Kumpulan Contoh Soal Integral

\begin{array}{|ll|}\hline \textbf{DEFINISI}&\\ &1.\quad \displaystyle \int f(x)\: dx=F(x)+C\quad \Leftrightarrow \quad f(x)=F'(x)\\ &2.\quad \displaystyle \int_{a}^{b}f(x)=F(b)-F(a)\\ \textbf{Rumus Umum}&\\ &1.\quad \displaystyle \int a\: dx=ax+C\\ &2.\quad \displaystyle \int x^{n}\: dx=\displaystyle \frac{1}{n+1}x^{n+1}+C;\quad n\neq -1\\ &3.\quad \displaystyle \int \frac{1}{x}\: dx=\ln \left | x \right |+C\\ &4.\quad \displaystyle \int e^{kx}\: dx=\displaystyle \frac{1}{k}e^{kx}+C;\quad k\neq 0\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{1}.&\textrm{\textbf{(UN 2015 Matematika IPA)} Hasil}\: \: \displaystyle \int 6x\left ( 1-x^{2} \right )^{4}\: \: dx\: \: \textrm{adalah ....} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad \displaystyle \frac{3}{5}\left ( 1+x^{2} \right )^{5}+C\\\\ &b.\quad \displaystyle \frac{2}{5}\left ( 1+x^{2} \right )^{5}+C\\\\ &c.\quad \displaystyle -\frac{1}{5}\left ( 1-x^{2} \right )^{5}+C\\\\ &d.\quad -\displaystyle \frac{2}{5}\left ( 1-x^{2} \right )^{5}+C\\\\ &e.\quad -\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C\end{array}.

Jawab:

\begin{array}{l|l}\\ \begin{aligned}\displaystyle &\int 6x\left ( 1-x^{2} \right )^{4}\: dx\\ &=\int \left ( 1-x^{2} \right )^{4}.6x\: dx\\ &\begin{cases} u& =1-x^{2} \\ & \\ du& =-2x\: \: \: dx\quad \Rightarrow \quad -3\: dx=6x\: dx \end{cases}\\ &\displaystyle \int u^{4}.\left ( -3\: du \right )\\ &=-3\displaystyle \int u^{4}\: du\\ &=-\displaystyle \frac{3}{5}u^{5}+C\\ &=-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C \end{aligned}&\begin{aligned}&\textrm{Atau dengan cara lain, yaitu:}\\ \displaystyle \int \left ( 1-x^{2} \right )^{4}\: 6x\: dx&=\int \left ( 1-x^{2} \right )^{2}.(-3).\left ( -2x\: dx \right )\\ &=-3\displaystyle \int \left ( 1-x^{2} \right )^{4}.\: \left ( -2x\: dx \right )\\ &=-3\displaystyle \int \left ( 1-x^{2} \right )^{4}\: .d\left ( 1-x^{2} \right )\\ &=-3\left [ \displaystyle \frac{\left ( 1-x^{2} \right )^{5}}{5} \right ]+C\\ &=-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C\\ &\\ &\\ & \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\begin{aligned}\displaystyle &\int \frac{dx}{2015x+2016}\end{aligned} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad \ln \left | 2015x+2016 \right |+C\\ &\\ &b.\quad \displaystyle \frac{1}{2014}\ln \left | 2015x+2016 \right |+C\\ &\\ &c.\quad \displaystyle \frac{1}{2015}\ln \left | 2015x+2016 \right |+C\\ &\\ &d.\quad \displaystyle \frac{1}{2016}\ln \left | 2015x+2016 \right |+C\\ &\\&e.\quad \displaystyle \frac{2015}{2016}\ln \left | 2015x+2016 \right |+C \end{array}.

Jawab:

\textrm{Misalkan}\: \: \begin{cases} u & =2015x+2016 \\ du & =2015\: dx\quad \Rightarrow \quad \displaystyle \frac{1}{2015}\: du=dx \end{cases}\\\\ \begin{aligned}\displaystyle \int \frac{dx}{2015x+2016}=\displaystyle \int \frac{1}{u}.\left ( \frac{1}{2015}\: du \right )=\displaystyle \frac{1}{2015}\int \frac{1}{u}\: du=\displaystyle \frac{1}{2015}\ln \left | 2015x+2016 \right |+C \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\begin{aligned}\displaystyle &\int \frac{\sin \displaystyle \frac{1}{x}}{x^{2}}\: \: dx= ....(\textbf{EBTANAS 2003})\end{aligned} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad \sin x^{2}+C\\ &b.\quad \cos x+C\\ &c.\quad \sin \displaystyle \frac{1}{x}+C\\ &d.\quad \cos \displaystyle \frac{1}{x}+C\\ &e.\quad \cos x^{2}+C \end{array}.

Jawab:

\textrm{Misalkan}\: \: \begin{cases} u & = \cos \displaystyle \frac{1}{x}\\ &\\ du & = -\sin \displaystyle \frac{1}{x}.\left ( \displaystyle \frac{-1}{x^{2}} \right )\: dx=\displaystyle \frac{\sin \displaystyle \frac{1}{x}}{x^{2}}\: \: dx \end{cases}\\\\ \displaystyle \int \frac{\sin \displaystyle \frac{1}{x}}{x^{2}}=\displaystyle \int du=u+C=\cos \displaystyle \frac{1}{x}+C.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui}\: \displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\: \textrm{jika} \: f(4)=19\: \textrm{ maka}\: f(1)=.... \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad 2\\ &b.\quad 3\\ &c.\quad 4\\ &d.\quad 5\\ &e.\quad 6\end{array}.

Jawab:

\textrm{Perlu diingat bahwa}\: \: \displaystyle \frac{d}{dx}f(x)=h(x)\: \Rightarrow \: \displaystyle \int h(x)\: dx=f(x)\\\\ \begin{array}{l|l}\\ \begin{aligned}\displaystyle \frac{df(x)}{dx}&=3\sqrt{x}\\ f(x)&=\displaystyle \int 3\sqrt{x}\: \: \: dx=\int 3x^{\frac{1}{2}}\: \: \: dx\\ f(x)&=\displaystyle \frac{3}{\frac{3}{2}}x^{\frac{3}{2}}+C\\ f(x)&=2x^{\frac{3}{2}}+C\\ f(x)&=2x\sqrt{x}+C \end{aligned}&\begin{aligned}\textrm{Diketahui}&\\f(4)&=19\\ 2(4)(\sqrt{4})+C&=19\\ 16+C&=19\\ C&=3\\ \textrm{Sehingga}&\quad f(x)=2x\sqrt{x}+3\\ \textrm{maka}&\quad f(1)=2.1.\sqrt{1}+3=2+3=5 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Jika}\: \displaystyle \frac{d}{dx}g(x)=f(x)\: \textrm{di mana} \: f(x)\: \textrm{kontinu dari a sampai b, maka}\: \displaystyle \int_{a}^{b}f(x).g(x)\: dx \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad 0\\ &b.\quad f(b)-f(a)\\ &c.\quad g(b)-g(a)\\ &d.\quad \displaystyle \frac{\left [ f(b) \right ]^{2}-\left [ f(a) \right ]^{2}}{2}\\ &e.\quad \displaystyle \frac{\left [ g(b) \right ]^{2}-\left [ g(a) \right ]^{2}}{2}\end{array}.

Jawab:

\begin{aligned}\displaystyle \int_{a}^{b} f(x).g(x)\: dx&=\displaystyle \int_{a}^{b} g(x).f(x)\: dx\\ &=\displaystyle \int_{a}^{b} g(x).\displaystyle \frac{d\left ( g(x) \right )}{dx}\: dx,&\textnormal{ingat bahwa}\quad \displaystyle \frac{d}{dx}g(x)=f(x),\: \textrm{f(x) kontinu dari a sampai b}\\ &=\displaystyle \int_{a}^{b} g(x)\: d\left ( g(x) \right )\\ &=\left [\displaystyle \frac{\left ( g(x) \right )^{2}}{2} \right ]_{a}^{b}\\ &=\displaystyle \frac{\left [ g(b) \right ]^{2}-\left [ g(a) \right ]^{2}}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{6}.&\begin{aligned}\displaystyle &-\int \frac{dx}{x^{2}+3x+2}\: \: dx= \end{aligned} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad \displaystyle \ln \left | \displaystyle \frac{x+1}{x+2} \right |+C\\ &\\ &b.\quad \displaystyle \ln \left | \displaystyle \frac{x+2}{x+1} \right |+C\\ &\\ &c.\quad \displaystyle \ln \left | x^{2}+3x+2 \right |+C\\ &\\ &d.\quad \arctan 2\left ( \displaystyle x+\frac{3}{2} \right )+C\\ &\\ &e.\quad -\displaystyle \arctan 2\left (\displaystyle x+ \frac{3}{2}\right )+C \end{array}.

Jawab:

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

\begin{array}{ll}\\ \fbox{7}.&\begin{aligned}\displaystyle &\int e^{ax}\: \cos bx\: \: dx= \end{aligned} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad \displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left ( b^{2}\sin bx+a^{2}\cos bx \right )+C\\ &\\ &b.\quad \displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left ( b^{2}\sin bx-a^{2}\cos bx \right )+C\\ &\\ &c.\quad \displaystyle \frac{e^{ax}}{a+b}\left ( b\sin bx+a\cos bx \right )+C\\ &\\ &d.\quad \displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left ( b\sin bx+a\cos bx \right )+C\\ &\\ &e.\quad \displaystyle \frac{e^{ax}}{a^{2}-b^{2}}\left ( b\sin bx-a\cos bx \right )+C \end{array}.

Jawab:

\begin{aligned}\displaystyle &\int e^{ax}\: \cos bx\: \: dx=\displaystyle \int \underset{\displaystyle \underset{u}{\mid }}{e^{ax}}.\underset{dv}{\underbrace{\cos bx\: \: dx}}=uv-\displaystyle \int v\: du \end{aligned}\\ \quad\quad\quad \begin{array}{lll}\\ \displaystyle u=e^{ax}&&dv=\cos bx\: dx\\ du=\displaystyle ae^{ax}&&\displaystyle \int dv=\int \cos bx\: dx\\ &&v=\displaystyle \frac{1}{b}\sin bx\: +C\\ \end{array}\\\\.

Selanjutnya

\begin{aligned}\displaystyle \int \underset{\displaystyle \underset{u}{\mid }}{e^{ax}}.\underset{\displaystyle \underset{dv}{\mid }}{\underbrace{\cos bx\: \: dx}}&=\underset{\displaystyle \underset{u}{\mid }}{e^{ax}}.\underset{v}{\underbrace{\displaystyle \frac{1}{b}\sin bx}}-\displaystyle \int \underset{v}{\underbrace{\left ( {\displaystyle \frac{1}{b}\sin bx} \right )}}.\displaystyle \underset{\displaystyle \underset{du}{\mid }}{\underbrace{ae^{ax} \: \: dx}}\\ &=\displaystyle \frac{e^{ax}}{b}.\sin bx-\displaystyle \frac{a}{b}\int e^{ax}.\sin bx\: \: dx,&\textnormal{dengan cara yang sama, maka}\\ &=\displaystyle \frac{e^{ax}}{b}.\sin bx-\displaystyle \frac{a}{b}\left [ -\displaystyle \frac{e^{ax}}{b}\cos bx+\displaystyle \frac{a}{b}\int e^{ax}\cos bx\: \: dx \right ]\\ &=\displaystyle \frac{e^{ax}}{b}\sin bx+\displaystyle \frac{a}{b^{2}}.e^{ax}\cos bx-\displaystyle \frac{a^{2}}{b^{2}}\int e^{ax}\cos bx\: \: dx\\ \left ( \displaystyle 1+ \frac{a^{2}}{b^{2}}\right )\displaystyle \int e^{ax}\cos bx\: \: dx&=\displaystyle \frac{e^{ax}}{b}\sin bx+\displaystyle \frac{a}{b^{2}}.e^{ax}\cos bx\\ \left ( \displaystyle \frac{a^{2}+b^{2}}{b^{2}} \right )\displaystyle \int e^{ax}\cos bx\: \: dx&=\displaystyle \frac{e^{ax}}{b}\sin bx+\displaystyle \frac{a}{b^{2}}.e^{ax}\cos bx\\ \displaystyle \int e^{ax}\cos bx\: \: dx&=\left ( \displaystyle \frac{b^{2}}{a^{2}+b^{2}} \right )\left [\displaystyle \frac{e^{ax}}{b}\sin bx+\displaystyle \frac{a}{b^{2}}.e^{ax}\cos bx \right ]\\ \displaystyle \int e^{ax}\cos bx\: \: dx&=\displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left [ b\sin bx+a\cos bx \right ]+C \end{aligned}.

\begin{array}{ll}\\ \fbox{8}.&\begin{aligned}\displaystyle &\int e^{ax}\: \sin bx\: \: dx= \end{aligned} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad \displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left ( a^{2}\sin bx+b^{2}\cos bx \right )+C\\ &\\ &b.\quad \displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left ( a^{2}\sin bx-b^{2}\cos bx \right )+C\\ &\\ &c.\quad \displaystyle \frac{e^{ax}}{a+b}\left ( a\sin bx+b\cos bx \right )+C\\ &\\ &d.\quad \displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\sin bx+b\cos bx \right )+C\\ &\\ &e.\quad \displaystyle \frac{e^{ax}}{a^{2}+b^{2}}\left ( a\sin bx-b\cos bx \right )+C \end{array}.

Jawab: e

Pembahasan diserahkan kepada pembaca yang budiman

\begin{array}{ll}\\ \fbox{9}.&\begin{aligned}\displaystyle &\int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi } \displaystyle \frac{d\theta }{\sqrt{\tan ^{2}\theta -\sin ^{2}\theta }}= \end{aligned} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad 1\\ &b.\quad -1\\ &c.\quad 0\\ &d.\quad \infty \\ &e.\quad \displaystyle \frac{1}{2} \end{array}.

Jawab:

\begin{aligned}\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi } \displaystyle \frac{d\theta }{\sqrt{\tan ^{2}\theta -\sin ^{2}\theta }}&=\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi }\displaystyle \frac{d\theta }{\sqrt{\displaystyle \frac{\sin ^{2}\theta }{\cos ^{2}\theta }-\sin ^{2}\theta} }=\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi }\displaystyle \frac{d\theta }{\sqrt{\displaystyle \frac{\sin ^{2}\theta }{\cos ^{2}\theta }-\frac{\sin ^{2}\theta \cos ^{2}\theta }{\cos ^{2}\theta }}}\\ &=\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi }\displaystyle \frac{d\theta }{\displaystyle \frac{\sin \theta }{\cos \theta }\sqrt{1-\cos ^{2}\theta }}\\ &=\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi }\displaystyle \frac{d\theta }{\displaystyle \frac{\sin \theta \times \sin \theta }{\cos \theta }}\\ &=\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi }\displaystyle \frac{\cos \theta }{\sin ^{2}\theta }\: \: d\theta \\ &\begin{cases} u & =\sin \theta \\ du & =\cos \theta \: d\theta \end{cases}\\ &=\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi }\displaystyle \frac{1}{u^{2}}\: du\\ &=\displaystyle \int_{\frac{1}{6}\pi }^{ \frac{1}{2}\pi } u^{-2}\: du\\ &=\left [\displaystyle -u^{-1} \right ]_{\frac{1}{6}\pi }^{\frac{1}{2}\pi }\\ &=\left [ \displaystyle -\frac{1}{u} \right ]_{30^{0}}^{90^{0}}\\ &=\left [ \displaystyle -\frac{1}{\sin \theta } \right ]_{30^{0}}^{90^{0}}\\ &=\left [ -\displaystyle \frac{1}{\sin 90^{0}} \right ]-\left [ -\displaystyle \frac{1}{\sin 30^{0}} \right ]\\ &=\left ( -\displaystyle \frac{1}{1} \right )-\left ( -\displaystyle \frac{1}{\frac{1}{2}} \right )=-1+2=1 \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{\textbf{(Olimpiade Matematika ITS (OMITS) SMA 2012)} Berikut adalah hubungan integrasi yang benar, \textit{kecuali} ....} \end{array}\\\\ \begin{array}{ll}\\ .\quad &a.\quad \displaystyle \int \csc \theta \: d\theta =-\ln \left | \csc \theta +\cot \theta \right |+C\\ &b.\quad \displaystyle \int \csc \theta \: d\theta =-\ln \left | \csc \theta -\cot \theta \right |+C\\ &c.\quad \displaystyle \int \csc \theta \: d\theta =\ln \left | \csc \theta -\cot \theta \right |+C\\ &d.\quad \displaystyle \int \sec \theta \: d\theta =\ln \left | \sec \theta +\tan \theta \right |+C\\ &e.\quad \displaystyle \int \tan \theta \: d\theta =\ln \left | \sec \theta \right |+C\end{array}.

Jawab: b

\begin{array}{|l|l|}\hline \begin{aligned}\displaystyle \int \tan \theta \: d\theta &=\displaystyle \int \frac{\sin \theta }{\cos \theta }\: d\theta \\ &=-\displaystyle \int \frac{d\left ( \cos \theta \right )}{\cos \theta }\\ &=-\ln \left | \cos \theta \right |+C,&\textnormal{atau}\\ &=\ln \left | \cos \theta \right |^{-1}+C\\ &=\ln \left | \displaystyle \frac{1}{\cos \theta } \right |+C\\ &=\ln \left | \sec \theta \right |+C \end{aligned}&\begin{aligned}\displaystyle \int \sec \theta \: d\theta &=\displaystyle \int \sec \theta \times \left (\displaystyle \frac{\sec \theta +\tan \theta }{\sec \theta +\tan \theta} \right )\: d\theta\\ &=\displaystyle \int \frac{\sec ^{2}\theta +\sec \theta .\tan \theta }{\sec \theta +\tan \theta }\: d\theta \\ &=\displaystyle \int \frac{d\left ( \sec \theta +\tan \theta \right )}{\sec \theta +\tan \theta}\\ &=\ln \left | \sec \theta +\tan \theta \right |+C\\ &\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|}\hline \begin{aligned}\displaystyle \int \csc \theta \: d\theta&=\displaystyle \int \csc \theta \times \left ( \displaystyle \frac{\csc \theta -\cot \theta }{\csc \theta -\cot \theta} \right )\: d\theta \\ &=\displaystyle \int \frac{\csc ^{2}\theta -\csc \theta .\cot \theta }{\csc \theta -\cot \theta}\: d\theta \\ &=\displaystyle \int \frac{d\left ( \csc \theta -\cot \theta \right )}{\csc \theta -\cot \theta}\\ &=\ln \left | \csc \theta -\cot \theta \right |+C\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned} &\begin{aligned}\displaystyle \int \csc \theta \: d\theta &=\ln \left | \csc \theta -\cot \theta \right |+C\\ &=\ln \left | \displaystyle \frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta } \right |+C\\ &=\ln \left | \displaystyle \frac{1-\cos \theta }{\sin \theta } \right |+C\\ &=\ln \left | \displaystyle \frac{1-\cos \theta }{\sin \theta }\times \left ( \displaystyle \frac{1+\cos \theta }{1+\cos \theta } \right ) \right |+C\\ &=\ln \left | \displaystyle \frac{\sin \theta }{1+\cos \theta } \right |+C\\ &=\ln \left | \displaystyle \frac{1}{\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }} \right |+C\\ &=\ln \left | \displaystyle \frac{1}{\csc \theta +\cot \theta } \right |+C\\ &=\ln \left | \csc \theta +\cot \theta \right |^{-1}+C\\ &=-\ln \left | \csc \theta +\cot \theta \right |+C \end{aligned}\\\hline \end{array}.

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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