\ \text{b}.16\\ \text{c}.112\\ \text{d}.128\\ \text{e}.{\displaystyle 168}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=3x^2+4\Rightarrow du=6xdx\\ & \text{Selanjutnya}\\ & {\displaystyle {\int }_0^{2}18x\sqrt{3x^2+4}dx={\displaystyle {\int }_0^{2}3\sqrt{u}dx}}\\ & ={\displaystyle {\int }_0^{2}3u^{{.}^{\frac{1}{2}}}dx=2u^{{.}^{\frac{3}{2}}}{|}_0^2}\\ & =2{(3x^2+4)}^{{.}^{\frac{3}{2}}}{|}_0^2\\ & =2{({3.2}^2+4)}^{{.}^{\frac{3}{2}}}-(2(0+4{)}^{{.}^{\frac{3}{2}}})\\ & =2.(16{)}^{{.}^{\frac{3}{2}}}={2.4}^3-{2.2}^3=128-16\\ & =112\end{array}\end{array}$ .

$\begin{array}{l}\\ 9.& \text{Hasil substitusi dari}u=x+1\text{pada}\\ & {\displaystyle {\int }_0^1{x}^2\sqrt{x+1}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{b}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{c}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{d}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\\ \text{e}.{\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x+1\Rightarrow x=u-1\Rightarrow dx=du\\ & \text{Selanjutnya}\\ & {\displaystyle {\int }_0^1(u-1{)}^2\sqrt{u}du}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Nilai dari}{\displaystyle {\int }_0^{1}5x(1-x{)}^{6}dx=....}\\ & \begin{array}{l}\text{a}.{\displaystyle \frac{75}{56}}\\ \text{b}.{\displaystyle \frac{10}{56}}\\ \text{c}.{\displaystyle \frac{5}{56}}\\ \text{d}.-{\displaystyle \frac{7}{56}}\\ \text{e}.-{\displaystyle \frac{10}{56}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui soal bentuk}\mathbf{\text{integral parsial}}\\ & \text{dengan metode}\mathbf{\text{Tanzalin}}\text{diperoleh}\\ & \begin{array}{|cc|}\hline \text{Diturunkan}& \text{Diintegralkan}\\ 5x& (1-x{)}^6\\ 5& -{\displaystyle \frac{1}{7}(1-x{)}^7}\\ 0& {\displaystyle \frac{1}{56}(1-x{)}^8}\\ \hline\end{array}\\ & \text{Sehingga}\\ & {\displaystyle {\int }_0^{1}5x(1-x{)}^{6}dx}\\ & =(5x)(-{\displaystyle \frac{1}{7}(1-x{)}^7})-(5)\left({\displaystyle \frac{1}{56}(1-x{)}^8}\right){|}_0^1\\ & =(0-0)-(0-\left({\displaystyle \frac{5}{56}}\right))\\ & ={\displaystyle \frac{5}{56}}\end{array}\end{array}$ 

Contoh Soal 1 Materi Integral Tentu Fungsi Aljabar

$\begin{array}{l}\\ 1.& \mathbf{\text{(UAN 2014)}}\\ & \text{Hasil}{\displaystyle {\int }_{-1}^2(x^3+3x^2+4x+5)dx=....}\\ & \begin{array}{ll}\text{a}.& 33{\displaystyle \frac{1}{4}}\\ \text{b}.& 33{\displaystyle \frac{3}{4}}\\ \text{c}.& 32{\displaystyle \frac{1}{4}}\\ \text{d}.& 31{\displaystyle \frac{3}{4}}\\ \text{e}.& 23{\displaystyle \frac{3}{4}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\int }_{-1}^2(x^3+3x^2+4x+5)dx\\ & ={\displaystyle \frac{x^4}{4}+x^3+2x^2+5x{|}_{-1}^2}\\ & =\left({\displaystyle \frac{2^4}{4}+2^3+{2.2}^2+5.2}\right)\\ & -(\frac{{(-1)}^4}{4}+{(-1)}^3+2.{(-1)}^2+5(-1))\\ & =33\frac{3}{4}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \mathbf{\text{(UAN 2003)}}\\ & \text{Jika}f(x)={(x-2)}^2-4\text{dan}g(x)=-f(x),\\ & \text{maka luas daerah yang di batasi kurva }\\ & f\text{dan}g\text{adalah}....\\ & \begin{array}{ll}\text{a}.& 10{\displaystyle \frac{2}{3}\text{satuan luas}}\\ \text{b}.& 21{\displaystyle \frac{1}{3}\text{satuan luas}}\\ \text{c}.& 22{\displaystyle \frac{2}{3}\text{satuan luas}}\\ \text{d}.& 42{\displaystyle \frac{2}{3}\text{satuan luas}}\\ \text{e}.& 45{\displaystyle \frac{1}{3}\text{satuan luas}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \end{array}$.

$.\begin{array}{rl}& \text{Alternatif 1}\\ & \begin{array}{rl}& {\displaystyle {\int }_0^4(g(x)-f(x))dx}\\ & ={\displaystyle {\int }_0^4(4x-x^2)-(x^2-4x)dx}\\ & ={\displaystyle {\int }_0^4(8x-2x^2)dx}\\ & ={\displaystyle {[4x^2-\frac{2}{3}{x}^3]}_0^4}\\ & ={\displaystyle ({4.4}^2-\frac{2}{3}{.4}^3)-({4.0}^2-\frac{2}{3}{.0}^3)}\\ & ={\displaystyle (64-\frac{2}{3}.64)-0}\\ & ={\displaystyle \frac{64}{3}=21\frac{1}{3}\text{satuan luas}}\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan menggunakan rumus}L={\displaystyle \frac{D\sqrt{D}}{6a^2}}\\ & \begin{array}{|c|}\hline \begin{array}{rl}& f(x)=g(x)\\ & f(x)=-f(x),\text{ingat}g(x)=-f(x)\\ & 2f(x)=0,\text{tidak boleh disederhanakan},\\ & 2\times ({(x-2)}^2-4)=0,\text{karena akan }\\ & 2\times (x^2-4x)=0\text{mempengaruhi hasil }\\ & 2x^2-8x=0,\text{akhir}\\ & \{\begin{array}{ll}a=2,b=-8& c=0\\ D=b^2-4ac,& D={(-8)}^2-4(2)(0)=64\end{array}\\ & L_{\mathbf{\text{daerah}}}={\displaystyle \frac{\mathbf{\text{D}}\sqrt{\mathbf{\text{D}}}}{6{\mathbf{\text{a}}}^2}}\\ & ={\displaystyle \frac{64\sqrt{64}}{6(2{)}^2}}\\ & ={\displaystyle \frac{64\times 8}{6\times 4}}\\ & ={\displaystyle \frac{64}{3}}\\ & =21{\displaystyle \frac{1}{3}}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \mathbf{\text{(UAN 2014)}}\\ & \text{Luas daerah yang diarsir pada gambar }\\ & \text{dapat dinyatakan dengan rumus}\end{array}$.

$.\begin{array}{l}\\ & \begin{array}{ll}\text{a}.& {\displaystyle {\int }_0^{4}4xdx-{\int }_2^4(2x-4)dx}\\ \text{b}.& {\displaystyle {\int }_0^{4}4xdx-{\int }_2^4(2x+4)dx}\\ \text{c}.& {\displaystyle {\int }_0^{4}2\sqrt{2}dx-{\int }_2^4(2x-4)dx}\\ \text{d}.& {\displaystyle {\int }_0^{4}2\sqrt{2}dx-{\int }_2^4(4-2x)dx}\\ \text{e}.& {\displaystyle {\int }_0^{4}2\sqrt{x}dx-{\int }_2^4(4+2x)dx}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\text{Luas}}_{\text{arsiran}}={\displaystyle {\int }_0^4{y}_{1}dx-{\int }_2^4{y}_{2}dx}\\ & ={\displaystyle {\int }_0^4\sqrt{4x}dx-{\int }_2^4(2x-4)dx}\\ & ={\displaystyle {\int }_0^{4}2\sqrt{x}dx-{\int }_2^4(2x-4)dx.}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \mathbf{\text{(UAN 2014)}}\\ & \text{Volume benda putar yang terbentuk dari daerah}\\ & \text{yang dikuadran I yang dibatasi oleh kurva}\\ & x=2\sqrt{3}{y}^2,\text{sumbu Y , dan lingkaran}{x}^2+y^2=1,\\ & \text{diputar mengelilingi sumbu Y adalah}....\\ & \begin{array}{ll}\text{a}.& {\displaystyle \frac{4}{60}\pi \text{satuan volum}}\\ \text{b}.& {\displaystyle \frac{17}{60}\pi \text{satuan volum}}\\ \text{c}.& {\displaystyle \frac{23}{60}\pi \text{satuan volum}}\\ \text{d}.& {\displaystyle \frac{44}{60}\pi \text{satuan volum}}\\ \text{e}.& {\displaystyle \frac{112}{60}\pi \text{satuan volum}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Volumenya jika diputar mengelilingi Sumbu }\\ & \text{Y adalah}:\\ & V=\pi {\displaystyle {\int }_0^{\frac{1}{2}}{x}_1^{2}dy+\pi {\int }_{\frac{1}{2}}^1{x}_2^{2}dy}\\ & \iff V=\pi {\int }_0^{\frac{1}{2}}{\left(2\sqrt{3}{y}^2\right)}^{2}dy+\pi {\int }_{\frac{1}{2}}^1(1-y^2)dy\\ & \iff V=\frac{12}{5}{y}^5\pi {|}_0^{\frac{1}{2}}+(y-\frac{1}{3}{y}^3)\pi {|}_{\frac{1}{2}}^1\\ & \iff V=\pi (\frac{12}{5}\times \frac{1}{32}+(1-\frac{1}{3})-(\frac{1}{2}-\frac{1}{3}\times \frac{1}{8}))\\ & \iff V=\frac{17}{60}\pi .\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jika}{\displaystyle \frac{d}{dx}g(x)=f(x)\text{di mana}f(x)}\\ & \text{kontinu dari a sampai b, }\\ & \text{maka}{\displaystyle {\int }_a^{b}f(x).g(x)dx}\\ & \begin{array}{l}\\ \text{a}.0\\ \text{b}.f(b)-f(a)\\ \text{c}.g(b)-g(a)\\ \text{d}.{\displaystyle \frac{{[f(b)]}^2-{[f(a)]}^2}{2}}\\ \text{e}.{\displaystyle \frac{{[g(b)]}^2-{[g(a)]}^2}{2}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle {\int }_a^{b}f(x).g(x)dx}\\ & ={\displaystyle {\int }_a^{b}g(x).f(x)dx}\\ & ={\displaystyle {\int }_a^{b}g(x).{\displaystyle \frac{d(g(x))}{dx}dx,}}\\ & \text{ingat bahwa}{\displaystyle \frac{d}{dx}g(x)=f(x),}\\ & \text{f(x) kontinu dari a sampai b}\\ & ={\displaystyle {\int }_a^{b}g(x)d(g(x))}\\ & ={\left[{\displaystyle \frac{{(g(x))}^2}{2}}\right]}_a^b\\ & ={\displaystyle \frac{{[g(b)]}^2-{[g(a)]}^2}{2}}\end{array}\end{array}$.

Penggunaan Integral Tentu Fungsi Aljabar

 A. Luas

Menghitung luas yang dibatasi oleh sebuah kurva dan sumbu X kita dapat menggunakan bantuan integral tentuk sebagaimana uraian sebelumnya

Perhatikan ilustrasi gambar berikut

$\begin{array}{|cc|}\hline \text{Di Atas Sumbu X}& \text{Di Bawah Sumbu X}\\ & -{\displaystyle {\int }_a^{b}f(x)dx}\\ {\displaystyle {\int }_a^{b}f(x)dx}& atau\\ & {\displaystyle {\int }_b^{a}f(x)dx}\\ \hline\end{array}$.

B. Volume Benda Putar

Adapun untuk volume diformulasikan dengan integral tentu berikut

$\boxed{{\displaystyle V=\pi {\displaystyle {\int }_a^b{(f(x))}^{2}dx=\pi {\displaystyle {\int }_a^b{y}^{2}dx}}}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah luas daerah bidang berikut dan }\\ & \text{tentukan pula volumenya seandainya bidang }\\ & \text{yang diarsir tersebut diputar terhadap sumbu X}\end{array}$.

$.\begin{array}{rl}& \begin{array}{rl}{L}_{\text{Arsiran}}& ={\displaystyle {\int }_1^{3}2xdx}\\ & ={\displaystyle {\left[x^2\right]}_1^3}\\ & ={\left(3\right)}^2-{\left(1\right)}^2\\ & =9-1\\ & =8\mathbf{\text{satuan luas}}\end{array}\\ & \begin{array}{rl}{V}_{\text{Benda putar}}& =\pi {\displaystyle {\int }_1^3{\left(y\right)}^{2}dx}\\ & =\pi {\displaystyle {\int }_1^3{\left(2x\right)}^{2}dx}\\ & =\pi {\displaystyle {\int }_1^{3}4x^{2}dx}\\ & =\pi {\left[{\displaystyle \frac{4x^3}{3}}\right]}_1^3\\ & =\pi \left({\displaystyle \frac{4\times 3^3}{3}}\right)-\pi \left({\displaystyle \frac{4\times 1^3}{3}}\right)\\ & =36\pi -{\displaystyle \frac{4}{3}\pi }\\ & =34{\displaystyle \frac{2}{3}\pi \mathbf{\text{satuan volum}}}\end{array}\end{array}$.

$$\begin{array}{l}\\ 2.& \text{Diketahui parabola}{f}_1(x)=a_1{x}^2+b_{1}x+c_1\\ & \text{dan}{f}_2(x)=a_2{x}^2+b_{2}x+c_2.\\ & \text{Titik potong kedua parabola tersebut }\\ & \text{dapat cari dengan}\\ & \\ & f_1(x)=f_2(x)\\ & \iff a_1{x}^2+b_{1}x+c_1=a_2{x}^2+b_{2}x+c_2\\ & \iff a{x}^2+bx+c=0.\\ & \\ & \text{Jika kedua parabola berpotongan di dua}\\ & \text{titik, tunjukkan bahwa luas daerah antara}\\ & \text{kedua parabola tersebut dapat }\\ & \text{dinyatakan dengan}{\displaystyle \mathbf{\text{L}}=\frac{\mathbf{\text{D}}\sqrt{\mathbf{\text{D}}}}{{\mathbf{\text{6a}}}^{\mathbf{\text{2}}}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & a{x}^2+bx+c=0\{\begin{array}{ll}& x_1={\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}}\\ & \\ & x_2={\displaystyle \frac{-b-\sqrt{b^2-4ac}}{2a}}\end{array}\\ & \begin{array}{rl}L& ={\displaystyle {\int }_{\frac{-b-\sqrt{b^2-4ac}}{2a}}^{\frac{-b+\sqrt{b^2-4ac}}{2a}}(a{x}^2+bx+c)dx={\left[{\displaystyle \frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx}\right]}_{\frac{-b-\sqrt{b^2-4ac}}{2a}}^{\frac{-b+\sqrt{b^2-4ac}}{2a}}}\\ & =\left[{\displaystyle \frac{a}{3}{\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)}^3+{\displaystyle \frac{b}{2}{\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)}^2+c\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)}}\right]\\ & -\left[{\displaystyle \frac{a}{3}{\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)}^3+{\displaystyle \frac{b}{2}{\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)}^2+c\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)}}\right]\\ & ={\displaystyle \frac{a}{24a^3}[({\sqrt{D}}^3-3{\sqrt{D}}^{2}b+3\sqrt{D}{b}^2-b^3)+({\sqrt{D}}^3+3{\sqrt{D}}^{2}b+3\sqrt{D}{b}^2+b^3)]}\\ & +{\displaystyle \frac{b}{8a^2}[(b^2-2b\sqrt{D}+{\sqrt{D}}^2)-(b^2+2b\sqrt{D}+{\sqrt{D}}^2)]+{\displaystyle \frac{c}{2a}[(-b+\sqrt{D})-(-b-\sqrt{D})]}}\\ & ={\displaystyle \frac{1}{24a^2}[2{\sqrt{D}}^3+6\sqrt{D}{b}^2]+{\displaystyle \frac{b}{8a^2}[-4b\sqrt{D}]+{\displaystyle \frac{c}{2a}\left[2\sqrt{D}\right]}}}\\ & ={\displaystyle \frac{{\sqrt{D}}^3}{12a^2}+\frac{b^2\sqrt{D}}{4a^2}-\frac{b^2\sqrt{D}}{2a^2}+\frac{c\sqrt{D}}{a}={\displaystyle \frac{D\sqrt{D}}{12a^2}+\frac{b^2\sqrt{D}}{4a^2}-\frac{b^2\sqrt{D}}{2a^2}+\frac{c\sqrt{D}}{a}}}\\ & ={\displaystyle \frac{\sqrt{D}}{12a^2}(D+3b^2-6b^2+12ac)}\\ & ={\displaystyle \frac{\sqrt{D}}{12a^2}[(b^2-4ac)-3b^2+12ac]}\\ & ={\displaystyle \frac{\sqrt{D}}{12a^2}[-2b^2+8ac]}\\ & =-{\displaystyle \frac{\sqrt{D}}{6a^2}[b^2-4ac]=-\frac{\sqrt{D}}{6a^2}\left[D\right]}\\ & =-\frac{D\sqrt{D}}{6a^2},\mathbf{\text{luas tidak mungkin negatif}}\\ & ={\displaystyle \frac{D\sqrt{D}}{6a^2}◼}\end{array}\end{array}$$.

$\begin{array}{l}\\ 3.& \text{Tentukan volume benda putar yang terbentuk, }\\ & \text{jika suatu daerah yang dibatasi oleh kurva}\\ & y^2=x\text{dan}y=x\text{diputar mengelilingi sumbu X}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Menentukan batas}}\\ & \begin{array}{|rl|}\hline \begin{array}{rl}y& =y\\ x^2& =x\\ x^2-x& =0\\ x(x-1)& =0\\ x=0\text{atau}x& =1\\ & \\ & \\ & \end{array}& \begin{array}{rl}V& =\pi {\displaystyle {\int }_a^b(y_1^2-y_2^2)dx}\\ & =\pi {\displaystyle {\int }_0^1(x-x^2)dx}\\ & =\pi {\left[{\displaystyle \frac{1}{2}{x}^2-\frac{1}{3}{x}^3}\right]}_0^1\\ & =\pi \left[{\displaystyle \frac{1}{2}-\frac{1}{3}}\right]\\ V& ={\displaystyle \frac{1}{6}\pi }\end{array}\\ \hline\end{array}\\ & \mathbf{\text{Keterangan lanjutan}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa;}\\ & y^2=x\Rightarrow y=\sqrt{x},\text{dianggap sebagai}{y}_1\\ & \text{Sehingga}{y}_1-\text{nya adalah}\sqrt{x}\\ & \text{dan}y=x\text{dianggap sebagai}{y}_2\\ & (y_1^2-y_2^2)=({\left(\sqrt{x}\right)}^2-{\left(x\right)}^2)=x-x^2\end{array}\\ & \text{Jadi, volume dari benda putar tersebut}\\ & \text{dalam satuan volum adalah}{\displaystyle \frac{1}{6}\pi }\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan volume benda putar yang terbentuk,}\\ & \text{jika suatu daerah yang dibatasi oleh kurva}\\ & y=2x,y=x,x=1,\text{dan}x=3\\ & \text{diputar mengelilingi sumbu X}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi gambar berikut}\end{array}$.

$.\begin{array}{rl}& \text{Langkah-Langkah penyelesaiannya adalah:}\\ & \begin{array}{|cc|}\hline \text{Batas}& \text{Menentukan Volumenya}\\ x=1\text{dan}x=3& \begin{array}{rl}V& ={\displaystyle \pi {\int }_a^b(f^2(x)-g^2(x))dx}\\ & ={\displaystyle \pi {\int }_1^3({\left(2x\right)}^2-{\left(x\right)}^2)dx}\\ & ={\displaystyle \pi {\int }_1^{3}3x^{2}dx}\\ & ={\displaystyle \pi {\left[x^3\right]}_1^3}\\ & ={\displaystyle \pi \left(3^3\right)-\pi \left(1^3\right)}\\ & =27\pi -1\pi \\ V& =26\pi \mathbf{\text{Satuan Volum}}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukan volume daerah yang dibatasi}\\ & \text{oleh lingkaran}{x}^2+y^2=4,\text{selang}\\ & -2\le x\le 2\text{dan}\text{diputar mengelilingi}\\ & \text{sumbu X}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikan ilustrasi berikut}\end{array}$.

$.\begin{array}{rl}& \text{Langkah-Langkah penyelesaiannya adalah:}\\ & \begin{array}{|cc|}\hline \text{Batas}& \text{Menentukan Volumenya}\\ x=-2\text{sampai}x=2& \begin{array}{rl}V& ={\displaystyle \pi {\int }_a^b{y}^{2}dx}\\ & ={\displaystyle \pi {\int }_{-2}^2(4-x^2)dx}\\ & ={\displaystyle \pi {[4x-{\displaystyle \frac{x^3}{3}}]}_{-2}^2}\\ & ={\displaystyle \pi (8-{\displaystyle \frac{8}{3}})-\pi (-8+{\displaystyle \frac{8}{3}})}\\ & ={\displaystyle \pi (8+8-\frac{8}{3}-\frac{8}{3})}\\ V& ={\displaystyle \frac{32}{3}\pi \mathbf{\text{Satuan Volum}}}\end{array}\\ \hline\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah luas daerah yang diarsir berikut}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa luas ellips}{\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}\\ & \text{adalah}\pi ab\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah volume benda putar yang terjadi}\\ & \text{jika daerah dari hasil putar tersebut mengelilingi }\\ & \text{sumbu X serta dibatasi oleh}\\ & \begin{array}{l}\text{a}.y=x+3,\text{sumbu x, garis x = 2, dan x = 4}\\ \text{b}.{\displaystyle y=\frac{1}{2}x+2,\text{sumbu x, garis x = 0, dan x = 4}}\\ \text{c}.{\displaystyle y=\sqrt{x+2},\text{sumbu x, garis x = 2, dan x = 4}}\\ \text{d}.y=4-2x,\text{sumbu x, garis x = 0, dan x = 4}\\ \text{e}.{\displaystyle x^2+y^2=16,\text{dan sumbu X}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukanlah volume benda putar yang}\\ & \text{terjadi jika daerah dari hasil putar tersebut }\\ & \text{mengelilingi sumbu X serta dibatasi oleh}\\ & \begin{array}{l}\text{a}.y=2x-x^2,\text{dan}y=0\\ \text{b}.{\displaystyle y^2=x,\text{dan}y=2}\\ \text{c}.{\displaystyle y=x^2,\text{dan}y=-x^2+4}\\ \text{d}.y=7-2x^2,\text{dan}y=x^2+4\\ \text{e}.{\displaystyle y=x^2,\text{dan}{y}^2=x}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007.  Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar ISI 2006. Jakarta: ESIS.

Inetgral Tentu Fungsi Aljabar

A. Integral Tentu

Sebelumnya sudah dibahasa mengenai integral tak tentu, di mana integral ini memiliki ciri selalu ada nilai konstantanya. Hal ini menunjukkan bahwa ada nilai yang belum terukur dengan jelas. Jika nantinya pada integral sudah ditentukan batas bawah dan batas atasnya, sehingga tidak akan diperlukan lagi nilai konstantanya. Adapun rumus formulasi dari integral tentu adalah sebagai berikut:

${\displaystyle {\int }_a^{b}f(x)dx={[F(x)]}_a^b=F(x){|}_a^b=F(b)-F(a)}$.

$\text{CONTOH SOAL}$.

$\begin{array}{rl}1.& {\int }_1^3(2x-1)dx\\ & ={[x^2-x]}_1^3\\ & =(9-3)-(1-1)=6\\ 2.& {\int }_0^5|x-1|+|x-2|dx\\ & =\frac{1}{2}(x-1)|x-1|{|}_0^5+\frac{1}{2}(x-2)|x-2|{|}_0^5\\ & =(\frac{1}{2}.4.4+\frac{1}{2}.3.3)-(\frac{1}{2}.-1.1+\frac{1}{2}.-2.2)\\ & =8+4\frac{1}{2}+\frac{1}{2}+2=15\end{array}$.

Anda dapat mengecek jawaban di atas dengan melmperhatikan gambar kemudian menyelesaikannya dengan prinsip segitiga, persegi, dan atau persegipanjang.

B. Sifat-Sifat Integral Tentu

$\begin{array}{rl}1.& {\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx\\ 2.& {\displaystyle {\int }_a^{b}k.f(x)dx=k{\int }_a^{b}f(x)dx}\\ 3.& {\displaystyle {\int }_a^b(f(x)\pm g(x))dx}\\ & ={\displaystyle {\int }_a^{b}f(x)dx\pm {\int }_a^{b}g(x)dx}\\ 4.& {\displaystyle {\int }_a^{b}f(x)dx}\\ & ={\displaystyle {\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx,dengana<c<b}\\ 5.& {\displaystyle {\int }_a^{a}f(x)dx=0}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Hitunglah hasil dari integral}{\displaystyle {\int }_1^3(2y+1)dy}\\ \\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\int }_1^3(2y+1)dy& =y^2+y{|}_1^3\\ & =(3^2+3)-(1^2+1)\\ & =12-2\\ & =10\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Hitunglah hasil dari integral}{\displaystyle {\int }_{-1}^1{x}^3-1dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\int }_{-1}^1& (x^3-1)dx\\ & =-({\displaystyle \frac{1}{4}{x}^4-x{|}_{-1}^1)}\\ & =-\left({\displaystyle \frac{1}{4}.{\left(1\right)}^4-1}\right)+\left({\displaystyle \frac{1}{4}.{(-1)}^4-(-1)}\right)\\ & =-\left({\displaystyle \frac{1}{4}-1}\right)+\left({\displaystyle \frac{1}{4}+1}\right)\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Hitunglah hasil dari integral}{\displaystyle {\int }_{-2}^1{(x+3)}^{2}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\int }_{-2}^1& {(x+3)}^{2}dx\\ & ={\displaystyle \frac{1}{3}{x}^2+3x^2+9x{|}_{-2}^1}\\ & =\left({\displaystyle \frac{1}{3}.{\left(1\right)}^3+{3.1}^2+9.1}\right)-\left({\displaystyle \frac{1}{3}.{(-2)}^3+3.(-2{)}^2+9.(-2)}\right)\\ & =\left({\displaystyle \frac{1}{3}+12}\right)-(-{\displaystyle \frac{8}{3}+12-18})\\ & ={\displaystyle \frac{9}{3}+18=3+18=21}\end{array}\end{array}$ .

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ & \text{Hitunglah nilai tiap integral berikut ini}\\ & \begin{array}{l}\\ 1.& {\displaystyle {\int }_0^{3}2xdx}& 10.& {\displaystyle {\int }_1^2\sqrt{x^5}dx}\\ 2.& {\displaystyle {\int }_2^{3}2x^{2}dx}& 11.& {\displaystyle {\int }_1^2{\displaystyle \frac{1}{x^2}dx}}\\ 3.& {\displaystyle {\int }_3^7{\displaystyle \frac{1}{2}{x}^{3}dx}}& 12.& {\displaystyle {\int }_4^{9}3\sqrt{x}dx}\\ 4.& {\displaystyle {\int }_0^3(x^2+2x-1)dx}& {\displaystyle }\\ 5.& {\displaystyle {\int }_0^3{(x+3)}^{2}dx}& \\ 6.& {\displaystyle {\displaystyle {\int }_{-1}^2(x+2)(x-1)dx}}& \\ 7.& {\displaystyle {\int }_1^3(t+t^2-3t^3)dt}& \\ 8.& {\displaystyle {\int }_1^4\left({\displaystyle \frac{x^4-x^3+\sqrt{x}-1}{x^2}}\right)dx}& \\ 9.& {\displaystyle {\int }_4^6(\sqrt{t}+t^2-\sqrt[3]{t^2})dt}\end{array}\end{array}$.