Integral

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Sebagai pengingat untuk Kelas XII IPA

\begin{tabular}{|c|p{6.0cm}|c|p{8.0cm}|}\hline \multicolumn{2}{|c|}{Standar Kompetensi}&\multicolumn{2}{|c|}{Kompetensi Dasar}\\\hline 1.&Menggunakan konsep integral dalam pemecahan masalah&1.1&Memahami konsep integal tak dan integral tentu\\\cline{3-4} &&1.2&Menghitung integral tak tentu dan integral tentu dari fungsi aljabar dan fungsi trigonometri yang sederhana\\\cline{3-4} &&1.3&Menggunakan integral untuk menghitung luas daerah di bawah kurva dan volume benda putar\\\hline\end{tabular}.

Untuk kelas XII IPS

\begin{tabular}{|c|p{6.0cm}|c|p{8.0cm}|}\hline \multicolumn{2}{|c|}{Standar Kompetensi}&\multicolumn{2}{|c|}{Kompetensi Dasar}\\\hline 1.&Menggunakan konsep integral dalam pemecahan masalah sederhana&1.1&Memahami konsep integal tak dan integral tentu\\\cline{3-4} &&1.2&Menghitung integral tak tentu dan integral tentu dari fungsi aljabar sederhana\\\cline{3-4} &&1.3&Menggunakan integral untuk menghitung luas daerah di bawah kurva\\\hline\end{tabular}.

A. MATERI INTEGRAL

\begin{array}{|l|c|}\hline &\int x^{n}\: dx=\displaystyle \frac{1}{n+1}\: x^{n+1}+C,\qquad n\neq -1\\\cline{2-2} \textrm{Integral}\: \textrm{tak}\: \textrm{tentu}&\int ax^{n}\: dx=\displaystyle \frac{a}{n+1}\: x^{n+1}+C,\qquad n\neq -1\\\cline{2-2} &\int a\: dx=ax+C\\\hline \textrm{Integral}\: \textrm{tentu}&\int_{a}^{b}f\left ( x \right )dx=\left [ F\left ( x \right ) \right ]_{a}^{b}=F\left ( b \right )-F\left ( a \right )\\\hline \end{array}  .

  • Teknik Pengintegralan

\begin{array}{|c|c|}\hline \textrm{dengan}\: \textrm{substitusi}\: \textrm{aljabar}&\int u^{n}\: u'dx=\int u^{n}\: du=\displaystyle \frac{1}{n+1}u^{n+1}+C,\quad n\neq -1\\\hline &\textrm{jika}\quad \sqrt{a^{2}-u^{2}},\qquad \textrm{maka}\quad u=a\sin \theta\\\cline{2-2} \textrm{dengan}\: \textrm{substitusi}\: \textrm{trigonometri}&\textrm{jika}\quad \sqrt{a^{2}+u^{2}},\qquad \textrm{maka}\quad u=a\tan \theta \\\cline{2-2} &\textrm{jika}\quad \sqrt{u^{2}-a^{2}},\qquad \textrm{maka}\quad u=a\sec \theta\\\hline \textrm{integral}\: \textrm{parsial}&\int u\: dv=uv-\int v\: du\\\hline \end{array}  .

  • Integral Tak Tentu Fungsi Trigonometri

.\qquad \begin{cases} \int \sin x\: dx & =-\cos x+C \\ \int \sin ax\: dx & =-\displaystyle \frac{1}{a}\cos ax+C \\ \int \sin \left ( ax+b \right )\: dx & =-\displaystyle \frac{1}{a}\cos \left ( ax+b \right )+C \\ \int \cos x\: dx & =\sin x+C \\ \int \cos ax\: dx & =\displaystyle \frac{1}{a}\sin ax+C \\ \int \cos \left ( ax+b \right )\: dx & =\displaystyle \frac{1}{a}\sin \left ( ax+b \right )+C \\ \int \tan x\: dx & =-\ln \left | \cos x \right |+C=\ln \left | \sec x \right |+C \\ \int \tan ax\: dx & =-\displaystyle \frac{1}{a}\ln \left | \cos ax \right |+C=\displaystyle \frac{1}{a}\ln \left | \sec ax \right |+C\\ \int \tan \left ( ax+b \right )\: dx & =-\displaystyle \frac{1}{a}\ln \left | \cos \left ( ax+b \right ) \right |+C=\displaystyle \frac{1}{a}\ln \left | \sec \left ( ax+b \right ) \right |+C \end{cases}   .

B. CONTOH SOAL INTEGRAL TAK TENTU

\begin{array}{ll}\\\fbox{1}&\textrm{Tentukanlah}\: \textrm{hasi}l\: \textrm{dari}\: \textrm{integral}\: \textrm{tak}\: \textrm{tentu}\: \textrm{dari}\\ &\textrm{a}.\quad \int \left (x+1 \right )dx\\ &\textrm{b}.\quad \int \left (x^{3}+x^{2}+3x+2015 \right )\: dx\\ &\textrm{c}.\quad \int \left ( ax+b \right )^{2} dx\\ &\textrm{d}.\quad \int \left ( 4x-5 \right )^{2} dx\\ &\textrm{e}.\quad \int 4\left ( x-5 \right )^{2} dx\\ &\textup{f}.\quad \int x\left ( x-5 \right )^{2} dx\\ &\textrm{g}.\quad \int \left ( \displaystyle 8x^{2}-\frac{1}{x^{2}}+2015 \right )dx\\ &\textrm{h}.\quad \int \displaystyle \left ( x-\frac{1}{x} \right )^{2} dx\\ &\textrm{i}.\quad \int \left (\displaystyle x+\frac{1}{x} \right )dx\\ &\textrm{j}.\quad \int \displaystyle \sqrt{8x^{3}}\: dx\\ &\textrm{k}.\quad \int \displaystyle x\sqrt{x}\: dx\\ &\textrm{l}.\quad \int \displaystyle x^{2}\sqrt{x}\: dx \end{array}    .

Jawab:

\begin{aligned}1.\: \: a.\quad \int \left ( x+1 \right )dx&=\int x\: dx+\int 1\: dx\: \: ,&\textnormal{ingat bahwa }\qquad x=x^{1}\\ &=\displaystyle \frac{1}{\left (1+1 \right )}x^{\left (1+1 \right )}+x+C\: \: ,&\textnormal{perhatikan bahwa pangkat x dari 1 menjadi bertambah 1}\\ &=\displaystyle \frac{1}{2}x^{2}+x+C\: \: ,&\textnormal{sehingga pangkat x menjadi 2} \end{aligned}.

\begin{aligned}1.\: \: b.\quad \int \left ( x^{3}+x^{2}+3x+2015 \right )dx&=\displaystyle \frac{1}{\left (3+1 \right )}x^{\left (3+1 \right )}+\frac{1}{\left (2+1 \right )}x^{\left (2+1 \right )}+\frac{3x^{\left (1+1 \right )}}{\left (1+1 \right )}+2015x+C\\ &=\displaystyle \frac{1}{4}x^{4}+\frac{1}{3}x^{3}+\frac{3}{2}x^{2}+2015x \end{aligned}    .

\begin{aligned}1.\: \: c.\quad \int \left ( ax+b \right )^{2}dx&=\int \left ( ax+b \right ).\left ( ax+b \right )dx\\ &=\int \left (a^{2}x^{2}+abx+abx+b^{2} \right )dx=\int \left (a^{2}x^{2}+2abx+b^{2} \right )dx\\ &=\displaystyle \frac{a^{2}x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}+\frac{2abx^{\left (1+1 \right )}}{\left (1+1 \right )}+b^{2}x+C\\ &=\displaystyle \frac{a^{2}x^{3}}{3}+\frac{2abx^{2}}{2}+b^{2}x+C\: ,&\textnormal{atau dapat juga ditulis}\\ &=\displaystyle \frac{a^{2}}{3}x^{3}+abx^{2}+b^{2}x+C\end{aligned}.

\begin{aligned}1.\: \: d.\quad \int \left ( 4x-5 \right )^{2}dx&=\int \left ( 4x-5 \right ).\left ( 4x-5 \right )dx\\ &=\int \left (16x^{2}-40x+25 \right )dx\\ &=\displaystyle \frac{16x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}-\frac{40x^{\left (1+1 \right )}}{\left (1+1 \right )}+25x+C\\ &=\displaystyle \frac{16x^{3}}{3}-\frac{40x^{2}}{2}+25x+C\: ,&\textnormal{atau dapat juga ditulis}\\ &=\displaystyle \frac{16}{3}x^{3}-20x^{2}+25x+C\end{aligned}.

\begin{aligned}1.\: \: e.\quad \int 4\left ( x-5 \right )^{2}dx&=\int 4.\left ( x-5 \right ).\left ( x-5 \right )dx\\ &=\int \left (4x^{2}-40x+25 \right )dx\\ &=\displaystyle \frac{4x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}-\frac{40x^{\left (1+1 \right )}}{\left (1+1 \right )}+25x+C\\ &=\displaystyle \frac{4}{3}x^{3}-20x^{2}+25x+C\end{aligned}.

\begin{aligned}1.\: \: f.\quad \int x\left ( x-5 \right )^{2}dx&=\int x.\left ( x-5 \right ).\left ( x-5 \right )dx\\ &=\int \left (x^{3}-10x^{2}+25x \right )dx\\ &=\displaystyle \frac{x^{\left ( 3+1 \right )}}{\left ( 3+1 \right )}-\frac{10x^{\left (2+1 \right )}}{\left (2+1 \right )}+\frac{25x^{\left ( 1+1 \right )}}{\left (1+1 \right )}+C\\ &=\displaystyle \frac{1}{4}x^{4}-\frac{10}{3}x^{3}+\frac{25}{2}x^{2}+C\end{aligned}.

\begin{aligned}1.\: \: g.\quad \int \left ( 8x^{2}-\displaystyle \frac{1}{x^{2}}+2015 \right )dx&=\int 8x^{2}\: dx-\int \displaystyle \frac{1}{x^{2}}\: dx+\int 2015\: dx\\ &=\int 8x^{2}\: dx-\int x^{-2}\: dx+\int 2015\: dx\\ &=\displaystyle \frac{8x^{\left ( 2+1 \right )}}{\left ( 2+1 \right )}-\frac{x^{\left (-2+1 \right )}}{\left (-2+1 \right )}+2015x+C\\ &=\displaystyle \frac{8}{3}x^{3}-\frac{x^{-1}}{\left (-1 \right )}+2015x+C\\ &=\displaystyle \frac{8}{3}x^{3}+x^{-1}+2015x+C\\ &=\displaystyle \frac{8}{3}x^{3}+\frac{1}{x}+2015x+C\end{aligned}.

\begin{aligned}1.\: \: h.\quad \int \left ( x-\displaystyle \frac{1}{x}\right )^{2}dx&=\int \left ( \displaystyle x-\frac{1}{x} \right ).\left ( x-\frac{1}{x} \right ) dx\\ &=\int \displaystyle \left (x^{2}-2+\frac{1}{x^{2}} \right )dx\\ &=\int \displaystyle x^{2}\: dx-\int 2\: dx+\int \frac{1}{x^{2}}\: dx\\ &=\displaystyle \frac{1}{3}x^{3}-2x-\frac{1}{x}+C\: ,&\textnormal{lihat pembahasan sebelumnya pada No. 1. g}\end{aligned}.

\begin{aligned}1.\: \: i.\quad \int \left ( x-\displaystyle \frac{1}{x}\right )dx&=\int x\: dx-\int \displaystyle \frac{1}{x}\: dx\\ &=\displaystyle \frac{1}{2}x^{2}-\ln x+C\end{aligned}.

\begin{aligned}1.\: \: j.\quad \int \displaystyle \sqrt{8x^{3}}\: dx&=\int \displaystyle \sqrt{8}.x^{ \frac{3}{2}}\: dx\\ &=\displaystyle \frac{\displaystyle \sqrt{8}. x^{\left ( \frac{3}{2}+1 \right )}}{\left ( \displaystyle \frac{3}{2}+1 \right )}+C\\ &=\displaystyle \frac{\sqrt{8}x^{\frac{5}{2}}}{\displaystyle \frac{5}{2}}+C\quad ,&\textnormal{ingat bahwa}\: \sqrt{8}=2\sqrt{2}\\ &=\displaystyle\frac{4\sqrt{2}.\sqrt{x^{5}}}{5}+C\: ,&\textnormal{ingat juga bahwa }\: \sqrt{x^{5}}=\sqrt{x^{4}.x}=x^{2}\sqrt{x}\\ &=\displaystyle \frac{4}{5}x^{2}\sqrt{2x}+C\end{aligned}.

\begin{aligned}1.\: \: k.\quad \int \displaystyle x\sqrt{x}\: dx&=\int \displaystyle x^{1}.x^{\frac{1}{2}}\: dx=\int x^{1+\frac{1}{2}}\: dx=\int x^{\frac{3}{2}}\: dx\\ &=\displaystyle \frac{x^{\left (\frac{3}{2}+1 \right )}}{\left (\displaystyle \frac{3}{2}+1 \right )}+C=\frac{x^{\frac{5}{2}}}{\displaystyle \left ( \frac{5}{2} \right )}+C\\ &=\displaystyle \frac{2}{5}\sqrt{x^{5}}+C\\ &=\displaystyle \frac{2}{5}x^{2}\sqrt{x}+C\end{aligned}.

\begin{aligned}1.\: \: l.\quad \int x^{2}\sqrt{x}\: dx&=\int x^{2}.x^{\frac{1}{2}}\: dx=\int x^{\frac{5}{2}}\: dx\\ &=\displaystyle \frac{1}{\displaystyle \left (\frac{5}{2}+1 \right )}x^{\left ( \frac{5}{2}+1 \right )}+C\\ &=\displaystyle \frac{1}{\left ( \frac{7}{2} \right )}x^{\left (\frac{7}{2} \right )}+C\\ &=\displaystyle \frac{2}{7}x^{3}\sqrt{x}+C \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Tentukanlah}\: \textrm{hasil}\: \textrm{integral}\: \textrm{tak}\: \textrm{tentu}\: \textrm{dari}\\ &\textrm{a}.\quad \int \sin 2x\: dx\\ &\textrm{b}.\quad \int \cos 3x\: dx\\ &\textrm{c}.\quad \int \tan 4x\: dx\\ &\textrm{d}.\quad \int \sin ^{2}x\: dx\\ &\textrm{e}.\quad \int \cos ^{2}2x\: dx\\ &\textrm{f}.\quad \int \tan ^{2}3x\: dx\\ &\textrm{g}.\quad \int \sin ^{3}4x\: dx\\ &\textrm{h}.\quad \int \cos ^{3}5x\: dx\\ &\textrm{i}.\quad \int \sin x\cos x\: dx\\ &\textrm{j}.\quad \int \left ( \sin x+\cos x\right )^{2}\: dx\\ &\textrm{k}.\quad \int \left ( x^{2}+\cos \left ( 2x+3 \right ) \right )\: dx \end{array}.

Jawab:

\begin{aligned}2.\: \: a.\quad \int \sin 2x\: dx&=-\displaystyle \frac{1}{2}\cos 2x+C \end{aligned}.

\begin{aligned}2.\: \: b.\quad \int \cos 3x\: dx&=\displaystyle \frac{1}{3}\sin 3x+C \end{aligned}.

\begin{aligned}2.\: \: c.\quad \int \tan 4x\: dx&=-\displaystyle \frac{1}{4}\ln \left | \cos 4x \right |+C \end{aligned}.

\begin{aligned}2.\: \: d.\quad \int \sin ^{2}x\: dx&=\int \left ( \displaystyle \frac{1-\cos 2x}{2} \right )\: dx=\int \displaystyle \frac{1}{2}\: dx-\int \displaystyle \frac{\cos 2x}{2}\: dx=\displaystyle \frac{1}{2}\int 1\: dx-\displaystyle \frac{1}{2}\int \cos 2x\: dx\\ &=\displaystyle \frac{1}{2}x-\displaystyle \frac{1}{2}\left ( \frac{1}{2}\sin 2x \right )+C\\ &=\displaystyle \frac{1}{2}x-\displaystyle \frac{1}{4}\sin 2x+C\end{aligned}.

\begin{aligned}2.\: \: e.\quad \int \cos ^{2}2x\: dx&=\int \left ( \displaystyle \frac{1+\cos 4x}{2} \right )\: dx=\displaystyle \frac{1}{2}\int 1\: dx+\displaystyle \frac{1}{2}\int \cos 4x\: dx\\ &=\displaystyle \frac{1}{2}x+\displaystyle \frac{1}{2}\left ( \frac{1}{4}\sin 4x \right )+C\\ &=\displaystyle \frac{1}{2}x+\displaystyle \frac{1}{8}\sin 4x+C\end{aligned}.

Sebagai pengingat    \LARGE\boxed{\cos 2x=2\cos ^{2}x-1=1-2\sin ^{2}x}.

\begin{aligned}2.\: \: f.\quad \int \tan ^{2}3x\: dx&=\int \left ( \sec ^{2}3x-1 \right )\: dx\\ &=\int \sec ^{2}3x\: dx-\int 1\: dx\\ &=\displaystyle \frac{1}{3}\tan 3x-x+C\end{aligned}.

\begin{aligned}2.\: \: g.\quad \int \sin ^{3}4x\: dx&=\int \left ( \displaystyle \frac{3}{4}\sin 4x-\displaystyle \frac{1}{4}\sin 12x \right )\: dx,&\textnormal{ingat bahwa:}\: \: \sin 3x=3\sin x-4\sin ^{3}x\\ &=\displaystyle \frac{3}{4}\int \sin 4x\: dx-\displaystyle \frac{1}{4}\int \sin 12x\: dx\\ &=\displaystyle \frac{3}{4}\left ( -\frac{\cos 4x}{4} \right )-\displaystyle \frac{1}{4}\left ( -\frac{\cos 12x}{12} \right )+C\\ &=-\displaystyle \frac{3}{14}\cos 4x+\displaystyle \frac{1}{48}\cos 12x+C\end{aligned}.

\begin{aligned}2.\: \: h.\quad \int \cos ^{3}5x\: dx&=\int \left ( \displaystyle \frac{3}{4}\cos 5x+\displaystyle \frac{1}{4}\cos 15x \right )\: dx,&\textnormal{ingat bahwa:}\: \: \cos 3x=4\cos ^{3}x-3\cos x\\ &=\displaystyle \frac{3}{4}\int \cos 5x\: dx+\displaystyle \frac{1}{4}\int \cos 15x\: dx\\ &=\displaystyle \frac{3}{4}\left ( \frac{\sin 5x}{5} \right )+\displaystyle \frac{1}{4}\left ( \frac{\sin 15x}{15} \right )+C\\ &=\displaystyle \frac{3}{20}\sin 5x+\displaystyle \frac{1}{60}\sin 15x+C\end{aligned}.

\begin{aligned}2.\: \:i.\quad \int \sin x\cos x\: dx&=\displaystyle \int \frac{1}{2}\sin 2x\: dx=-\displaystyle \frac{1}{4}\cos 2x+C \end{aligned}.

\begin{aligned}2.\: \:j\quad \int \left (\sin x+\cos x \right )^{2}\: dx&=\displaystyle \int \left ( \sin x+\cos x \right )\left ( \sin x+\cos x \right )\: dx\\ &=\int \left (\sin ^{2}x+2\sin x\cos x+\cos ^{2}x \right )\: dx\\ &=\int \left (\sin ^{2}x+\cos ^{2}x+2\sin x\cos x \right )\: dx=\int \left ( 1+\sin 2x \right )\: dx\\ &=x-\displaystyle \frac{1}{2}\cos 2x+C \end{aligned}.

\begin{aligned}2.\: \:k\quad \int \left ( x^{2}+\cos \left ( 2x+3 \right ) \right )\: dx&=\displaystyle \frac{1}{3}x^{3}+\displaystyle \frac{1}{2}\sin \left ( 2x+3 \right )+C \end{aligned}.

\begin{array}{ll}\fbox{3}.&\textrm{Tentukan}\: \textrm{integral}\: \textrm{tak}\: \textrm{tentu}\: \textrm{dari}\\ &\textrm{a}.\quad \int \sin x\cos x\: dx\\ &\textrm{b}.\quad \int \sin x\sin 2x\: dx\\ &\textrm{c}.\quad \int \cos x\cos 3x\: dx\\ &\textrm{d}.\quad \int \sin x\cos 2x\: dx\\ &\textrm{e}.\quad \int \sin 2x\cos 2x\: dx\\ &\textrm{f}.\quad \int \cos x\sin 3x\: dx\\ &\textrm{g}.\quad \int \sin x\sin 2x\sin 3x\: dx\\ &\textrm{h}.\quad \int \cos x\cos 2x\cos 3x\: dx\end{array}.

Jawab:

\begin{aligned}3.\: \: a.\quad \int \sin x\cos x\: dx&=\int \displaystyle \frac{1}{2}\left ( \sin \left ( x+x \right )+\sin \left ( x-x \right ) \right )\: dx\\ &\textnormal{Kita dapat menggunakan formula}\: \begin{cases} 2\sin A\sin B & = \cos \left ( A-B \right )-\cos \left ( A+B \right )\\ 2\cos A\cos B & = \cos \left ( A-B \right )+\cos \left ( A+B \right )\\ 2\sin A\cos B & = \sin \left ( A+B \right )+\sin \left ( A-B \right )\\ 2\cos A\sin B & = \sin \left ( A+B \right )-\sin \left ( A-B \right ) \end{cases}\\ &=\displaystyle \frac{1}{2}\int \sin 2x\: dx\\ &=-\displaystyle \frac{1}{2}\left ( \frac{1}{2}\cos 2x \right )+C\\ &=-\displaystyle \frac{1}{4}\left ( 1-2\sin ^{2}x \right )+C\\ &=\displaystyle \frac{1}{2}\sin ^{2}x+C-\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{1}{2}\sin ^{2}x+C\\ &\textnormal{sebagai alternatif jawaban pada No. 2. i} \end{aligned}.

\begin{aligned}3.\: \: b.\quad \int \sin x\sin 2x\: dx&=\int \displaystyle \frac{1}{2}\left ( \cos \left ( x-2x \right )-\cos \left ( x+2x \right ) \right )\: dx\\ &=\displaystyle \frac{1}{2}\int \left (\cos \left ( -x \right )-\cos 3x \right )\: dx\: &\textnormal{ingat}\: \: \cos \left ( -x \right )=\cos x\\ &=\displaystyle \frac{1}{2}\left ( \int \cos x\: dx-\int \cos 3x\: dx \right )\\ &=\displaystyle \frac{1}{2}\left ( \sin x-\displaystyle \frac{1}{3}\sin 3x \right )+C \end{aligned}.

\begin{aligned}3.\: \: c.\quad \int \cos x\cos 3x\: dx&=\int \displaystyle \frac{1}{2}\left ( \cos \left ( x-3x \right )+\cos \left ( x+3x \right ) \right )\: dx\\ &=\displaystyle \frac{1}{2}\int \left (\cos \left ( -2x \right )+\cos 4x \right )\: dx\\ &=\displaystyle \frac{1}{2}\int \left ( \cos 2x+\cos 4x \right )\: dx\\ &=\displaystyle \frac{1}{2}\left ( \displaystyle \frac{1}{2}\sin 2x+\displaystyle \frac{1}{4}\sin 4x \right )+C\\ &=\displaystyle \frac{1}{4}\sin 2x+\displaystyle \frac{1}{8}\sin 4x+C\end{aligned}.

\begin{aligned}3.\: \: g.\quad \int \sin x\sin 2x\sin 3x\: dx&=\int \displaystyle \frac{1}{2}\left ( 2\sin x\sin 2x \right )\sin 3x\: dx\\ &=\displaystyle \frac{1}{2}\int \left ( \cos x-\cos 3x \right )\sin 3x\: dx\\ &=\displaystyle \frac{1}{2}\left (\int \cos x\sin 3x\: dx-\int \cos 3x\sin 3x\: dx \right )\\ &=\displaystyle \frac{1}{2}.\frac{1}{2}\int \left ( \sin \left ( x+3x \right )-\sin \left ( x-3x \right ) \right )\: dx-\displaystyle \frac{1}{2}.\frac{1}{2}\int \left ( \sin \left ( 3x+3x \right )-\sin \left ( 3x-3x \right ) \right )\: dx\\ &=\displaystyle \frac{1}{4}\int \sin 4x\: dx-\displaystyle \frac{1}{4}\int -\sin 2x\: dx-\displaystyle \frac{1}{4}\int \sin 6x\: dx\\ &=-\displaystyle \frac{1}{16}\cos 4x-\displaystyle \frac{1}{8}\cos 2x+\displaystyle \frac{1}{24}\cos 6x+C\end{aligned}.

Untuk jawaban soal-soal lainnya silahkan dibuat sebagai latihan mandiri

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui}\: f'\left ( x \right )=6x^{2}-2x+6\: \textrm{dan}\: \textrm{nilai}\: \textrm{fungsi}\: f\left ( 2 \right )=-7.\: \textrm{Tentukan}\: \textrm{rumus}\: \textrm{fungsi}\: \textrm{tersebut}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}f\left ( x \right )&=\int f'\left ( x \right )\: dx\\ &=\int \left ( 6x^{2}-2x+6 \right )\: dx\\ &=2x^{3}-x^{2}+6x+C \end{aligned}\\\\ &\textrm{Karena}\: f\left ( 2 \right )=-7,\: \textrm{maka}\\\\ &\begin{aligned}f\left ( 2 \right )&=2.2^{3}-2^{2}+6.2+C\\ \Leftrightarrow -7&=16-4+12+C\\ \Leftrightarrow C&=-31 \end{aligned}\\\\ &\textrm{Jadi},\: f\left ( x \right )=2x^{3}-x^{2}+6x-31 \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Diketahui}\: \textrm{gradien}\: \textrm{suatu}\: \textrm{kurva}\: \textrm{adalah}\: \: \displaystyle \frac{dy}{dx}=2x-2.\: \textrm{Tentukan}\: \textrm{persamaan}\: \textrm{kurva}\: \textrm{tersebut}\: \textrm{jika}\: \textrm{melalui}\: \textrm{titik}\: \left ( 3,2 \right )\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}\displaystyle \frac{dy}{dx}&=2x-2\\ dy&=\left ( 2x-2 \right )\: dx\\ \int dy&=\int \left ( 2x-2 \right )\: dx\\ y&=x^{2}-2x+C\end{aligned}\\\\ &\textrm{Karena}\: \textrm{kurva}\: \textrm{melalui}\: \left ( 3,2 \right ),\: \textrm{maka}\\\\ &\begin{aligned}2&=3^{2}-2.3+C\\ C&=-1 \end{aligned}\\\\ &\textrm{Jadi},\: \textrm{persamaan}\: \textrm{kurvanya}\: \textrm{adalah}\: y=x^{2}-2x-1 \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Carilah}\: y=f(x)\: \textrm{jika}\: \textrm{diketahui}\\ &\textrm{a}.\quad {f}''\left ( x \right )=2,\: {f}'\left ( 2 \right )=5,\: \textrm{dan}\: f\left ( 2 \right )=10.\\ &\textrm{b}.\quad {f}''\left ( x \right )=x^{2},\: {f}'\left ( 0 \right )=6,\: \textrm{dan}\: f\left ( 0 \right )=3.\\ &\textrm{c}.\quad {f}''\left ( x \right )=\displaystyle x^{-\frac{3}{2}},\: {f}'\left ( 4 \right )=2,\: \textrm{serta}\: f\left ( 0 \right )=0.\\ &\textrm{d}.\quad \displaystyle \frac{d^{2}y}{dx^{2}}=4-6x,\: \textrm{untuk}\: \: x=2,\: \displaystyle \frac{dy}{dx}=-4\: \: \textrm{dan}\: y=7.\\ &\textrm{e}.\quad \displaystyle \frac{d^{2}y}{dx^{2}}=25x-2,\: \textrm{untuk}\: \: x=2,\: \displaystyle \frac{dy}{dx}=25\: \: \textrm{dan}\: y=20.\\ &\textrm{f}.\quad \displaystyle \frac{d^{2}y}{dx^{2}}=6x-4,\: \textrm{kurva}\: \textrm{melalui}\: \textrm{titik}\: \left ( -1,2 \right )\: \textrm{dan}\: \left ( 1.4 \right ).\end{array}.

Jawab:

\begin{aligned}6.\: \: a.\qquad\qquad {f}''\left ( x \right )&=2\\ \int {f}''\left ( x \right )dx&=\int 2\: dx\\ {f}'\left ( x \right )&=2x+C_{1},&\textnormal{diketahui dari soal}\: {f}'\left ( 2 \right )=5,\: \textrm{maka}\\ 2x+C_{1}&={f}'\left ( x \right )\\ 2\left ( 2 \right )+C_{1}&={f}'\left ( 2 \right )=5\\ C_{1}&=5-4=1\\ \textrm{sehingga}\: \: {f}'\left ( x \right )&=2x+1 \end{aligned}\\\\ \textrm{Selanjutnya dengan cara yang kurang lebih sama kita akan mendapatkan f(x), yaitu}\\ \begin{aligned}{f}'\left ( x \right )&=2x+1\\ \int {f}'\left ( x \right )dx&=\int \left ( 2x+1 \right )\: dx\\ f\left ( x \right )&=x^{2}+x+C,&\textnormal{diketahui juga dari soal bahwa f(2)=10},\: \textrm{maka}\\ x^{2}+x+C&=f\left ( x \right )\\ \left (2 \right )^{2}+2+C&=f\left ( 2 \right )=10\\ C&=10-6=4\\ \textrm{sehingga}\: \:f\left ( x \right )&=x^{2}+x+4 \end{aligned}\\\\ \therefore f\left ( x \right )=x^{2}+x+4        .

\begin{aligned}6\: \: d.\quad \displaystyle \frac{d^{2}y}{dx^{2}}&=4-6x\\ \int \displaystyle \frac{d^{2}y}{dx^{2}}&=\int \left ( 4-6x \right )dx\\ \displaystyle \frac{dy}{dx}&=4x-3x^{2}+C_{1},&\textnormal{di soal diketahui}\: \: \displaystyle \frac{dy}{dx}=-4,\: x=2\: \textrm{maka} \\ -4&=4\left ( 2 \right )-3\left ( 2 \right )^{2}+C_{1}\\ C_{1}&=0\end{aligned}\\ \textrm{Sehingga}\: \: \displaystyle \frac{dy}{dx}=-3x^{2}+4x\\\\ \begin{aligned}\displaystyle \frac{dy}{dx}&=-3x^{2}+4x\\ \int \displaystyle \frac{dy}{dx}&=\int \left ( -3x^{2}+4x \right )\: dx\\ y&=-x^{3}+2x^{2}+C,&\textnormal{diketahui juga bahwa y=7, x=2},\: \textrm{maka}\\ 7&=-\left ( 2 \right )^{3}+2\left ( 2 \right )^{2}+C\\ C&=7\\ \textrm{Sehingga}\: y&=-x^{3}+2x^{2}+7 \end{aligned}\\\\ \therefore y=-x^{3}+2x^{2}+7          .

Untuk Soal yang lain silahkan dikerjakan sebagai laithan mandiri

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

\textrm{Tentukan}\: \textrm{integralkan}\: \textrm{tak}\: \textrm{tentu}\: \textrm{berikut}\\ \begin{array}{lllll}\\ 1.&a.& \int dx& k.&\int -2x^{-5}\: dx\\ &b.&\int 3\: dz&l.&\int \left (m^{\frac{1}{3}}-\displaystyle \frac{1}{m^{\frac{1}{2}}} \right )\: dm\\ &c.&\int \left ( 3x^{2}-2 \right )\: \: dx&m.&\int \left ( t^{2}-\displaystyle \frac{4}{t^{2}} \right )\: dt\\ &d.&\int \left (x^{5}-3x+4 \right )\: dx&n.&\int \left ( x+5 \right )\left ( x-7 \right )\: dx\\ &e.&\int \left ( 4x^{3}-6x^{2}+3x-6 \right )\: dx&o.&\int \displaystyle \frac{x^{3}+5x^{2}-4}{x^{2}}\: dx\\ &f.&\int \left (6-2t+3t^{2}-8t^{3} \right )\: dt&p.&\int \sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}\: dx\\ &g.&\int y\: dx&q.&\int x^{-6}\left ( x\sqrt{x}-x^{-2} \right )\: dx\\ &h.&\int x\: dy&r.&\int \left ( n^{2}-\displaystyle \frac{2}{n^{2}} \right )\: dn\\ &i.&\int \left ( x+y+z \right )\: dx&s.&\int \left ( y\sqrt{y}-\displaystyle \frac{1}{y\sqrt{y}} \right )\: dy\\ &j.&\int \displaystyle \frac{3}{x^{2}\sqrt{x}}\: dx&t.&\int \left ( x-1 \right )\left ( x-2 \right )\left ( x-3 \right )\: dx\end{array}.

\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \textrm{pada}\: \textrm{tiap}\: \textrm{titik}\: \left ( x,y \right )\: \textrm{gradien}\: \textrm{sebuah}\: \textrm{kurva}\: \textrm{ditentukan}\: \textrm{oleh}\: \: \displaystyle \frac{dy}{dx}=1-2x.\\ & \textrm{Jika}\: \textrm{nilai}\: \textrm{maksimum}\:\: y\:\: \textrm{adalah}\: \displaystyle \frac{25}{4},\: \textrm{maka}\: \textrm{persamaan}\: \textrm{kurva}\: \textrm{tersebut}\: \textrm{adalah}.... \end{array}.

\begin{tabular}{ll}\\ 3.&lihat pembahasan pada contoh soal No.3\\ &Silahkan kerjakan soal-soal yang belum dibahas\end{tabular}.

\begin{tabular}{ll}\\ 4.&lihat pembahasan pada contoh soal No.6\\ &Silahkan kerjakan juga soal-soal yang belum dibahas\end{tabular}.

C. CONTOH SOAL INTEGRAL TENTU

\begin{array}{ll}\\ \fbox{1}.&\textrm{Hitunglah\: hasil\: dari\: integral-integral\: berikut}\\ &\textrm{a}.\quad \int_{1}^{3}\left ( 2y+1 \right )\: dy\\ &\textrm{b}.\quad \int_{-1}^{1}x^{3}-1\: dx\\ &\textrm{c}.\quad \int_{-1}^{1}\left ( 7x^{3}-3 \right )\: dx\\ &\textrm{d}.\quad \int_{-2}^{1}\left ( m+3 \right )^{2}\: dm\\ &\textrm{e}.\quad \int_{0}^{ \frac{\pi }{2}}\sin \theta \: d\theta \\ &\textrm{f}.\quad \int_{-\pi }^{\pi }\left ( 1+\cos x \right )\: dx \end{array}.

Jawab:

\begin{aligned}1.\: \: a.\quad \int_{1}^{3}\left ( 2y+1 \right )\: dy&=y^{2}+y\: |_{1}^{3}\\ &=\left ( 3^{2}+3 \right )-\left ( 1^{2}+1 \right )\\ &=12-2\\ &=10 \end{aligned}.

\begin{aligned}1.\: \: b.\quad \int_{-1}^{1}\left ( x^{3}-1 \right )\: 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}.\left ( -1 \right )^{4}-\left ( -1 \right ) \right )\\ &=\left ( \displaystyle \frac{1}{4}-1 \right )-\left ( \displaystyle \frac{1}{4}+1 \right )\\ &=-2 \end{aligned}.

\begin{aligned}1.\: \: e.\quad \int_{0}^{\frac{\pi }{2}}\sin \theta \: d\theta &=-\cos \theta \: |_{0}^{\frac{\pi }{2}}\\ &=\left ( -\cos \displaystyle \frac{\pi }{2} \right )-\left ( -\cos 0 \right )\\ &=\left ( 0 \right )-\left ( -1 \right )\\ &=1 \end{aligned}.

\begin{aligned}1.\: \: f.\quad \int_{-\pi }^{\pi }\left ( 1+\cos x \right )\: dx&=\left ( x+\sin x \right )|_{-\pi }^{\pi }\\ &=\left ( \pi +\sin \pi \right )-\left ( -\pi +\sin \left ( -\pi \right ) \right )\\ &=\left ( \pi +0 \right )-\left ( -\pi -0 \right )\\ &=2\pi \end{aligned}.

\begin{aligned}2.\: \: \textrm{Nilai}&\: \int_{0}^{\frac{\pi }{3}}\sin 2x\: dx=....\\ a.\quad \frac{3}{4}&\qquad b.\quad \frac{1}{2}\qquad c.\quad\frac{1}{3}\qquad d.\quad \frac{1}{4}\qquad e.\quad 0\qquad\qquad \textbf{(UN\: 2006)}\end{aligned}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\int_{0}^{\frac{\pi }{3}}\sin 2x\: dx&=-\displaystyle \frac{1}{2}\cos 2x|_{0}^{\frac{\pi }{3}}\\ &=\left ( -\displaystyle \frac{\cos \frac{2\pi }{3}}{2} \right )-\left ( -\displaystyle \frac{\cos 0}{2} \right )\\ &=\left ( -\displaystyle \frac{\frac{1}{2}}{2} \right )-\left ( -\displaystyle \frac{1}{2} \right )\\ &=\left ( -\displaystyle \frac{1}{4} \right )-\left ( -\displaystyle \frac{1}{2} \right )\\ &=\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}\\ &=\displaystyle \frac{1}{4} \end{aligned}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

\begin{array}{ll}\\ 1.\quad \textrm{Hitunglah\: nilai\: tiap\: integral\: berikut\: ini} \end{array}\\ \begin{array}{lllll}\\ .\quad&a.&\int_{0}^{3}2x\: dx&i.&\int_{1}^{2}\sqrt{x^{5}}\: dx\\ &b.&\int_{2}^{3}2x^{2}\: dx&j.&\int_{1}^{2}\displaystyle \frac{1}{x^{2}}\: dx\\ &c.&\int_{3}^{7}\displaystyle \frac{1}{2}x^{3}\: dx&k.&\int_{4}^{9}3\sqrt{x}\: dx\\ &d.&\int_{0}^{3}\left ( x^{2}+2x-1 \right )\: dx&l.&\int_{4}^{6}\left ( \sqrt{t}+t^{2}-\sqrt[3]{t^{2}} \right )\: dt\\ &e.&\int_{0}^{3}\left ( x+3 \right )^{2}\: dx&m.&\int_{0}^{\pi }\cos x\: dx\\ &f.&\int_{-1}^{2}\left ( x+2 \right )\left ( x-1 \right )\: dx&n.&\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin x\cos x\: dx\\ &g.&\int_{1}^{3}\left ( t+t^{2}-3t^{3} \right )\: dt&o.&\int_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\sin x\: dx\\ &h.&\int_{1}^{4}\left ( \displaystyle \frac{x^{4}-x^{3}+\sqrt{x}-1}{x^{2}} \right )\: dx&p.&\int_{0}^{\frac{\pi }{6}}\left ( \sin 2x+\cos 3x \right )\: dx \end{array}.

\begin{aligned}2.\: \: \textrm{Nilai}\: \int_{\frac{\pi }{6}}^{\frac{\pi }{3}}&\left ( 3\cos x-5\sin x \right )\: dx=....\textbf{(EBTANAS\: 1997)} \end{aligned}.

Sumber Referensi

  1. Kuntarti, Sulistiyono dan Sri Kurnianingsih. 2007. Matematika  SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: esis.
  2. Noormandiri, Endar Sucipto. 2003. Buku Pelajaran Matematika SMU untuk Kelas 3 Program IPA. Jakarta: Erlangga.
  3. Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas XII IPA untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.

 

 

 

 

 

 

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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