Contoh Soal 2 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{l}\\ 6.& {\displaystyle \int x^3\sqrt{x}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{2}{9}{x}^4\sqrt{x}+C}\\ \text{b}.{\displaystyle \frac{9}{2}{x}^4\sqrt{x}+C}\\ \text{c}.{\displaystyle \frac{1}{9}{x}^4\sqrt{x}+C}\\ \text{d}.{\displaystyle 9x^4\sqrt{x}+C}\\ \text{e}.{\displaystyle x^4\sqrt{x}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^3\sqrt{x}dx=\int x^3.x^{\frac{1}{2}}dx}\\ & =\int x^{3\frac{1}{2}}dx=\int x^{\frac{7}{2}}dx={\displaystyle \frac{x^{\frac{7}{2}+1}}{{\displaystyle \frac{7}{2}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{9}{2}}}{{\displaystyle \frac{9}{2}}}+C={\displaystyle \frac{2}{9}{x}^{4\frac{1}{2}}+C={\displaystyle \frac{2}{9}{x}^4\sqrt{x}+C}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& {\displaystyle \int x\sqrt{x\sqrt[3]{x^2}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{17}{6}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{b}.{\displaystyle \frac{6}{17}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{c}.{\displaystyle x^2\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{d}.{\displaystyle \frac{6}{17}x\sqrt{x\sqrt[3]{x^2}}+C}\\ \text{e}.{\displaystyle \frac{1}{2}x\sqrt{x\sqrt[3]{x^2}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x\sqrt[3]{x^2}}dx}\\ & ={\displaystyle \int x{(x.x^{\frac{2}{3}})}^{\frac{1}{2}}dx}\\ & ={\displaystyle \int x^{1+\frac{1}{2}+\frac{2}{6}}dx}\\ & ={\displaystyle \int x^{\frac{11}{6}}dx={\displaystyle \frac{x^{\frac{11}{6}+1}}{{\displaystyle \frac{11}{6}+1}}+C}}\\ & ={\displaystyle \frac{x^{\frac{17}{6}}}{{\displaystyle \frac{17}{6}}}+C={\displaystyle \frac{6}{17}{x}^2.x^{\frac{5}{6}}+C}}\\ & ={\displaystyle \frac{6}{17}{x}^2{(x.x^{\frac{2}{3}})}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{17}{x}^2\sqrt{x\sqrt[3]{x^2}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& {\displaystyle \int x^2\sqrt{x^2\sqrt[3]{x^3}}dx=....}\\ & \begin{array}{l}\\ \mathbf{\text{a}}.{\displaystyle x^3\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{b}}.{\displaystyle \frac{27}{6}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{c}}.{\displaystyle \frac{6}{27}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{d}}.{\displaystyle \frac{6}{21}{x}^2\sqrt{x^2\sqrt[3]{x^3}}+C}\\ \mathbf{\text{e}}.{\displaystyle \frac{6}{21}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^2\sqrt{x^2\sqrt[3]{x^3}}dx=\int x^2{(x^2.x^1)}^{\frac{1}{2}}dx}\\ & =\int x^{2+\frac{2}{2}+\frac{1}{2}}dx=\int x^{\frac{7}{2}}dx={\displaystyle \frac{x^{\frac{7}{2}+1}}{{\displaystyle \frac{7}{2}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{9}{2}}}{{\displaystyle \frac{9}{2}}}+C}\\ & ={\displaystyle \frac{2}{9}{x}^{\frac{9}{2}}+C}\\ & ={\displaystyle \frac{2.3}{9.3}{x}^3.x^{\frac{3}{2}}+C}\\ & ={\displaystyle \frac{6}{27}{x}^3{\left(x^3\right)}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{27}{x}^3\sqrt{x^2.x}+C}\\ & ={\displaystyle \frac{6}{27}{x}^3\sqrt{x^2\sqrt[3]{x^3}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& {\displaystyle \int x^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{b}.{\displaystyle \frac{6}{23}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{c}.{\displaystyle \frac{23}{6}{x}^4\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{d}.{\displaystyle \frac{23}{6}{x}^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\\ \text{e}.{\displaystyle \frac{3}{4}{x}^3\sqrt{{\displaystyle \frac{1}{x}\sqrt[3]{x^2}}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x^3\sqrt{\frac{1}{x}\sqrt[3]{x^2}}dx=\int x^3{(x^{-1}.x^{\frac{2}{3}})}^{\frac{1}{2}}dx}\\ & =\int x^{3-\frac{1}{2}+\frac{1}{3}}dx=\int x^{\frac{17}{6}}dx\\ & ={\displaystyle \frac{x^{\frac{17}{6}+1}}{{\displaystyle \frac{17}{6}+1}}+C={\displaystyle \frac{x^{\frac{23}{6}}}{{\displaystyle \frac{23}{6}}}+C}}\\ & ={\displaystyle \frac{6}{23}{x}^{\frac{24-1}{6}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4.x^{-\frac{1}{6}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4.{\left(x^{-\frac{1}{3}}\right)}^{\frac{1}{2}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4\sqrt{x^{-1+\frac{2}{3}}}+C}\\ & ={\displaystyle \frac{6}{23}{x}^4\sqrt{\frac{1}{x}\sqrt[3]{x^2}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& {\displaystyle \int x\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle x^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{b}.{\displaystyle \frac{13}{8}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{c}.{\displaystyle \frac{8}{13}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{d}.{\displaystyle \frac{1}{2}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\\ \\ \text{e}.{\displaystyle \frac{8}{5}{x}^2\sqrt{{\displaystyle \frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}dx}\\ & =\int x{\left(x^{-1}{\left(x{\left(x^{-1}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}\right)}^{\frac{1}{2}}dx\\ & =\int x(x^{-\frac{1}{2}}.x^{\frac{1}{4}}.x^{-\frac{1}{8}})dx\\ & =\int x^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}}dx=\int x^{\frac{5}{8}}dx\\ & ={\displaystyle \frac{x^{\frac{5}{8}+1}}{{\displaystyle \frac{5}{8}+1}}+C}\\ & ={\displaystyle \frac{x^{\frac{13}{8}}}{{\displaystyle \frac{13}{8}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^{\frac{16-3}{8}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2.x^{-\frac{3}{8}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{x^{-\frac{3}{4}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{x^{-1+\frac{1}{4}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}.{\left(x^{\frac{1}{2}}\right)}^{\frac{1}{2}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}\sqrt{x^{1-\frac{1}{2}}}}+C}\\ & ={\displaystyle \frac{8}{13}{x}^2\sqrt{\frac{1}{x}\sqrt{x\sqrt{\frac{1}{x}}}}+C}\end{array}\end{array}$

Contoh Soal 1 Materi Integral Tak Tentu Fungsi Aljabar

$\begin{array}{l}\\ 1.& {\displaystyle \int x^{2}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{3}{x}^3+C}\\ \text{b}.{\displaystyle \frac{1}{4}{x}^6+C}\\ \text{c}.{\displaystyle \frac{1}{3}{x}^6+C}\\ \text{d}.{\displaystyle \frac{1}{6}{x}^3+C}\\ \text{e}.{\displaystyle \frac{2}{3}{x}^3+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{2}dx={\displaystyle \frac{x^{2+1}}{2+1}+C={\displaystyle \frac{1}{3}{x}^3+C}}}\end{array}$

$\begin{array}{l}\\ 2.& {\displaystyle \int x^{-2}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -2x^{-1}+C}\\ \text{b}.{\displaystyle -x^{-1}+C}\\ \text{c}.{\displaystyle -\frac{1}{2}{x}^{-2}+C}\\ \text{d}.{\displaystyle -\frac{1}{3}{x}^{-3}+C}\\ \text{e}.{\displaystyle -3x^{-3}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{-2}dx={\displaystyle \frac{x^{-2+1}}{-2+1}+C={\displaystyle \frac{1}{-1}{x}^{-1}+C=-x^{-1}+C}}}\end{array}$.

$\begin{array}{l}\\ 3.& {\displaystyle \int x^{{.}^{\frac{1}{3}}}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{3}{4}{x}^{\frac{4}{3}}+C}\\ \text{b}.{\displaystyle x^{\frac{4}{3}}+C}\\ \text{c}.{\displaystyle \frac{3}{4}{x}^{\frac{2}{3}}+C}\\ \text{d}.{\displaystyle x^{-\frac{2}{3}}+C}\\ \text{e}.{\displaystyle \frac{3}{4}{x}^{-\frac{2}{3}}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int x^{{.}^{\frac{1}{3}}}dx={\displaystyle \frac{x^{\frac{1}{3}+1}}{{\displaystyle \frac{1}{3}+1}}+C={\displaystyle \frac{1}{{\displaystyle \frac{4}{3}}}{x}^{{.}^{\frac{4}{3}}}+C={\displaystyle \frac{3}{4}{x}^{{.}^{\frac{4}{3}}}+C.}}}}\end{array}$.

$\begin{array}{l}\\ 4.& {\displaystyle \int \frac{1}{x^3}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle -\frac{1}{2x^2}+C}\\ \text{b}.{\displaystyle -\frac{2}{x^2}+C}\\ \text{c}.{\displaystyle \frac{1}{3x^4}+C}\\ \text{d}.{\displaystyle \frac{3}{x^4}+C}\\ \text{e}.{\displaystyle -\frac{1}{4x^3}+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \int \frac{1}{x^3}dx=\int x^{-3}dx}\\ & ={\displaystyle \frac{x^{-3+1}}{-3+1}+C=\frac{x^{-2}}{-2}+C=-\frac{1}{2x^2}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& {\displaystyle \int \frac{1}{3}{x}^{3}dx=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{3}{x}^4+C}\\ \text{b}.{\displaystyle \frac{1}{4}{x}^4+C}\\ \text{c}.{\displaystyle x^4+C}\\ \text{d}.{\displaystyle \frac{1}{12}{x}^4+C}\\ \text{e}.{\displaystyle \frac{4}{3}{x}^4+C}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & {\displaystyle \int \frac{1}{3}{x}^{3}dx={\displaystyle \frac{1}{3}.\frac{x^{3+1}}{3+1}+C={\displaystyle \frac{1}{12}{x}^4+C}}}\end{array}$.

Penggunaan Integral Tak Tentu

Penggunaan Integral Tak Tentu

Penggunaan integral tak tentu ini dapat digunakan dalam menentukan suatu fungsi jika turunan dari fungsi tersebut diberikan. Selain itu untuk menentukan posisi, kecepatan, percepatan suatu benda pada waktu tertentu.

$\begin{array}{rl}& \text{Misalkan}\\ & \bullet s\text{adalah menunjukkan posisi}\\ & \bullet v\text{adalah menunjukkan kecepatan}\\ & \bullet a\text{adalah menunjukkan percepatan}\\ & \bullet t\text{adalah menunjukkan waktu}\\ \\ & \mathbf{\text{Perhatikan hubungan berikut}}\\ & v={\displaystyle \frac{ds}{dt}\iff ds=vdt}\\ & \iff {\displaystyle \int ds=\int vdt}\\ & \iff s={\displaystyle \int vdt}\\ \\ & \mathbf{\text{Demikian juga hubungan berikut}}\\ & a={\displaystyle \frac{dv}{dt}\iff dv=adt}\\ & \iff {\displaystyle \int dv=\int adt}\\ & \iff v={\displaystyle \int adt}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan}y,\text{Jika}{\displaystyle \frac{dy}{dx}=2022x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & {\displaystyle \frac{dy}{dx}=2022x\iff dy=2022xdx}\\ & \iff {\displaystyle \int dy=\int 2022xdx}\\ & \iff y={\displaystyle \frac{2022}{2}{x}^2+C}\\ & \iff y=1011x^2+C\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui fungsi turunan pertama kurva}\\ & \text{adalah}{\displaystyle \frac{dy}{dx}=2x-2.\text{Jika kurva melalui}}\\ & \text{titik}(3,2),\text{tentukan persamaan dari kurva}\\ & \text{tersebut}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & {\displaystyle \frac{dy}{dx}=2x-2}\\ & \iff dy=(2x-2)dx\\ & \iff {\displaystyle \int dy=\int (2x-2)dx}\\ & \iff y=x^2-2x+C\\ & \text{Karena kurva melalui}(3,2),\text{maka}\\ & (2)=(3{)}^2-2(3)+C\iff 2=3+C\\ & \iff C=-1\\ & \text{Jadi},y=x^2-2x-1\text{atau}\\ & f(x)=x^2-2x-1\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui bahwa}{f}^{″}(x)=x^2.\text{Jika}{f}^{\prime }(0)=6\\ & \text{dan}f(0)=3,\text{tentukanlah}f(x)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & {\displaystyle f^{″}(x)=x^2}\\ & \iff {\displaystyle \int f^{″}(x)dx={\displaystyle \int x^{2}dx}}\\ & \iff {\displaystyle f^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+C}}\\ & \text{Karena}{f}^{\prime }(x)=6,\text{maka}\\ & f^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+C\iff 6={\displaystyle \frac{1}{3}(0{)}^3+C}}\\ & \iff C=6\\ & \text{Sehingga}{f}^{\prime }(x)={\displaystyle \frac{1}{3}{x}^3+6}\\ & \text{Selanjutnya}{\displaystyle \int f^{\prime }(x)dx={\displaystyle \int {\displaystyle \frac{1}{3}{x}^3+6dx}}}\\ & f(x)={\displaystyle \frac{1}{12}{x}^4+6x+C}\\ & \text{Dan juga karena}f(0)=3,\text{maka}\\ & f(0)={\displaystyle \frac{1}{12}(0{)}^4+6(0)+C=3}\\ & C=3,\text{sehingga diperoleh}\\ & f(x)={\displaystyle \frac{1}{12}{x}^4+6x+3}\\ & \text{Jadi},f(x)={\displaystyle \frac{1}{12}{x}^4+6x+3}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui}{f}^{\prime }\left(x\right)=6x^2-2x+6\text{dan}\\ & \text{nilai}\text{fungsi}f\left(2\right)=-7.\text{Tentukanlah}\\ & \text{rumus}\text{fungsi}\text{tersebut}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f\left(x\right)& =\int f^{\prime }\left(x\right)dx\\ & =\int (6x^2-2x+6)dx\\ & =2x^3-x^2+6x+C\end{array}\\ \\ & \text{Karena}f\left(2\right)=-7,\text{maka}\\ \\ & \begin{array}{rl}f\left(2\right)& ={2.2}^3-2^2+6.2+C\\ \iff -7& =16-4+12+C\\ \iff C& =-31\end{array}\\ \\ & \text{Jadi},f\left(x\right)=2x^3-x^2+6x-31\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan}f(x),\text{Jika diketahui}\\ & \text{a}.f^{\prime }(x)=12x^3\text{dan}f(1)=8\\ & \text{b}.f^{\prime }(x)=x^2-2x+3\text{dan}f(3)=9\\ & \text{c}.f^{″}(x)=2,f^{\prime }(2)=2\text{dan}f(2)=10\\ & \text{d}.f^{″}(x)=x^2,f^{\prime }(0)=6\text{dan}f(0)=3\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan persamaan kurva}f(x)\\ & \text{disetiap titik}(x,y)\text{yang memenuhi}\\ & \text{syarat berikut}\\ & \text{a}.{\displaystyle \frac{dy}{dx}=4x+1\text{dan kurva melalui}(0,2)}\\ & \text{b}.{\displaystyle \frac{dy}{dx}={\displaystyle \frac{1}{x^2}+3\text{dan kurva melalui}(1,4)}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui}2x-y=3\text{merupakan garis}\\ & \text{kurva di titik}(1,-1).\text{Jika di tiap titik}\\ & \text{pada kurva berlaku}{\displaystyle y^{″}=2x^2-3x+1}\\ & \text{tentukan persamaan kurva tersebut}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui kecepatan sebuah sepeda}\\ & \text{diformulasian dengan}{\displaystyle \frac{ds}{dt}=3t^2+4t-1}\\ & \text{Jika}s\text{menyatakan jarak yang ditempuh}\\ & \text{dalam satuan meter},t\text{menyatakan waktu}\\ & \text{dengan}s=5\text{untuk}t=2,\text{tentukanlah}\\ & \text{jarak yang ditempuh pengendara dalam}\\ & \text{waktu 5 menit}\end{array}$.

Teknik Pengintegralan (Bagian 2)

2. Integral Parsial

2. 1 Integral Parsial

Jika teknik pada no.1 pada pembahasan sebelumnya tidak dapat digunakan, maka kemungkinan adalah dengan menggunakan teknik yang satunya ini, yaitu teknik integral parsial. Adapun untuk teknik integral ini diformulasikan dengan bentuk rumus

${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Perhatian kembali soal berikut}\\ & {\displaystyle \int x\sqrt{x-1}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misalkan}u=x-1\\ & du=1dx\iff du=dx\end{array}\\ & \text{Dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int (x-1+1)\sqrt{x-1}dx\\ & =\int (\underset{u}{\underbrace{(x-1)}}+1)\sqrt{\underset{u}{\underbrace{x-1}}}dx\\ & =\int (u+1)\sqrt{u}du\\ & =\int (u\sqrt{u}+\sqrt{u})du\\ & =\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})du\\ & ={\displaystyle \frac{1}{(\frac{3}{2}+1)}{u}^{(\frac{3}{2}+1)}+{\displaystyle \frac{1}{(\frac{1}{2}+1)}{u}^{(\frac{1}{2}+1)}+C}}\\ & ={\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}\\ \\ & \text{Alternatif 2}\\ & \begin{array}{l}\\ {\displaystyle \int x\sqrt{x-1}dx=\int \underset{{\displaystyle \underset{u}{\mid }}}{x}.\underset{{\displaystyle \underset{dv}{\mid }}}{\underbrace{\sqrt{x-1}dx}}}& \\ \begin{array}{cccc}u=x& & & dv=\sqrt{x-1}dx\\ du=dx& & & {\displaystyle \int dv=\int \sqrt{x-1}dx}\\ & & & v=\int {(x-1)}^{\frac{1}{2}}dx\\ & & & v={\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\end{array}\end{array}\\ & \text{Dengan integral parsial}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int \underset{{\displaystyle \underset{u}{\mid }}}{x}.\underset{{\displaystyle \underset{dv}{\mid }}}{\underbrace{\sqrt{x-1}dx}}=u.v-\int v.du\\ & =x.\left({\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\right)-\int \left({\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}\right)dx\\ & ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{2}{3}\times \frac{2}{5}{(x-1)}^{\frac{5}{2}}+C}}\\ & ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\end{array}\end{array}\end{array}$.

$\begin{array}{rl}& \text{Perhatikan bahwa}\\ & \begin{array}{|cc|}\hline \text{Hasil dengan Substitusi}& \text{Hasil dengan Integral Parsial}\\ {\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}& {\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\\ \hline\end{array}\\ & \text{Jika kita sejajarkan dengan ruas }\\ & \text{yang berbeda, maka}\\ & \begin{array}{rl}{\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}& ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}{(x-1)}^{\frac{5}{2}}+C}}\\ {\displaystyle \frac{2}{5}(x-1){(x-1)}^{\frac{3}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}}}& ={\displaystyle \frac{2x}{3}{(x-1)}^{\frac{3}{2}}-{\displaystyle \frac{4}{15}(x-1){(x-1)}^{\frac{3}{2}}}}\\ \left({\displaystyle \frac{2}{5}(x-1)+\frac{2}{3}}\right){(x-1)}^{\frac{3}{2}}& =\left({\displaystyle \frac{2x}{3}-\frac{4}{15}(x-1)}\right){(x-1)}^{\frac{3}{2}}\\ {\displaystyle \frac{2x}{5}-\frac{2}{5}+\frac{2}{3}}& ={\displaystyle \frac{2x}{3}-\frac{4x}{15}+\frac{4}{15}}\\ {\displaystyle \frac{2x}{5}+\frac{4}{15}}& ={\displaystyle \frac{2x}{5}+\frac{4}{15}}\\ \text{ruas kiri}& =\text{ruas kanan}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah integral dari}\int (x+2)dx\\ & \text{dengan cara}\\ & \text{a})\text{substitusi}\\ & \text{b})\text{parsial}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \mathbf{\text{Cara substitusi}}\\ & \int (x+2)dx=........?\\ & \text{Misalkan}m=x+2\\ & dm=dx\\ & \text{maka},\\ & \int (x+2)dx=\int \underset{\begin{array}{c}|\\ \mathbf{\text{m}}\end{array}}{\underbrace{(x+2)}}.\underset{\begin{array}{c}|\\ \mathbf{\text{dm}}\end{array}}{\underbrace{dx}}\\ & =\frac{1}{2}{m}^2+C=\frac{1}{2}(x+2{)}^2+C\\ & =\frac{1}{2}(x^2+4x+4)+C=\frac{1}{2}{x}^2+2x+\underset{\begin{array}{c}|\\ \mathbf{\text{C}}\end{array}}{\underbrace{2+C}}\\ & =\frac{1}{2}{x}^2+2x+C\end{array}\\ \\ & \begin{array}{rl}& \mathbf{\text{Cara parsial}}\\ & \int (x+2)dx=\int \underset{\begin{array}{c}|\\ \mathbf{\text{u}}\end{array}}{\underbrace{1}}.\underset{\begin{array}{c}|\\ \mathbf{\text{dv}}\end{array}}{\underbrace{(x+2)dx}}\\ & \{\begin{array}{c}u=1\to du=0\\ \\ v=\underset{\begin{array}{c}|\\ =\frac{1}{2}{x}^2+2x+C\end{array}}{\int dv}\leftarrow dv=(x+2)dx\end{array}\\ & =\mathbf{\text{u.v}}-\int \mathbf{\text{v.du}}\\ & =1.(\frac{1}{2}{x}^2+2x+C)-\int (\frac{1}{2}{x}^2+2x+C).0\\ & =\frac{1}{2}{x}^2+2x+C\end{array}\end{array}$.

2. 2 Aturan Tanzalin

Sumber Referensi

$\begin{array}{rl}& \text{Tentukanlah hasil integral berikut}\\ & {\displaystyle \int 2x^2(x-4{)}^{2}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Bentuk}{\displaystyle \int 2x^2(x-4{)}^{2}dx\text{dianggap sebagai}}\\ & {\displaystyle \int udv\text{dengan}u\text{adalah bagian yang mudah}}\\ & \text{kita diferensialkan, maka}u\text{kita pilihkan}\\ & \text{yaitu}u=2x^2\\ & \text{Selanjutnya dengan}\mathbf{\text{aturan Tanzalin}}\\ & \text{sebagai berikut}\end{array}$.

$\begin{array}{|cc|}\hline \text{Diderensialkan}& \text{Diintegralkan}\\ \begin{array}{rl}& +2x^2\\ & -4x\\ & +4\\ & -0\\ & \end{array}& \begin{array}{rl}& (x-4{)}^4\\ & {\displaystyle \frac{1}{5}(x-4{)}^5}\\ & {\displaystyle \frac{1}{30}(x-4{)}^6}\\ & {\displaystyle \frac{1}{210}(x-4{)}^7}\end{array}\\ \hline\end{array}$.

Hasil dari integral teknik ini adalah:

$\begin{array}{rl}& {\displaystyle \int 2x^2(x-4{)}^{4}dx}\\ & =(+2x^2)\left({\displaystyle \frac{1}{5}(x-4{)}^5}\right)+(-4x)\left({\displaystyle \frac{1}{30}{(x-4)}^6}\right)\\ & +(+4)\left({\displaystyle \frac{1}{210}(x-4{)}^7}\right)+C\\ & ={\displaystyle \frac{2}{5}{x}^2(x-4{)}^5-{\displaystyle \frac{2}{15}x(x-4{)}^6+{\displaystyle \frac{2}{105}(x-4{)}^7+C}}}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{rl}& \text{1. Selesaikan soal berikut ini}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int {\displaystyle x^2\sqrt{x+7}dx}}& \text{d}.& {\displaystyle \int -(2x^2+1)\sqrt{3-x}dx}\\ & \text{b}.& {\displaystyle \int 2x^2\sqrt{3-x}dx}& \text{e}.& {\displaystyle \int x^5(2x+1{)}^{6}dx}\\ & \text{c}.& {\displaystyle \int 3x^2\sqrt{2x+1}dx}& \text{f}.& {\displaystyle \int {\displaystyle \frac{2x^6}{\sqrt[3]{3x-1}}dx}}\end{array}\end{array}$ .

$\begin{array}{rl}& \text{2. Selesaikan soal berikut dengan Metode Tanzalin}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int -(2x^2+1)\sqrt{3-x}dx}\\ & \text{b}.& {\displaystyle \int x^5(2x+1{)}^{6}dx}\\ & \text{c}.& {\displaystyle \int {\displaystyle \frac{2x^6}{\sqrt[3]{3x-1}}dx}}\\ & \text{d}.& {\displaystyle \int {\displaystyle 3(x-2{)}^4.\sqrt[3]{x}dx}}\\ & \text{e}.& {\displaystyle \int {\displaystyle \frac{(x^4-3)}{\sqrt{1-x}}dx}}\\ & \text{f}.& {\displaystyle \int {\displaystyle x^3\sqrt{1-2x}dx}}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Teknik Pengintegralan (Bagian 1)

1. Integral Substitusi

Ada beberapa bentuk integral yang terkadang pengintegralannya membutuhkan teknik tertentu. Di antara bentuk tertentu itu adalah dengan substitusi, yaitu:

${\displaystyle \int u^n.u^{\prime }dx={\displaystyle \int u^{n}du={\displaystyle \frac{1}{n+1}{u}^{n+1}+C}}}$ .

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {(x^2+3)}^{20}2xdx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& x^2& +& 3& ,& \text{maka}\\ du& =& 2x& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& \int {(x^2+3)}^{20}2xdx=\int u^{20}du\\ & =\frac{1}{21}{u}^{21}+C=\frac{1}{21}{(x^2+3)}^{21}+C\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {(x^4-x^2)}^5(16x^3-8x)dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& x^4& -& x^2& ,& \text{maka}\\ du& =& 4x^3& -& 2x& dx\\ 4du& =& 16x^3& -& 8x& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int u^5.4du=\frac{4}{6}{u}^6+C}\\ & ={\displaystyle \frac{2}{3}{(x^4-x^2)}^6+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int \frac{x+2}{x^2+4x+4}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccccc}u& =& x^2& +& 4x& +& 4& ,& \text{maka}\\ du& =& 2x& +& 4& dx\\ \frac{1}{2}du& =& x& +& 2& dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int \frac{1}{u}.\frac{1}{2}du={\displaystyle \frac{1}{2}\int \frac{1}{u}du}}\\ & ={\displaystyle \frac{1}{2}\ln u+C=\frac{1}{2}\ln (x^2+4x+4)+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int \frac{e^{3y}}{{(1-2e^{3y})}^2}dy}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan}\\ & \{\begin{array}{ccccccc}u& =& 1& -& 2e^{3y}& ,& \text{maka}\\ du& =& -& 6e^{3y}& dy& ,\\ -\frac{1}{6}du& =& e^{3y}& dy\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int -\frac{1}{6}\frac{1}{u^2}du=-\frac{1}{6}\int \frac{du}{u^2}}\\ & =-{\displaystyle \frac{1}{6}(-1)\left(u^{-1}\right)+C}\\ & ={\displaystyle \frac{1}{6u}+C}\\ & ={\displaystyle \frac{1}{6}.\frac{1}{1-2e^{3y}}+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int 12x{(x^2+3)}^{5}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{r}\text{Misalkan}\\ u& =x^2+3\\ du& =2xdx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int 12x{(x^2+3)}^{5}dx}\\ & =\int {(x^2+3)}^5.6.2xdx\\ & =6\int \underset{u^5}{\underbrace{{(x^2+3)}^5}}.\underset{du}{\underbrace{2xdx}}\\ & =6\int u^{5}du\\ & =6.{\displaystyle \frac{u^{5+1}}{5+1}+C}\\ & =u^6+C\\ & ={(x^2+3)}^6+C\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int x\sqrt{x-1}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}u=x-1\\ & du=1dx\iff du=dx\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& {\displaystyle \int x\sqrt{x-1}dx}\\ & =\int (x-1+1)\sqrt{x-1}dx\\ & =\int (\underset{u}{\underbrace{(x-1)}}+1)\sqrt{\underset{u}{\underbrace{x-1}}}dx\\ & =\int (u+1)\sqrt{u}du\\ & =\int (u\sqrt{u}+\sqrt{u})du\\ & =\int (u^{\frac{3}{2}}+u^{\frac{1}{2}})du\\ & ={\displaystyle \frac{1}{(\frac{3}{2}+1)}{u}^{(\frac{3}{2}+1)}+{\displaystyle \frac{1}{(\frac{1}{2}+1)}{u}^{(\frac{1}{2}+1)}+C}}\\ & ={\displaystyle \frac{2}{5}{(x-1)}^{\frac{5}{2}}+{\displaystyle \frac{2}{3}{(x-1)}^{\frac{3}{2}}+C}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {\displaystyle \frac{x^5}{x^6+a}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x^6+a\\ & \iff du=6x^{5}dx\\ & \iff {\displaystyle \frac{1}{6}du=x^{5}dx}\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}& \int {\displaystyle \frac{x^5}{x^6+a}dx}\\ & =\int {\displaystyle \frac{1}{x^6+a}.x^{5}dx}\\ & =\int {\displaystyle \frac{1}{\underset{u}{\underbrace{x^6+a}}}.\underset{\frac{1}{6}du}{\underbrace{x^{5}dx}}}\\ & ={\displaystyle \frac{1}{6}\int \frac{1}{u}du}\\ & ={\displaystyle \frac{1}{6}\ln \left|u\right|+C}\\ & ={\displaystyle \frac{1}{6}\ln |x^6+a|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Selesaikan integral berikut}\\ & {\displaystyle \int {\displaystyle (x^2+1){(x^3+3x+2)}^{5}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Misalkan}\\ & u=x^3+3x+2\\ & \iff du=(3x^2+3)dx\\ & \iff du=3(x^2+1)dx\\ & \iff {\displaystyle \frac{1}{3}du=(x^2+1)dx}\end{array}\\ & \text{sehingga dengan integral substitusi}\\ & \begin{array}{rl}\int & (x^2+1){(x^3+3x+2)}^{5}dx\\ & =\int {(x^3+3x+2)}^5.(x^2+1)dx\\ & =\int \underset{u^5}{\underbrace{{(x^3+3x+2)}^5}}.\underset{\frac{1}{3}du}{\underbrace{(x^2+1)dx}}\\ & ={\displaystyle \frac{1}{3}\int u^{5}du}\\ & ={\displaystyle \frac{1}{3}.{\displaystyle \frac{u^{5+1}}{5+1}+C}}\\ & ={\displaystyle \frac{1}{18}.u^6+C}\\ & ={\displaystyle \frac{1}{18}{(x^3+3x+2)}^6+C}\end{array}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{rl}& \text{Selesaikan soal berikut ini}\\ & \begin{array}{l}\\ & \text{a}.& {\displaystyle \int {\displaystyle \frac{4x^6+3x^5-8}{x^5}dx}}& \text{f}.& {\displaystyle \int {\displaystyle \frac{{(\sqrt{x}+4)}^3}{\sqrt{x}}dx}}\\ & \text{b}.& {\displaystyle \int {\displaystyle \frac{x^2}{\sqrt{4-x^3}}dx}}& \text{g}.& {\displaystyle \int {\displaystyle \frac{x+3}{\sqrt[5]{{(x^2+6x-1)}^2}}dx}}\\ & \text{c}.& {\displaystyle \int x^3{(x^4+10)}^{{.}^{-\frac{2}{3}}}dx}& \text{h}.& {\displaystyle \int x^5\sqrt{x^6+1}dx}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Integral Fungsi Aljabar

A. Pengertian

Pengintegralan dari suatu fungsi $f(x)$ berbentuk  $\int f(x)dx$ dapat disebut sebagai integral tak tentu dari fungsi  $f(x)$ dan jika  $F(x)$ adalah anti turunan dari  $f(x)$, maka  $F(x)dx=F(x)+C$.

$\begin{array}{rl}& \text{Dengan}\\ & \begin{array}{rl}F(x)& =\text{fungsi integral dari}f(x)\\ f(x)& =\text{fungsi yang diintegralkan}\\ C& =\mathbf{\text{Konstanta}}\end{array}\end{array}$.

B. Rumus Dasar Integral tak Tentu Fungsi Ajabar

Berikut rumus dasar yang perlu diingat

$\begin{array}{rl}\bullet & \int dx=x+C\\ \bullet & \int k.\left({\displaystyle f(x)}\right)dx=k\int f(x)dx\\ \bullet & \int (f(x)\pm g(x))dx=\int f(x)dx+\int g(x)dx\\ \bullet & \int a{x}^{n}dx={\displaystyle \frac{a}{n+1}{x}^{n+1}+C}\end{array}$.

Sebagai rumus-rumus integral yang lain adalah sebagai berikut

$\begin{array}{l}\\ \bullet & {\displaystyle \int a{x}^{n}dx=\frac{a}{n+1}.x^{n+1}+C,\text{dengan}n\ne -1}\\ \bullet & {\displaystyle \int adx=ax+C}\\ \bullet & {\displaystyle \int \frac{1}{x}dx=\int x^{-1}dx=\ln x+C}\\ \bullet & {\displaystyle \int \left|x\right|dx=\frac{1}{2}x\left|x\right|+C}\\ \bullet & {\displaystyle \int \ln xdx=x\ln x-x+C}\\ \bullet & {\displaystyle \int e^{x}dx=e^x+C}\\ \bullet & {\displaystyle \int a^{x}dx=\frac{a^x}{\ln a}+C}\\ \bullet & {\displaystyle \int e^{ax}dx=\frac{1}{a}.e^{ax}+C}\\ \bullet & {\displaystyle \int (x^m+x^n+...+x^p)dx}\\ & ={\displaystyle \int x^{m}dx+\int x^{n}dx+...+\int x^{p}d}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int x^{5}dx}\\ & \text{b}.{\displaystyle \int 2022x^{5}dx}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{1}{x^{2022}}dx}}\\ & \text{d}.{\displaystyle \int 2022ydy}\\ & \text{e}.{\displaystyle \int e^{2022x}dx}\\ & \text{f}.{\displaystyle \int {2022}^{x}dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int x^{5}dx=\frac{1}{5+1}.x^{5+1}+C=\frac{1}{6}.x^6+C}\\ & \text{b}.{\displaystyle \int 2022x^{5}dx=\frac{2022}{5+1}.x^{5+1}+C=337x^6+C}\\ & \text{c}.{\displaystyle \int \frac{1}{x^{2022}}dx=\int x^{-2022}dx}\\ & ={\displaystyle \frac{1}{-2022+1}.x^{-2022+1}+C=-\frac{1}{2021}.x^{-2021}+C}\\ & =-{\displaystyle \frac{1}{2021x^{2021}}+C}\\ & \text{d}.{\displaystyle \int 2022ydy=\frac{2022}{1+1}{y}^{1+1}+C=1011y^2+C}\\ & \text{e}.{\displaystyle \int e^{2022x}dx=\frac{1}{2022}.e^{2022x}+C}\\ & \text{f}.{\displaystyle \int {2022}^{x}dx=\frac{{2022}^x}{\ln 2022}+C}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int (3x^2-x+2-\frac{1}{x}+\frac{3}{x^2})dx}\\ & \text{b}.{\displaystyle \int (x^2-2xy+y^2)dx}\\ & \text{c}.{\displaystyle \int |x-1|+|x-2|dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.{\displaystyle }& \int (3x^2-x+2-\frac{1}{x}+\frac{3}{x^2})dx\\ & ={\displaystyle \int 3x^{2}dx-\int xdx+2\int dx-\int \frac{1}{x}dx+3\int \frac{1}{x^2}dx}\\ & ={\displaystyle \frac{3}{2+1}{x}^{2+1}-\frac{1}{1+1}{x}^{1+1}+2x-\ln x+3\left(\frac{1}{-2+1}{x}^{-2+1}\right)+C}\\ & ={\displaystyle \frac{2}{3}{x}^3-\frac{1}{2}{x}^2+2x-\ln x-\frac{3}{x}+C}\end{array}\\ & \begin{array}{rl}\text{b}.{\displaystyle }& \int (x^2-2xy+y^2)dx\\ & ={\displaystyle \int x^{2}dx-2y\int xdx+y^2\int dx}\\ & ={\displaystyle \frac{1}{2+1}{x}^{2+1}-\frac{2y}{1+1}{x}^{1+1}+y^2.x+C}\\ & ={\displaystyle \frac{1}{3}{x}^3-x^{2}y+x{y}^2+C}\end{array}\\ & \begin{array}{rl}\text{c}.{\displaystyle }& \int |x-1|+|x-2|dx\\ & ={\displaystyle \frac{(x-1)}{2}|x-1|+\frac{(x-2)}{2}|x-2|+C}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 2x^{{.}^{\frac{2}{3}}}dx}\\ & \text{b}.{\displaystyle \int \frac{1}{3}\sqrt[4]{x^3}dx}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{x^4-4x^3}{x^2}dx}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int 2x^{{.}^{\frac{2}{3}}}dx={\displaystyle \frac{2}{{\displaystyle \frac{2}{3}+1}}{x}^{{.}^{\frac{2}{3}+1}}+C={\displaystyle \frac{6}{5}{x}^{{.}^{\frac{5}{3}}}+C}}}\\ & \text{b}.{\displaystyle \int \frac{1}{3}\sqrt[4]{x^3}dx={\displaystyle \frac{1}{3}{x}^{{.}^{\frac{3}{4}}}dx=\left({\displaystyle \frac{1}{3}}\right).{\displaystyle \frac{1}{{\displaystyle \frac{3}{4}+1}}{x}^{{.}^{\frac{3}{4}+1}}+C}}}\\ & =\left({\displaystyle \frac{1}{3}}\right){\displaystyle \frac{4}{7}{x}^{{.}^{\frac{7}{4}}}+C={\displaystyle \frac{4}{21}{x}^{{.}^{\frac{7}{4}}}+C}}\\ & \text{c}.{\displaystyle \int {\displaystyle \frac{x^4-4x^3}{x^2}dx=\int x^2-4xdx}}\\ & ={\displaystyle \frac{1}{2+1}{x}^{2+1}-{\displaystyle \frac{4}{1+1}{x}^{1+1}+C}}\\ & ={\displaystyle \frac{1}{3}{x}^3-2x^2+C}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{\sqrt{x}(6x^2-2x)}{x}dx}}\\ & \text{b}.{\displaystyle \int (12x^2-4x)(2x+1)dx}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{\sqrt{x}(6x^2-2x)}{x}dx={\displaystyle \int {\displaystyle \frac{x^{{.}^{\frac{1}{2}}}(6x^2-2x)}{x}dx}}}}\\ & ={\displaystyle \int {\displaystyle \frac{6x^{{.}^{\frac{5}{2}}}-2x^{{.}^{\frac{3}{2}}}}{x}dx={\displaystyle \int 6x^{{.}^{\frac{3}{2}}}-2x^{{.}^{\frac{1}{2}}}dx}}}\\ & ={\displaystyle \frac{6}{{\displaystyle \frac{3}{2}+1}}{x}^{{.}^{\frac{3}{2}+1}}-\frac{2}{{\displaystyle \frac{1}{2}+1}}{x}^{{.}^{\frac{1}{2}+1}}+C}\\ & ={\displaystyle \frac{12}{5}{x}^{{.}^{\frac{5}{2}}}-{\displaystyle \frac{4}{3}{x}^{{.}^{\frac{3}{2}}}+C={\displaystyle \frac{12}{5}{x}^2\sqrt{x}-{\displaystyle \frac{4}{3}x\sqrt{x}+C}}}}\\ & \text{b}.{\displaystyle \int (12x^2-4x)(2x+1)dx}\\ & ={\displaystyle \int (24x^3+4x^2-4x)dx}\\ & ={\displaystyle \frac{24}{3+1}{x}^{3+1}+\frac{4}{2+1}{x}^{2+1}-\frac{4}{1+1}{x}^{1+1}dx}\\ & =6x^4+{\displaystyle \frac{4}{3}{x}^3-2x^2+C}\end{array}$.

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 1.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 2022dx}\\ & \text{b}.{\displaystyle \int \frac{dx}{2022}}\\ & \text{c}.{\displaystyle \int 2022xdx}\\ & \text{d}.{\displaystyle \int -2022x^{2}dx}\\ & \text{e}.{\displaystyle \int (x+2022)dx}\\ & \text{f}.{\displaystyle \int (-2022x^3+2023x^2-2024)dx}\\ & \text{g}.{\displaystyle \int x\sqrt{x}dx}\\ & \text{h}.{\displaystyle \int \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\sqrt[6]{x}}}}}dx}\\ & \text{i}.{\displaystyle \int \frac{2022}{\sqrt[3]{x^2}}dx}\\ & \text{j}.{\displaystyle \int \frac{2022x}{\sqrt[3]{x^5}}dx}\\ & \text{k}.{\displaystyle \int (2022-2020t+t^2)dt}\\ & \text{l}.{\displaystyle \int (\frac{3}{t^3}+\frac{2}{t^2}+2022)dt}\\ & \text{m}.{\displaystyle \int (\sqrt{t}+\frac{1}{2\sqrt{t}})dt}\\ & \text{n}.{\displaystyle \int (a{y}^4+b{y}^2)dy}\\ & \text{o}.{\displaystyle \int (4a{x}^3+3b{x}^2+2cx+1)dx}\\ & \text{p}.{\displaystyle \int \frac{x^2+2022}{x^2}dx}\\ & \text{q}.{\displaystyle \int (e^x+e^{-x})dx}\\ & \text{r}.{\displaystyle \int e^{2023x}dx}\\ & \text{s}.{\displaystyle \int \frac{dx}{e^{2023x}}dx}\\ & \text{t}.{\displaystyle \int \left(\sqrt{{10}^x}\right)dx}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int (2022x-2202)dx}\\ & \text{b}.{\displaystyle \int (x^2-2x-8)dx}\\ & \text{c}.{\displaystyle \int \sqrt{2x}dx}\\ & \text{d}.{\displaystyle \int x^2(x+2)(x-1)dx}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int 3x(2x-{\displaystyle \frac{1}{x}})dx}\\ & \text{b}.{\displaystyle \int {\displaystyle \frac{(x^2-4)}{x^2}dx}}\\ & \text{c}.{\displaystyle \int {(2x-{\displaystyle \frac{1}{x^2}})}^{2}dx}\\ & \text{d}.{\displaystyle \int x^2\left({\displaystyle \frac{(x+4)(x-3)}{x}}\right)dx}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Selesaikan integral berikut}\\ & \text{a}.{\displaystyle \int {\displaystyle \frac{2x^3-3x}{\sqrt{x}}dx}}\\ & \text{b}.{\displaystyle \int {\displaystyle \frac{2}{\sqrt{x}}{(1-\sqrt{x^2})}^{2}dx}}\\ & \text{c}.{\displaystyle \int x^2{(\sqrt{x}+{\displaystyle \frac{1}{\sqrt{x}}})}^{2}dx}\\ & \text{d}.{\displaystyle \int {\displaystyle \frac{x^2{(2+\sqrt[3]{x^4})}^2}{\sqrt{x}}dx}}\end{array}$.

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono dan Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Probram IPA Standar ISI 2006. Jakarta: ESIS
2. Sharma, S.N., dkk. 2017. Jelajah Matematika SMA Kelas XI Program Wajib. Jakarta: YUDHISTIRA.
3. Tung, K.Y. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

Contoh Soal 4 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 16.& \text{Jika grafik fungsi}f(x)=5+15x+9x^2+x^3\\ & \text{naik untuk}x\text{yang memenuhi}....\\ & \begin{array}{l}\\ \text{a}.x<1\text{atau}x>5& \text{d}.x<-5\text{atau}x>-1\\ \text{b}.-{\displaystyle 1<x<5}& \text{e}.-5<x<1\\ \text{c}.{\displaystyle -5<x<-1}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & f(x)=x^3+9x^2+15x+5.\\ & \text{Dikatakan fungsi}f\text{naik, maka}\\ & f{}^{\prime }(x)>0\\ & \iff 3x^2+18x+15>0,\text{tiap ruas dibagi 3}\\ & \iff x^2+6x+5>0\\ & \iff (x+1)(x+5)>0\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Sebuah bola dilempar ke atas secara vertikal.}\\ & \text{Jika lintasan bola pada saat}t\text{detik adalah}\\ & h(t)=-{\displaystyle \frac{1}{4}{t}^4+\frac{2}{3}{t}^3+4t^2+5\text{m},}\\ & \text{maka tinggi maksimum yang}\\ & \text{dicapai oleh bola tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{67}{3}}& & \text{d}.{\displaystyle \frac{133}{3}}\\ \\ \text{b}.{\displaystyle \frac{123}{3}}& \text{c}.{\displaystyle \frac{128}{3}}& \text{e}.{\displaystyle \frac{143}{3}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Perh}& \text{atikan bahwa lintasan bola saat dilempar }\\ \text{verti}& \text{kal dituliskan dengan fungsi}\\ h(t)& =-{\displaystyle \frac{1}{4}{t}^4+\frac{2}{3}{t}^3+4t^2+5.}\\ \text{mak}& \text{a tinggi maksimum akan dicapai bola }\\ \text{saat}& h{}^{\prime }(t)=0,\\ \text{Sela}& \text{njutnya},\\ h{}^{\prime }(t)& =0\\ -t^3+& 2t^2+8t=0\\ -t(t^2& -2t-8)=0\\ t(t-& 4)(t+2)=0\{\begin{array}{ll}t& =0\\ t& =4\\ t& =-2\end{array}\\ \text{kita}& \text{ambil yang bernilai positif untuk }\\ & t\text{yaitu}t=4.\\ t=4& \to h(4)=-{\displaystyle \frac{1}{4}(4{)}^4+\frac{2}{3}(4{)}^3+4(4{)}^2+5}\\ h(4)& =-64+{\displaystyle \frac{128}{3}+64+5}\\ & ={\displaystyle \frac{143}{3}m}\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

Contoh Soal 3 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 11.& \text{Jika}f(x)={\displaystyle (x^2+2)\sqrt{x^2+x+3}}\\ & \text{maka}f{}^{\prime }(2)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 18}& & \text{d}.13\\ \\ \text{b}.{\displaystyle 17}& \text{c}.{\displaystyle 15}& \text{e}.12\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle (x^2+2)\sqrt{x^2+x+3}}\\ f{}^{\prime }(x)& =(2x)\sqrt{x^2+x+3}\\ & +(x^2+2).{\displaystyle \frac{1}{2}{(x^2+x+3)}^{{}^{-\frac{1}{2}}}.(2x+1)}\\ & =2x\sqrt{x^2+x+3}+{\displaystyle \frac{(x^2+2)(2x+1)}{2\sqrt{x^2+x+3}}}\\ f{}^{\prime }(2)& =2(2)\sqrt{(2{)}^2+(2)+3}+{\displaystyle \frac{((2{)}^2+2)(2(2)+1)}{2\sqrt{(2{)}^2+(2)+3}}}\\ & =4\sqrt{9}+{\displaystyle \frac{6.5}{2\sqrt{9}}}\\ & =4.3+{\displaystyle \frac{6.5}{2.3}}\\ & =12+5\\ & =17\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Jika rusuk suatu kubus bertambah panjang }\\ & \text{dengan laju 7}cm/detik,\text{maka laju}\\ & \text{bertambahnya volume pada saat rusuk }\\ & \text{panjangnya 15}cm\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 675c{m}^3/detik}& \text{d}.4725c{m}^3/detik\\ \text{b}.{\displaystyle 1575c{m}^3/detik}& \text{e}.23625c{m}^3/detik\\ \text{c}.{\displaystyle 3375c{m}^3/detik}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Laju }& \text{pertambahan volumenya}:\\ & \{\begin{array}{ll}\bullet \frac{\mathrm{d}s}{\mathrm{d}t}& =7cm/detik\\ \bullet V& =s^3\to \text{d}V=3s^2\text{d}s\text{atau}\\ \frac{\mathrm{d}V}{\mathrm{d}s}& =3s^{2}c{m}^3/cm\to s=15cm\end{array}\\ \frac{\mathrm{d}V}{\mathrm{d}t}& =\frac{\mathrm{d}V}{\mathrm{d}t}\\ & =\frac{\mathrm{d}V}{\mathrm{d}s}\times \frac{\mathrm{d}s}{\mathrm{d}t}\\ & =3s^{2}c{m}^3/cm\times 7cm/detik\\ & =3(15{)}^2\times 7c{m}^3/detik\\ & =4725c{m}^3/detik\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{persamaan garis singgung di}x=1\\ & \text{pada kurva}y=x^3-3x^2+1\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.y=-3x+2& \text{d}.y=3x-2\\ \text{b}.y=-3x+4& \text{e}.y=-3x+3\\ \text{c}.y=3x-4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|c|}\hline \text{Titik singgung di}x=1\\ \begin{array}{rl}y& =x^3-3x^2+1\\ y& =(1{)}^3-3(1{)}^2+1\\ & =1-3+1\\ & =-1\\ & \text{titik}(a,b)=(1,-1)\end{array}\\ \text{Gradien garis singgung di}x=1\\ \begin{array}{rl}y{}^{\prime }=m{}_{{}_{x=1}}& =3x^2-6x\\ & =3(1{)}^2-6(1)\\ & =3-6\\ & =-3\end{array}\\ \text{Persamaan garis singgung}\\ \begin{array}{rl}y& =m(x-a)+b\\ & =-3(x-1)+(-1)\\ & =-3x+3-1\\ & =-3x+2\end{array}\\ \hline\end{array}\\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Suatu kurva}y=x^3+2a{x}^2+b\\ & \text{Sebuah garis}y=-9x-2\text{menyinggung}\\ & \text{kurva di titik dengan}x=1,\\ & \text{maka nilai}a\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.-3& & \text{d}.3\\ \text{b}.-{\displaystyle \frac{1}{3}}& \text{c}.{\displaystyle \frac{1}{3}}& \text{e}.8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Gradien garis singgung}\\ & y{}_{{}_{{}_{dix=1}}}\{\begin{array}{ll}y& =x^3+2a{x}^2+b\\ & \to m=y{}^{\prime }=3x^2+4ax\\ \\ y& =-9x-2\to m=y^{\prime }=-9\end{array}\\ & \text{Sehingga},\\ & m=m=y{}^{\prime }\\ & -9=3x^2+4ax\\ & -9=3(1{)}^2+4a(1)\\ & -9=3+4a\\ & -4a=3+9\\ & a={\displaystyle \frac{12}{-4}}\\ & =-3\\ & \end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Jika grafik fungsi}f(x)=x^3+a{x}^2+bx+c\\ & \text{hanya turun untuk interval}-1<x<5,\\ & \text{maka nilai}a+b\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.-21& & \text{d}.21\\ \text{b}.-{\displaystyle 9}& \text{c}.{\displaystyle 9}& \text{e}.24\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}:\\ & f(x)=x^3+a{x}^2+bx+c\\ & \text{grafik fungsi turun}\text{berarti}:f{}^{\prime }(x)<0\\ & f{}^{\prime }(x)<0\\ & \iff 3x^2+2ax+b<0\\ & \iff (x+1)(x-5)<0\\ & \text{ini maksud pada interval}\\ & -1<x<5\text{pada soal}\\ & x^2-4x-5<0\\ & \text{dikalikan dengan 3 supaya terjadi persamaan}\\ & 3x^2-3.4x-3.5=3x^2+2ax+b<0\\ & 3x^2+2(-6)x+(-15)=3x^2+2ax+b<0\\ & \{\begin{array}{ll}a& =-6\\ b& =-15\end{array}\\ & \text{Sehingga},\\ & a+b=-6+(-15)=-21\end{array}\end{array}$.

Contoh Soal 2 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 6.& \text{Jika}W=\sin 2t,\text{maka}{\displaystyle \frac{dW}{dt}=....}\\ & \begin{array}{l}\\ \text{a}.\cos 2t& \text{d}.2t\cos 2t+\sin 2t\\ \text{b}.2\cos 2t& \text{e}.\sin 2t-t\cos 2t\\ \text{c}.\sin 2t+t\cos 2t\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & W=\sin 2t\\ & W=\sin u\text{dengan}u=2t\\ & {\displaystyle \frac{dW}{dt}={\displaystyle \frac{dW}{du}.\frac{du}{dt}}}\\ & =\cos u.2\\ & =2\cos u\\ & =2\cos 2t\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}f(x)={\displaystyle \frac{2x+4}{1+\sqrt{x}},\text{maka}{\displaystyle f{}^{\prime }(4)=....}}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{4}}& & \text{d}.1\\ \text{b}.{\displaystyle \frac{3}{7}}& \text{c}.{\displaystyle \frac{3}{5}}& \text{e}.4\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{2x+4}{1+\sqrt{x}}}\\ & ={\displaystyle \frac{U}{V}}\\ f{}^{\prime }(x)& ={\displaystyle \frac{U{}^{\prime }.V-U.V{}^{\prime }}{V^2}}\\ & ={\displaystyle \frac{(2)(1+\sqrt{x})-(2x+4).(1.{(1+\sqrt{x})}^0.{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}})}{{(1+\sqrt{x})}^2}}\\ & \text{ingat}\sqrt{x}=x^{{}^{\frac{1}{2}}}\\ f{}^{\prime }(4)& ={\displaystyle \frac{2(1+\sqrt{4})-(2.4+4).\frac{1}{2}.4^{{}^{-\frac{1}{2}}}}{{(1+\sqrt{4})}^2}}\\ & ={\displaystyle \frac{2(1+2)-(12).\frac{1}{2}.\frac{1}{2}}{(1+2{)}^2}}\\ & \text{ingat juga}{4}^{{}^{-\frac{1}{2}}}={\left(2^2\right)}^{{}^{-\frac{1}{2}}}=2^{{}^{-1}}={\displaystyle \frac{1}{2^1}=\frac{1}{2}}\\ & ={\displaystyle \frac{6-3}{9}}\\ & ={\displaystyle \frac{1}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Jika}f(x)={\displaystyle \frac{1+\sin x}{\cos x},\text{maka}f{}^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{2}}& & \text{d}.2\\ \text{b}.{\displaystyle -\frac{1}{2}}& \text{c}.{\displaystyle \frac{3}{4}}& \text{e}.-2\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{1+\sin x}{\cos x}}\\ f{}^{\prime }(x)& ={\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}}\\ & \text{ingat bahwa}{\sin }^{2}x+{\cos }^{2}x=1\\ & ={\displaystyle \frac{1+\sin x}{{\cos }^{2}x}}\\ f{}^{\prime }\left({\displaystyle \frac{1}{6}\pi }\right)& ={\displaystyle \frac{1+\sin \left({\displaystyle \frac{1}{6}\pi }\right)}{{\cos }^2\left({\displaystyle \frac{1}{6}\pi }\right)}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{2}}}{{\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}^2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{3}{4}}}}\\ & ={\displaystyle \frac{3}{2}\times \frac{4}{3}}\\ & =2\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}f(x)={\displaystyle 3x^2-2ax+7}\\ & \text{dan}f{}^{\prime }(1)=0,\text{maka}f{}^{\prime }(2)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 1}& & \text{d}.6\\ \text{b}.{\displaystyle 2}& \text{c}.{\displaystyle 4}& \text{e}.8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle 3x^2-2ax+7}\\ f{}^{\prime }(x)& =6x-2a\\ f{}^{\prime }(1)& =0\\ 6(1)-2a& =0\\ 6& =2a\\ 3& =a\\ \text{sehingga}& \\ f{}^{\prime }(x)& =6x-6\\ \text{maka},& \\ f{}^{\prime }(2)& =6.2-6\\ & =12-6\\ & =6\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika}f(x)={\displaystyle (6x-3{)}^3(2x-1)}\\ & \text{maka}f{}^{\prime }(1)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle 18}& & \text{d}.162\\ \\ \text{b}.{\displaystyle 24}& \text{c}.{\displaystyle 54}& \text{e}.216\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& ={\displaystyle (6x-3{)}^3(2x-1)}\\ & =(3.(2x-1){)}^3(2x-1{)}^1\\ & =3^3.(2x-1{)}^{3+1}\\ & =27(2x-1{)}^4\\ f{}^{\prime }(x)& =4.27(2x-1{)}^{4-1}.2\\ & =216.(2x-1{)}^3\\ f{}^{\prime }(1)& =216.(2.1-1{)}^3\\ & =216.1\\ & =216\end{array}\end{array}$.

Contoh Soal 1 Turunan Fungsi Aljabar

 $\begin{array}{l}\\ 1.& \text{Diketahui}f{}^{\prime }(2)=\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{x^3-8}{x-2},}\\ & \text{maka fungsi}f(x)=....\\ & \begin{array}{l}\\ \text{a}.12& & \text{d}.x^3\\ \text{b}.2& \text{c}.x& \text{e}.x-8\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Turunan fungsi f di}x=c\text{adalah}\\ & f{}^{\prime }(c)=\underset{x\to c}{\text{Lim}}{\displaystyle \frac{f(x)-f(c)}{x-c},\text{maka}}\\ & \text{turunan fungsi f di}x=2\text{adalah}\\ & f{}^{\prime }(2)=\underset{x\to 2}{\text{Lim}}{\displaystyle \frac{x^3-8}{x-2}}\\ & \text{sehingga akan didapa}\text{tkan fungsi}f\text{nya }\\ & \text{yaitu}f(x)=x^3\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}a\ne 0,\text{maka nilai dari}\\ & \underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}=....}\\ & \begin{array}{l}\\ \text{a}.3a{\displaystyle \sqrt[3]{a}}& & \text{d}.{\displaystyle \frac{1}{2a}\sqrt[3]{a}}\\ \text{b}.2a{\displaystyle \sqrt[3]{a}}& \text{c}.0& \text{e}.{\displaystyle \frac{1}{3a}\sqrt[3]{a}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Alternatif 1}\\ & \begin{array}{rl}f{}^{\prime }(a)& =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{(\sqrt[3]{x}-\sqrt[3]{a})(\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2})}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{1}{(\sqrt[3]{x^2}+\sqrt[3]{xa}+\sqrt[3]{a^2})}}\\ & ={\displaystyle \frac{1}{(\sqrt[3]{a^2}+\sqrt[3]{a.a}+\sqrt[3]{a^2})}}\\ & ={\displaystyle \frac{1}{3\sqrt[3]{a^2}}}\\ & ={\displaystyle \frac{1}{3\sqrt[3]{a^2}}\times {\displaystyle \frac{\sqrt[3]{a}}{\sqrt[3]{a}}}}\\ & ={\displaystyle \frac{1}{3a}\sqrt[3]{a}}\end{array}\\ & \text{Alternatif 2 (dengan aturan L'Hopital)}\\ & \begin{array}{rl}\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}}& =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{f(x)}{g(x)}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{f{}^{\prime }(x)}{g{}^{\prime }(x)}}\\ \underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \sqrt[3]{x}-{\displaystyle \sqrt[3]{a}}}}{x-a}}& =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{x^{{}^{\frac{1}{3}}}-a^{{}^{\frac{1}{3}}}}{x-a}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{1}{3}{x}^{{}^{\frac{1}{3}-1}}-0}}{1-0}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{1}{3}{x}^{{}^{-\frac{2}{3}}}}}{1}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{1}{3\sqrt[3]{x^2}}\times \frac{\sqrt[3]{x}}{\sqrt[3]{x}}}\\ & =\underset{x\to a}{\text{Lim}}{\displaystyle \frac{1}{3x}\sqrt[3]{x}}\\ & ={\displaystyle \frac{1}{3a}\sqrt[3]{a}}\end{array}\end{array}$ .

$\begin{array}{l}\\ 3.& \text{Jika}f(x)={\displaystyle \frac{1}{\sqrt{x}}\text{maka nilai dari}}\\ & -2f{}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{x\sqrt{x}}}& & \text{d}.{\displaystyle -\frac{1}{2x\sqrt{x}}}\\ \text{b}.{\displaystyle x\sqrt{x}}& \text{c}.{\displaystyle -\frac{1}{2\sqrt{x}}}& \text{e}.{\displaystyle -2x\sqrt{x}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dike}& \text{tahui}\\ f(x)& ={\displaystyle \frac{1}{\sqrt{x}}={\displaystyle \frac{1}{x^{{}^{\frac{1}{2}}}}=x^{{}^{-\frac{1}{2}}}}}\end{array}\\ & \begin{array}{|cc|}\hline y=a{x}^n\to y{}^{\prime }=na{x}^{n-1}& \begin{array}{rl}& \\ y& ={\displaystyle \frac{U}{V}\to y{}^{\prime }={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}}\\ & \end{array}\\ \begin{array}{rl}f{}^{\prime }(x)& =-{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}-1}}}\\ & =-{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{3}{2}}}}\\ & =-{\displaystyle \frac{1}{2x^{{}^{\frac{3}{2}}}}}\\ & =-{\displaystyle \frac{1}{2x^1.x^{{}^{\frac{1}{2}}}}}\\ & =-{\displaystyle \frac{1}{2x\sqrt{x}}}\end{array}& \begin{array}{rl}f{}^{\prime }(x)& ={\displaystyle \frac{0.\sqrt{x}-1.\frac{1}{2}{x}^{{}^{\frac{1}{2}-1}}}{{\left(\sqrt{x}\right)}^2}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}}{x}}\\ & =-{\displaystyle \frac{1}{2x{x}^{{}^{\frac{1}{2}}}}}\\ & =-{\displaystyle \frac{1}{2x\sqrt{x}}}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Turunan pertama dari}y={\displaystyle \sqrt[n]{x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1}{n}{x}^{{}^{\frac{1}{n}}}}& & \text{d}.(n-1)\sqrt[n-1]{x}\\ \text{b}.{\displaystyle \frac{1}{n}{x}^{{}^{\frac{1-n}{n}}}}& \text{c}.{\displaystyle \frac{1}{n-1}{x}^{{}^{n-1}}}& \text{e}.\sqrt[n-1]{x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}y& =\sqrt[n]{x}={\displaystyle x^{{}^{\frac{1}{n}}},\text{maka}}\\ y{}^{\prime }& ={\displaystyle \frac{1}{n}{x}^{{}^{\frac{1}{n}-1}}={\displaystyle \frac{1}{n}{x}^{{}^{\frac{1-n}{n}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Turunan ke}-n\text{dari}y={\displaystyle \frac{1}{x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle n!.x^{-(n+1)}}& \\ \text{b}.{\displaystyle (n+1)!.x^{-(n+1)}}& \\ \text{c}.{\displaystyle (-1{)}^{n}n!.x^{-(n+1)}}\\ \text{d}.(-1{)}^{n+1}n!.x^{-(n+1)}\\ \text{e}.(-1{)}^{n+1}(n+1)!.x^{-(n+1)}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|ccl|}\hline \text{Fungsi}& y={\displaystyle \frac{1}{x}}& y=x^{-1}\\ y^{\prime }={\displaystyle \frac{dy}{dx}}& -x^{-2}& (-1{)}^1.1!.x^{-(1+1)}\\ y^{″}={\displaystyle \frac{d^{2}y}{d{x}^2}}& 2x^{-3}& (-1{)}^2.2!.x^{-(2+1)}\\ y^{‴}={\displaystyle \frac{d^{3}y}{d{x}^3}}& -6x^{-4}& (-1{)}^3.3!.x^{-(3+1)}\\ y^{IV}={\displaystyle \frac{d^{4}y}{d{x}^4}}& 24x^{-5}& (-1{)}^4.4!.x^{-(4+1)}\\ y^V={\displaystyle \frac{d^{5}y}{d{x}^5}}& \cdots & \cdots \\ ⋮& ⋮& ⋮\\ y^n={\displaystyle \frac{d^{n}y}{d{x}^n}}& \cdots & (-1{)}^n.n!.x^{-(n+1)}\\ \hline\end{array}\end{array}$.

Menyelesaikan Masalah Berkaitan Keekstriman Fungsi dan Penggunaan Turunan Kedua Fungsi Aljabar.

Lanjutan Materi Penggunaan Turunan Fungsi (Silahkan Lihat materi sebelumnya di sini)

$\begin{array}{l}\\ 5.& \text{Diketahui jumlahdua bilangan adalah 30}\\ & \text{Jika perkalian salah satu bilangan dengan}\\ & \text{kuadrat bilangan lainnya mencapai nilai}\\ & \text{maksimum, tentukanlah}:\\ & \text{a}.\text{kedua bilangan tersebut}\\ & \text{b}.\text{nilai maksimumnya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Misalkan salah satu bilangan adalah}x,\\ & \text{maka bilangan yang lainnya adalah}:30-x\\ & \text{Dan misalkan pula perkalian ini dirumuskan}\\ & \text{dengan}p(x),\text{maka}\\ & \begin{array}{rl}\text{a}.p(x)& =(30-x)x^2=30x^2-x^3\\ \text{syara}& \text{t ekstrim maksimum}{p}^{\prime }(x)=0\\ 60x-& 3x^2=0\iff 3x(20-x)=0\\ x_1=& 0\text{atau}{x}_2=20\\ \text{Jadi},& \text{nilai agar maksimum}x=20\text{dan}\\ \text{nilai}& x\text{yang lain adalah}:30-20=10\\ \text{b}.\text{Dan}& \text{nilai maksimumnya adalah}:\\ p(20)& =(30-20){.20}^2=10.400=4000(\mathbf{\text{mak}})\end{array}\end{array}$ .

$\begin{array}{l}\\ 6.& \text{Sebuah peluru ditembakkan ke atas. Dan}\\ & \text{tinggi yang dapat dicapai peluru adalah}\\ & h\text{meter dalam waktu}t\text{detik yang dapat}\\ & \text{rumuskan dengan}h(t)=120t-5t^2\\ & \text{Tentukanlah}\\ & \text{a}.t\text{agar}h\text{maksimum}\\ & \text{b}.\text{tinggi}h\text{maksimum}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.\text{syara}& \text{at ekstrim maksimum}{h}^{\prime }(x)=0\\ 120-& 10t=0\iff t=12\\ \text{Jadi},& \text{tinggi maksimum dicapai saat}t=12\\ \text{b}.\text{Dan}& \text{tinggi maksimumnya adalah}:\\ h(12)& =120(12)-5(12{)}^2=720\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& (\mathbf{\text{OSK 2018}})\\ & \text{Diketahui bilangan real}x\text{dan}y\\ & \text{yang memenuhi}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Nilai minimum}{\displaystyle \frac{x}{2y-x}+\frac{2y}{2x-y}\text{adalah}....}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misal}t={\displaystyle \frac{x}{y}}\\ & \text{Misalkan juga}f(t)={\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}\\ & \text{maka}f(t)={\displaystyle \frac{2t^2-3t+4}{-2t^2+5t-2}}\\ & \bullet \text{Agar minimum, maka}{f}^{\prime }(t)=0\\ & \text{Sehingga}\\ & f^{\prime }(t)=4t^2+8t-14=0\\ & \iff t_{1,2}=-1\pm {\displaystyle \frac{3}{2}\sqrt{2}}\\ & \text{Pilih yang positif, yaitu}t=-1+{\displaystyle \frac{3}{2}\sqrt{2}}\\ & \bullet \text{Dengan proses substistusi harga}t\\ & \text{di atas, maka akan didapatkan }\\ & \text{nilai}f(t)=1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{Menurut bentuk}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{jelas bahwa baik}2y-x\text{dan}2x-y\\ & \text{keduanya}\mathbf{\text{positif}}\\ & \text{Lihat tabel berikut}\\ & \begin{array}{|ccc|}\hline \text{Bentuk}& \text{Pengecekan 1}& \text{Pengecekan 2}\\ \begin{array}{rl}& {\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Jelas bahwa}\\ & x,y\ne 0\\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat}(\times y)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(y)<\frac{x}{y}(y)<2(y)}\\ & \iff {\displaystyle \frac{1}{2}y<x<2y}\\ & \text{Jelas bahwa}\\ & 2y-x>0\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat dibali posisinya}\\ & {\displaystyle \frac{1}{2}<\frac{y}{x}<2}\\ & \text{Saat}(\times x)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(x)<\frac{y}{x}(x)<2(x)}\\ & \iff {\displaystyle \frac{1}{2}x<y<2x}\\ & \text{Jelas bahwa}\\ & 2x-y>0\end{array}\\ \hline\end{array}\\ & \text{Saat masing-masing}\\ & \bullet {\displaystyle \frac{x}{2y-x}={\displaystyle \frac{1}{3}+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\text{dan}}}\\ & \bullet {\displaystyle \frac{2y}{2x-y}={\displaystyle \frac{2}{3}+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}}\\ & \text{Dengan ketaksamaan AM-GM diperoleh}\\ & \begin{array}{rl}{\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}& =1+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)\\ & \ge 1+2\sqrt{\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}\\ & =1+2\sqrt{{\displaystyle \frac{8}{9}}}\\ & =1+2\left({\displaystyle \frac{2}{3}}\right)\sqrt{2}\\ & =1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Jika}x,y,z\text{adalah sisi sebuah segitiga}\\ & \text{Tunjukkan bahwa nilai minimum}\\ & {\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\text{adalah 3}}\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Alternatif 1}\\ & \begin{array}{rl}& \text{Misalkan}{\displaystyle \frac{y}{x}=p,\frac{z}{y}=q,\text{dan}{\displaystyle \frac{z}{x}=pq,}}\\ & \text{maka soal dapat kita ubah menjadi}\\ & {\displaystyle \frac{1}{p}+\frac{1}{q}+pq}\\ & \text{Kita tahu bahwa}{(\sqrt{{\displaystyle \frac{1}{p}}}-\sqrt{{\displaystyle \frac{1}{q}}})}^2\ge 0\\ & \iff {\displaystyle \frac{1}{p}+\frac{1}{q}\ge 2\sqrt{{\displaystyle \frac{1}{pq}}}}\\ & \text{Kita sesuaikan soal, yaitu}\\ & {\displaystyle \frac{1}{p}+\frac{1}{q}+pq\ge pq+2\sqrt{{\displaystyle \frac{1}{pq}}}}\\ & \text{Misalkan}\sqrt{{\displaystyle \frac{1}{pq}}}=a,{\displaystyle \frac{1}{pq}=a^2,pq={\displaystyle \frac{1}{a^2}}}\\ & \text{dengan}a\ge 0,\text{maka}\\ & {\displaystyle \frac{1}{p}+\frac{1}{q}+pq={\displaystyle \frac{1}{a^2}+2a=f(a)}}\\ & \text{Syarat ekstrim minimum}:f^{\prime }(a)=0\\ & -2a^{-3}+2=0\iff 2=2a^{-3}\iff 1=a^{-3}\\ & \iff 1={\displaystyle \frac{1}{a^3}\iff a^3=1\iff a=1}\\ & \text{Sehingga nilai minimumnya saat}a=1\\ & \text{dengan nilai}f(1)={\displaystyle \frac{1}{1^2}+2.1=3}\\ & \text{Jadi, nilai nilai minimum}{\displaystyle \frac{1}{p}+\frac{1}{q}+pq=3}\\ & \text{atau}{\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\text{adalah 3 atau juga}}\\ & \text{dapat dituliskan dengan}{\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\ge 3}\end{array}\\ & \text{Alternatif 2}\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\\ & {\displaystyle \frac{x}{y}+\frac{y}{x}+\frac{x}{z}\ge 3\sqrt[3]{{\displaystyle \frac{x}{y}.\frac{y}{x}.\frac{x}{z}}}=3.\sqrt[3]{1}=3}\end{array}$

DAFTAR PUSTAKA

1. Kartrini, Suprapto, Subandi, dan Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
2. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
3. Widodo, T. 2018. Booklet OSN SMA 2018: Soal dan Solusi OSK, OSP, OSN SMA Bidang Matematika. 

SUMBER WEBSITE

1. Pythagoras pada: https://pyth.eu/uploads/user/ArchiefPDF/Pyth35-56.pdf 

Penggunaan Turunan Fungsi (Lanjutan Materi Turunan Fungsi Aljabar)

 Penggunaan Turunan Fungsi Aljabar ini nantinya terdapat di antaranya pada:

• Persamaan garis singgung
• Fungsi naik dan fungsi turun
• Menggambar grafik fungsi aljabar
• Maksimum dan minimum fungsi
• Teorema L'Hopital (dibaca : Lopital)
• Titik Stasioner/Titik kritis/Titik Ekstrim (titik maksmum, titik minimum, dan titik belok)
• Kecepatan dan percepatan

Perhatikanlah tabel berikut

$\begin{array}{|cll|}\hline \text{No}& \text{Turunan Pertama}& \text{Turunan Pertama}\\ 1.& \text{Gradien garis singgung}& \begin{array}{rl}m& =f{}^{\prime }(x)\\ & =\underset{h\to 0}{\text{Lim}}={\displaystyle \frac{f(x+h)-f(x)}{h}}\end{array}\\ 2.& \text{Fungsi naik dan turun}& y=f(x)\{\begin{array}{ll}{f}^{\prime }(x)>0,& \\ \text{ fungsi naik }& \\ \\ f^{\prime }(x)<0,& \\ \text{ fungsi turun }& \end{array}\\ 3.& \text{Jarak, kec, percepatan}& y=s(t)\{\begin{array}{ll}s(t)& \text{ jarak}\\ s{}^{\prime }(t)& \text{ kecepatan }\\ s{}^{″}(t)& \text{ percepatan}\end{array}\\ 4.\text{a}& \text{Stasioner}& \begin{array}{rl}\text{Maksimum}:& \\ \to f& {}^{″}(k)<0\\ \text{titik mak}& (k,f(k))\end{array}\\ 4.\text{b}& \text{Stasioner}& \begin{array}{rl}\text{Minimum}:& \\ \to f& {}^{″}(k)>0\\ \text{titik min}& (k,f(k))\end{array}\\ 4.\text{c}& \begin{array}{rl}& \text{Syarat stasioner}\\ & f^{\prime }(x)=0,\to x\\ & \text{dengan}x=k\end{array}& \begin{array}{rl}\text{Belok}:& \\ \to f& {}^{″}(k)=0\\ \text{titik belok}& (k,f(k))\end{array}\\ 5.& \begin{array}{rl}& \text{Limit fungsi}\\ & \text{bentuk tak tentu}\end{array}& \begin{array}{rl}& \text{Aturan L'Hopital}\\ & \\ & \underset{x\to h}{\text{Lim}}{\displaystyle \frac{f(x)}{g(x)}=\underset{x\to h}{\text{Lim}}{\displaystyle \frac{f{}^{\prime }(x)}{g{}^{\prime }(x)}}}\\ & \\ & \text{untuk hasil limit}\\ & \text{bentuk}\frac{0}{0}\text{atau}\frac{\mathrm{\infty }}{\mathrm{\infty }}\end{array}\\ \text{No}& \text{Turunan Kedua}& \text{Turunan Kedua}\\ 6.& f^{″}={\displaystyle \frac{d^{2}y}{d{x}^2}}& \begin{array}{rl}\bullet & \text{Belok}\\ \bullet & \text{Percepatan}\\ \bullet & \text{Maksimum}\\ \bullet & \text{Minimum}\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah persamaan garis singgung }\\ & \text{pada kurva}y=x^2+2x-8\text{di titik yang}\\ & \text{berabsis}1\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahui},\text{persamaan sebuah kurva adalah:}\\ & y=x^2+2x-8\\ & \end{array}\\ & \begin{array}{|cc|}\hline \text{Titik singgung}& \text{Gradien},x=1\\ \begin{array}{rl}\text{absis}x& =1,\\ \text{maka}y& =(1{)}^2+2(1)-8\\ & =1+2-8\\ & =-5\\ & \\ \text{di titik}& (a,b)=(1,-5)\end{array}& \begin{array}{rl}{\displaystyle \frac{dy}{dx}=m}& =2x+2\\ & =2(1)+2\\ & =4\\ & \\ & \end{array}\\ \text{Persamaan garis singgung}& \text{Kesimpulan}\\ \begin{array}{rl}y& =m(x-a)+b\\ & =4(x-1)+(-5)\\ & =4x-4-5\\ & =4x-9\end{array}& \begin{array}{rl}& \text{Sehingga},\text{PGS adalah:}\\ & y=4x-9\mathbf{\text{atau}}\\ & y-4x+9=0\\ & \end{array}\\ \hline\end{array}\\ & \begin{array}{rl}& \text{Jadi},\text{persamaan garis singgungnya adalah:}\\ & y=4x-9\mathbf{\text{atau}}y-4x+9=0\\ \\ & \text{Dan berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah interval di mana kurva }\\ & \text{fungsi}f(x)=x^3+3x^2-9x+5\\ & \text{a}.\text{naik}\\ & \text{b}.\text{turun}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui fungsi}\\ f(x)& =x^3+3x^2-9x+5\\ f{}^{\prime }(x)& =3x^2+6x-9\\ & =3(x+3)(x-1)\\ & \end{array}\\ & \begin{array}{|ll|}\hline \text{naik};f{}^{\prime }(x)>0& \text{turun};f{}^{\prime }(x)<0\\ \begin{array}{rl}& \\ & 3(x+3)(x-1)>0\\ & \end{array}& \begin{array}{rl}& \\ & 3(x+3)(x-1)<0\\ & \end{array}\\ \begin{array}{rl}& \\ \text{naik},& x<-3\text{atau}x>1\\ & \end{array}& \begin{array}{rl}& \\ \text{turun},& -3<x<1\\ & \end{array}\\ \hline\end{array}\end{array}$ .

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai stasioner fungsi}\\ & f(x)=x^3+3x^2-9x+5\text{dan }\\ & \text{tentukan pula jenisnya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Diketahu fungsi}\\ & f(x)=x^3+3x^2-9x+5\\ & \text{Syarat stasioner}{f}^{\prime }(x)=0,\text{maka}\\ & f{}^{\prime }(x)=3x^2+6x-9=0\\ & 0=3(x+3)(x-1)\\ & x=-3\text{atau}x=1\\ & \text{Dan untuk nilai dan titik stasionernya}:\\ & f(-3)=(-3{)}^3+3(-3{)}^2-9(-3)+5=32\\ & \to (-3,32)\text{adalah titik balik maksimum}\\ \\ & f(1)=(1{)}^3+3(1{)}^2-9(1)+5=0\\ & \to (1,0)\text{adalah titik balik minimum}\\ \\ & \text{Dan berikut ilustrasi gambarnya}\\ & \text{untuk soal no.2 dan 3}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Masih sama dengan soal seperti pada }\\ & \text{No. sebelumnya yaitu fungsi}\\ & f(x)=x^3+3x^2-9x+5.\\ & \text{Tentukanlah koordinat titik beloknya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}f(x)& =x^3+3x^2-9x+5\\ f{}^{\prime }(x)& =3x^2+6x-9\\ f{}^{″}(x)& =6x+6\\ \text{Proses}& \text{mencari titik beloknya}\\ f{}^{″}(x)& =0\\ 6x+6& =0\\ 6x& =-6\\ x& =-1\\ & \end{array}\\ & \begin{array}{|ccccl|}\hline \text{Interval}& f(x)& f{}^{\prime }(x)& f{}^{″}(x)& \text{Keterangan}\\ x<-1& & & -& \text{grafik cekung ke bawah}\\ x=-1& 16& -12& 0& \text{grafik memiliki titik belok}\\ x>-1& & & +& \text{grafik cekung ke atas}\\ \hline\end{array}\\ & \text{Koordinat titik beloknya}:(-1,16)\\ \\ & \text{Berikut ilustrasi gambarnya}\end{array}$.

Aturan Rantai pada Turunan Pertama dan Turunan Kedua Fungsi Aljabar (Lanjutan Materi Turunan Fungsi Aljabar)

 Aturan Rantai

$\begin{array}{rl}& \\ \text{Untuk}:& \\ & \{\begin{array}{ll}a& \in \mathbb{R}\\ n& \in \mathbb{Q}\\ c& \text{konstanta}\\ U& =g(x)\\ V& =h(x)\end{array}\\ & \end{array}$.

$\begin{array}{|l|}\hline \mathbf{\text{Sifat-Sifat}}\\ \begin{array}{rl}y& =c\to y^{\prime }=0\\ y& =c.U\to y^{\prime }=c.U^{\prime }\\ y& =U\pm V\to y^{\prime }=U^{\prime }\pm V^{\prime }\\ y& =U.V\to y^{\prime }=U^{\prime }.V+U.V^{\prime }\\ y& ={\displaystyle \frac{U}{V}\to y^{\prime }={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}}\end{array}\\ \mathbf{\text{Fungsi Aljabar}}\\ \begin{array}{rl}y& =a.x^n\to y{}^{\prime }=n.a.x^{(n-1)}\\ y& =a.U^n\to y{}^{\prime }=n.a.U^{(n-1)}.U^{\prime }\end{array}\\ \mathbf{\text{Fungsi Trigonometri}}\\ \begin{array}{rlrl}y& =a\sin U\to & y^{\prime }& =(a\cos U).U^{\prime }\\ y& =a\cos U\to & y^{\prime }& =(-a\sin U).U^{\prime }\\ y& =a\tan U\to & y^{\prime }& =(a{\sec }^{2}U).U^{\prime }\end{array}\\ \mathbf{\text{Aturan rantai }}\\ \begin{array}{rl}& \text{pada turunan untuk}y=f(u),\text{jika}\\ & \text{untuk}u\text{merupakan fungsi}x,\text{maka}:\\ & y{}^{\prime }=f{}^{\prime }(x).u^{\prime }\text{atau}{\displaystyle \frac{dy}{dx}={\displaystyle \frac{dy}{du}.\frac{du}{dx}}}\\ & \end{array}\\ \hline\end{array}$.

Turunan Kedua Fungsi Aljabar

$\begin{array}{rl}& \\ \text{Suatu fungsi}& \\ & f:x\to y\\ & \end{array}$.

$\begin{array}{|cc|}\hline \text{Notasi}& \text{Proses}\\ \begin{array}{rl}& \\ & \{\begin{array}{ll}f{}^{\prime }(x)& ={\displaystyle \frac{dy}{dx}}\\ \\ f{}^{″}(x)& ={\displaystyle \frac{d^{2}y}{d{x}^2}}\end{array}\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \\ y& =a{x}^n\\ y^{\prime }& =n.a.x^{n-1}\\ y^{″}& =(n-1).n.a.x^{n-2}\\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai dari}{\displaystyle \frac{dy}{dx}}\\ & \text{a}.y=\sqrt{x}\\ & \text{b}.y=\sqrt{3+\sqrt{x}}\\ & \text{c}.y=\sqrt{3+\sqrt{3+\sqrt{x}}}\\ & \text{d}.y=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{x}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\text{a}.y& =\sqrt{x}\\ & =x^{{}^{\frac{1}{2}}}\\ y^{\prime }& ={\displaystyle \frac{1}{2}.x^{{}^{\frac{1}{2}-1}}}\\ & ={\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{2x^{{}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{1}{2\sqrt{x}}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}\text{b}.y& =\sqrt{3+\sqrt{x}}\\ & ={(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}}}\\ y^{\prime }& ={\displaystyle \frac{1}{2}{(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}-1}}.{\displaystyle \frac{1}{2}{x}^{{}^{\frac{1}{2}-1}}}}\\ & ={\displaystyle \frac{1}{4}{(3+x^{{}^{\frac{1}{2}}})}^{{}^{-\frac{1}{2}}}.x^{{}^{-\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{4{(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}}}.x^{{}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{1}{4\sqrt{3+\sqrt{x}}.\sqrt{x}}}\\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah turunannya}\\ & \begin{array}{l}\\ \text{a}.& f(x)=2\sin x\cos x& \text{k}.& f(x)=4\sin {(5x+\pi )}^3\\ \text{b}.& f(x)=x^2.\sin x& \text{l}.& f(x)={\cos }^3(x+5)& \\ \text{c}.& f(x)=3{\cos }^{2}x& \text{m}.& f(x)={\sin }^2(\pi -3x)& \\ \text{d}.& f(x)=3{\sin }^{2}x-x^3& \text{n}.& f(x)={\displaystyle \frac{{\sin }^{2}x}{\cos x}}& \\ \text{e}.& f(x)=5{\sin }^{4}x& \text{o}.& f(x)={\displaystyle \frac{\sin x^4}{x^2}}& \\ \text{f}.& f(x)=\sin \left({\displaystyle \frac{x+3}{x-2}}\right)\\ \text{g}.& f(x)=\cos (x^2-6x)\\ \text{h}.& f(x)={\displaystyle \frac{1}{\sin x}}\\ \text{i}.& f(x)={\displaystyle \frac{\sin x}{1+\cos x}}\\ \text{j}.& f(x)={\displaystyle \frac{\sin x-\cos x}{\sin x+\cos x}}\end{array}\end{array}$.

$\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ 2.\text{a}& \text{Alternatif 1}\\ & f(x)=2\sin x\cos x\\ & \begin{array}{rl}f(x)& =\sin 2x\\ f{}^{\prime }(x)& =\cos 2x.(2)\\ & =2\cos 2x\end{array}\\ \\ & \text{Alternatif 2}\\ & f(x)=y=\sin 2x\\ & \{\begin{array}{ll}y=\sin u& \to {\displaystyle \frac{dy}{du}=\cos u}\\ \\ u=2x& \to {\displaystyle \frac{du}{dx}=2}\end{array}\\ & \text{sehingga}\\ & \begin{array}{rl}y& =\sin 2x\\ {\displaystyle \frac{dy}{dx}}& ={\displaystyle \frac{dy}{du}.\frac{du}{dx}}\\ & =\cos u.2\\ & =2\cos u\\ & =2\cos 2x\end{array}\end{array}\end{array}$. 

$\begin{array}{l}\\ 2.\text{b}& f(x)=x^2\sin x\\ & \begin{array}{rl}f{}^{\prime }(x)& =2x^{2-1}.(\sin x)+x^2.(\cos x)\\ & =2x\sin x+x^2\cos x\end{array}\\ \text{c}& f(x)=3{\cos }^{2}x\\ & f^{\prime }(x)=2.3.{\cos }^{2-1}x.(-\sin x)\\ & =-6\sin x\cos x\\ & =-3\sin 2x\\ \text{d}& f(x)=3{\sin }^{2}x-x^3\\ & f^{\prime }(x)=2.3.{\sin }^{2-1}x.(\cos x)-3x^{3-1}\\ & =6\sin x\cos x-3x^2\\ & =3\sin 2x-3x^2\\ \text{e}& f(x)=5{\sin }^{4}x\\ & f^{\prime }(x)=4.5.{\sin }^{4-1}x.(\cos x)\\ & =20{\sin }^{3}x\cos x\\ & \text{atau boleh juga}\\ & =20{\sin }^{2}x\sin x\cos x\\ & =20{\sin }^{2}x\sin 2x=20\sin 2x{\sin }^{2}x\end{array}$ .

$\begin{array}{l}\\ 2.\text{m}& y=f(x)={\sin }^2(\pi -3x)\\ & \begin{array}{rl}f{}^{\prime }(x)& =2{\sin }^{2-1}(\pi -3x).\cos (\pi -3x).(-3)\\ & =-6\sin (\pi -3x)\cos (\pi -3x)\\ & =-3.2\sin (\pi -3x)\cos (\pi -3x)\\ & =-3\sin 2(\pi -3x)\\ & =-3\sin (2\pi -6x)\end{array}\\ \\ & \text{atau boleh juga}\\ & \begin{array}{rl}y=& f(x)={\sin }^2(\pi -3x)\\ & \{\begin{array}{ll}y=u^2& \to {\displaystyle \frac{dy}{du}=2u.}\\ \\ u=\sin w& \to {\displaystyle \frac{du}{dw}=\cos w}\\ \\ w=\pi -3x& \to {\displaystyle \frac{dw}{dx}=-3}\end{array}\\ {\displaystyle \frac{dy}{dx}}& ={\displaystyle \frac{dy}{du}.\frac{du}{dw}.\frac{dw}{dx}}\\ & =\left(2u\right).(\cos w).(-3)\\ & =-3.(2\sin w).(\cos w)\\ & =-3\sin 2w\\ & =-3\sin (2\pi -6x)\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai dari}{\displaystyle \frac{d^{2}y}{d{x}^2}}\\ & \text{a}.y=\sqrt{x}\\ & \text{b}.y=\sqrt{3+\sqrt{x}}\\ & \text{c}.y=15x^3-4x\\ & \text{d}.y={\displaystyle \frac{x^4}{3}+\frac{x^2}{2}+x+\sqrt{x}+1+{\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^2}+\frac{3}{x^4}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Lihat pembahasan soal no.1, yaitu }\\ & y=\sqrt{x}\\ & y^{\prime }={\displaystyle \frac{1}{2\sqrt{x}}}\end{array}\\ & \text{maka}\\ & \begin{array}{rl}3.\text{a}.y& =\sqrt{x}\\ y^{\prime }& ={\displaystyle \frac{dy}{dx}={\displaystyle \frac{1}{2\sqrt{x}}={\displaystyle \frac{1}{2}{x}^{-\frac{1}{2}}}}}\\ y^{″}& ={\displaystyle \frac{d^{2}y}{d{x}^2}=(-{\displaystyle \frac{1}{2}}){\displaystyle \frac{1}{2}.x^{-\frac{1}{2}-1}}}\\ & =-{\displaystyle \frac{1}{4x^{\frac{1}{2}+1}}=-{\displaystyle \frac{1}{4x\sqrt{x}}}}\end{array}\end{array}$.

$.\begin{array}{|l|}\hline \begin{array}{rl}& 3.\text{d}\\ & \begin{array}{rl}y& ={\displaystyle \frac{x^4}{3}+\frac{x^2}{2}+x+\sqrt{x}+1+{\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^2}+\frac{3}{x^4}}}\\ & \parallel \\ & \mathbf{\text{Turunan pertama}}\\ & \Downarrow \\ {\displaystyle \frac{dy}{dx}=y{}^{\prime }}& ={\displaystyle \frac{4x^3}{3}+x+1+{\displaystyle \frac{1}{2\sqrt{x}}+0-\frac{1}{2x\sqrt{x}}-\frac{1}{x^2}-\frac{2}{x^3}-\frac{12}{x^5}}}\\ & ={\displaystyle \frac{4x^3}{3}+x+1+{\displaystyle \frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}-\frac{1}{x^2}-\frac{2}{x^3}-\frac{12}{x^5}}}\\ & \parallel \\ & \mathbf{\text{Turunan kedua}}\\ & \Downarrow \\ {\displaystyle \frac{d^{2}y}{d{x}^2}=y{}^{″}}& =4x^2+1+0-{\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^2\sqrt{x}}+\frac{2}{x^3}+\frac{6}{x^4}+\frac{60}{x^6}}\\ & =4x^2+1-{\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^2\sqrt{x}}+\frac{2}{x^3}+\frac{6}{x^4}+\frac{60}{x^6}}\end{array}\\ & \end{array}\\ \hline\end{array}$.

$\text{LATIHAN SOAL}$.

Silahkan kerjakan soal yang belum diselesaikan atau dijawab

DAFTAR PUSTAKA

1. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU
2. Kartrini, Suprapto, Subandi, dan Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
3. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: ESIS
4. Sunardi, Waluyo, S., Sutrisno, Subagya. 2004. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.