Soal-Soal Matematika/Integral

Kaidah umum

1. $\int af(x)\,dx=a\int f(x)\,dx\qquad {\mbox{(}}a{\mbox{ konstan)}}\,\!$
2. $\int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx$
3. $\int [f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx$
4. $\int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int \left[f'(x)\left(\int g(x)\,dx\right)\right]\,dx$
5. $\int [f(x)]^{n}f'(x)\,dx={[f(x)]^{n+1} \over n+1}+C\qquad {\mbox{(untuk }}n\neq -1{\mbox{)}}\,\!$
6. $\int {f'(x) \over f(x)}\,dx=\ln {\left|f(x)\right|}+C$
7. $\int {f'(x)f(x)}\,dx={1 \over 2}[f(x)]^{2}+C$
8. $\int f(x)\,dx=F(x)+C$
9. $\int _{a}^{b}f(x)\,dx=F(b)-F(a)$

rumus sederhana

$\int \,dx=x+C$

$\int x^{n}\,dx={\frac {x^{n+1}}{n+1}}+C\qquad {\mbox{ jika }}n\neq -1$

$\int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\mbox{ jika }}n\neq -1$

$\int {dx \over x}=\ln {\left|x\right|}+C$

$\int {dx \over {a^{2}+x^{2}}}={1 \over a}\arctan {x \over a}+C$

Eksponen dan logaritma

$\int e^{x}\,dx=e^{x}+C$

$\int a^{x}\,dx={\frac {a^{x}}{\ln {a}}}+C$

$\int \ln {x}\,dx=x\ln {x}-x+C$

$\int \,^{b}\!\log {x}\,dx=x\cdot \,^{b}\!\log x-x\cdot \,^{b}\!\log e+C$

Trigonometri

$\int \sin {x}\,dx=-\cos {x}+C$

$\int \cos {x}\,dx=\sin {x}+C$

$\int \tan {x}\,dx=\ln {\left|\sec {x}\right|}+C$

$\int \cot {x}\,dx=-\ln {\left|\csc {x}\right|}+C$

$\int \sec {x}\,dx=\ln {\left|\sec {x}+\tan {x}\right|}+C$

$\int \csc {x}\,dx=-\ln {\left|\csc {x}+\cot {x}\right|}+C$

Hiperbolik

$\int \sinh x\,dx=\cosh x+C$

$\int \cosh x\,dx=\sinh x+C$

$\int \tanh x\,dx=\ln |\cosh x|+C$

$\int \coth x\,dx=\ln |\sinh x|+C$

$\int {\mbox{sech}}\,x\,dx=\arctan(\sinh x)+C$

$\int {\mbox{csch}}\,x\,dx=\ln \left|\tanh {x \over 2}\right|+C$

jenis integral

integral biasa

Berikut contoh penyelesaian cara biasa.

$\int x^{3}-cos3x+e^{5x+2}-{\frac {1}{2x+5}}dx$

$={\frac {1}{3+1}}x^{3+1}-{\frac {sin3x}{3}}+{\frac {e^{5x+2}}{5}}-{\frac {ln(2x+5)}{2}}+C$

$={\frac {x^{4}}{4}}-{\frac {sin3x}{3}}+{\frac {e^{5x+2}}{5}}-{\frac {ln(2x+5)}{2}}+C$

integral substitusi

Berikut contoh penyelesaian cara substitusi.

$\int {\frac {\ln(x)}{x}}\,dx$

$t=\ln(x),\,dt={\frac {dx}{x}}$

Dengan menggunakan rumus di atas,

${\begin{aligned}&\;\int {\frac {\ln(x)}{x}}\,dx\\=&\;\int t\,dt\\=&\;{\frac {1}{2}}t^{2}+C\\=&\;{\frac {1}{2}}\ln ^{2}x+C\end{aligned}}$

integral parsial

Cara 1

Rumus

Integral parsial diselesaikan dengan rumus berikut.

$\int u\,dv=uv-\int v\,du$

Berikut contoh penyelesaian cara parsial dengan rumus.

$\int x\sin(x)\,dx$

$u=x,\,du=1\,dx,\,dv=\sin(x)\,dx,\,v=-\cos(x)$

Dengan menggunakan rumus di atas,

${\begin{aligned}&\;\int x\sin(x)\,dx\\=&\;(x)(-\cos(x))-\int (-\cos(x))(1\,dx)\\=&\;-x\cos(x)+\int \cos(x)\,dx\\=&\;-x\cos(x)+\sin(x)+C\end{aligned}}$

Cara 2

Tabel

Untuk ${\textstyle \int u\,dv}$, berlaku ketentuan sebagai berikut.

Tanda	Turunan	Integral
+	$u$	$dv$
-	${\frac {du}{dx}}$	$v$
+	${\frac {d^{2}u}{dx^{2}}}$	$\int v\,dx$

Berikut contoh penyelesaian cara parsial dengan tabel.

$\int x\sin(x)\,dx$

Tanda	Turunan	Integral
+	$x$	$\sin(x)$
-	$1$	$-\cos(x)$
+	$0$	$-\sin(x)$

Dengan tabel di atas,

${\begin{aligned}&\;\int x\sin(x)\,dx\\=&\;(x)(-\cos(x))-(1)(-\sin(x))+C\\=&\;-x\cos(x)+\sin(x)+C\end{aligned}}$

integral pecahan parsial

Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional).

$\int {\frac {dx}{x^{2}-4}}$

Pertama, pisahkan pecahan tersebut.

${\begin{aligned}=&\;{\frac {1}{x^{2}-4}}\\=&\;{\frac {1}{(x+2)(x-2)}}\\=&\;{\frac {A}{x+2}}+{\frac {B}{x-2}}\\=&\;{\frac {A(x-2)+B(x+2)}{x^{2}-4}}\\=&\;{\frac {(A+B)x-2(A-B)}{x^{2}-4}}\end{aligned}}$

Kita tahu bahwa ${\textstyle A+B=0}$ dan ${\textstyle A-B=-{\frac {1}{2}}}$ dapat diselesaikan, yaitu ${\textstyle A=-{\frac {1}{4}}}$ dan ${\textstyle B={\frac {1}{4}}}$.

${\begin{aligned}&\;\int {\frac {dx}{x^{2}-4}}\\=&\;\int -{\frac {1}{4(x+2)}}+{\frac {1}{4(x-2)}}\,dx\\=&\;{\frac {1}{4}}\int {\frac {1}{x-2}}-{\frac {1}{x+2}}\,dx\\=&\;{\frac {1}{4}}(\ln |x-2|-\ln |x+2|)+C\\=&\;{\frac {1}{4}}\ln |{\frac {x-2}{x+2}}|+C\end{aligned}}$

integral substitusi trigonometri

Bentuk	Trigonometri
${\sqrt {a^{2}-b^{2}x^{2}}}$	$x={\frac {a}{b}}\sin(\alpha )$
${\sqrt {a^{2}+b^{2}x^{2}}}$	$x={\frac {a}{b}}\tan(\alpha )$
${\sqrt {b^{2}x^{2}-a^{2}}}$	$x={\frac {a}{b}}\sec(\alpha )$

Berikut contoh penyelesaian cara substitusi trigonometri.

$\int {\frac {dx}{x^{2}{\sqrt {x^{2}+4}}}}$

$x=2\tan(A),\,dx=2\sec ^{2}(A)\,dA$

Dengan substitusi di atas,

${\begin{aligned}&\;\int {\frac {dx}{x^{2}{\sqrt {x^{2}+4}}}}\\=&\;\int {\frac {2\sec ^{2}(A)\,dA}{(2\tan(A))^{2}{\sqrt {4+(2\tan(A))^{2}}}}}\\=&\;\int {\frac {2\sec ^{2}(A)\,dA}{4\tan ^{2}(A){\sqrt {4+4\tan ^{2}(A)}}}}\\=&\;\int {\frac {2\sec ^{2}(A)\,dA}{4\tan ^{2}(A){\sqrt {4(1+\tan ^{2}(A))}}}}\\=&\;\int {\frac {2\sec ^{2}(A)\,dA}{4\tan ^{2}(A){\sqrt {4\sec ^{2}(A)}}}}\\=&\;\int {\frac {2\sec ^{2}(A)\,dA}{(4\tan ^{2}(A))(2\sec(A))}}\\=&\;\int {\frac {\sec(A)\,dA}{4\tan ^{2}(A)}}\\=&\;{\frac {1}{4}}\int {\frac {\sec(A)\,dA}{\tan ^{2}(A)}}\\=&\;{\frac {1}{4}}\int {\frac {\cos(A)}{\sin ^{2}(A)}}\,dA\end{aligned}}$

Substitusi berikut dapat dibuat.

$\int {\frac {\cos(A)}{\sin ^{2}(A)}}\,dA$

$t=\sin(A),\,dt=\cos(A)\,dA$

Dengan substitusi di atas,

${\begin{aligned}&\;{\frac {1}{4}}\int {\frac {\cos(A)}{\sin ^{2}(A)}}\,dA\\=&\;{\frac {1}{4}}\int {\frac {dt}{t^{2}}}\\=&\;{\frac {1}{4}}\left(-{\frac {1}{t}}\right)+C\\=&\;-{\frac {1}{4t}}+C\\=&\;-{\frac {1}{4\sin(A)}}+C\end{aligned}}$

Ingat bahwa ${\textstyle \sin(A)={\frac {x}{\sqrt {x^{2}+4}}}}$ berlaku.

${\begin{aligned}=&\;-{\frac {1}{4\sin(A)}}+C\\=&\;-{\frac {\sqrt {x^{2}+4}}{4x}}+C\end{aligned}}$

jenis integral lainnya

panjang busur

Sumbu x

$S=\int _{x_{1}}^{x_{2}}{\sqrt {1+(f'(x))^{2}}}\,dx$

Sumbu y

$S=\int _{y_{1}}^{y_{2}}{\sqrt {1+(f'(y))^{2}}}\,dy$

luas daerah

Satu kurva

Sumbu x

$L=\int _{x_{1}}^{x_{2}}f(x)\,dx$

Sumbu y

$L=\int _{y_{1}}^{y_{2}}f(y)\,dy$

Dua kurva

Sumbu x

$L=\int _{x_{1}}^{x_{2}}(f(x_{2})-f(x_{1}))\,dx$

Sumbu y

$L=\int _{y_{1}}^{y_{2}}(f(y_{2})-f(y_{1}))\,dy$

atau juga $L={\frac {D{\sqrt {D}}}{6a^{2}}}$

luas permukaan benda putar

Sumbu x sebagai poros

$L_{p}=2\pi \int _{x_{1}}^{x_{2}}f(x)\,ds$

dengan

$ds={\sqrt {1+(f'(x))^{2}}}\,dx$

Sumbu y sebagai poros

$L_{p}=2\pi \int _{y_{1}}^{y_{2}}f(y)\,ds$

dengan

$ds={\sqrt {1+(f'(y))^{2}}}\,dy$

volume benda putar

Satu kurva

Sumbu x sebagai poros

$V=\pi \int _{x_{1}}^{x_{2}}(f(x))^{2}\,dx$

Sumbu y sebagai poros

$V=\pi \int _{y_{1}}^{y_{2}}(f(y))^{2}\,dy$

Dua kurva

Sumbu x sebagai poros

$V=\pi \int _{x_{1}}^{x_{2}}((f(x_{2}))^{2}-(f(x_{1}))^{2})\,dx$

Sumbu y sebagai poros

$V=\pi \int _{y_{1}}^{y_{2}}((f(y_{2}))^{2}-(f(y_{1}))^{2})\,dy$

integral lipat

Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga.

contoh

• Tentukan luas (tak tentu) dengan persamaan garis $y=x$ dan batas-batas sumbu y dengan cara integral!

${\begin{aligned}L&=\int x\,dx\\L&={\frac {1}{2}}x^{2}\\L&={\frac {1}{2}}xy\quad (y=x)\end{aligned}}$

• Tentukan luas (tak tentu) dengan persamaan garis $y=x^{2}$ dan batas-batas sumbu y dengan cara integral!

${\begin{aligned}L&=\int x^{2}\,dx\\L&={\frac {1}{3}}x^{3}\\L&={\frac {1}{3}}xy\quad (y=x^{2})\end{aligned}}$

• Tentukan luas (tak tentu) dengan persamaan garis $y={\sqrt {x}}$ dan batas-batas sumbu y dengan cara integral!

${\begin{aligned}L&=\int {\sqrt {x}}\,dx\\L&={\frac {2}{3}}x^{\frac {3}{2}}\\L&={\frac {2}{3}}xy\quad (y={\sqrt {x}})\end{aligned}}$

• Buktikan luas persegi $L=s^{2}$ dengan cara integral!

Dengan posisi $y=s$ dan titik (s, s),

${\begin{aligned}L&=\int _{0}^{s}s\,dx\\L&=sx|_{0}^{s}\\L&=ss-0\\L&=s^{2}\end{aligned}}$

• Buktikan luas persegi panjang $L=pl$ dengan cara integral!

Dengan posisi $y=l$ dan titik (p, l),

${\begin{aligned}L&=\int _{0}^{p}l\,dx\\L&=lx|_{0}^{p}\\L&=pl-0\\L&=pl\end{aligned}}$

• Buktikan luas segitiga $L={\frac {at}{2}}$ dengan cara integral!

Dengan posisi $y={\frac {-tx}{a}}+t$ dan titik (a, t),

${\begin{aligned}L&=\int _{0}^{a}\left({\frac {-tx}{a}}+t\right)\,dx\\L&=\left.{\frac {-tx^{2}}{2a}}+tx\right|_{0}^{a}\\L&={\frac {-ta^{2}}{2a}}+ta-0+0\\L&={\frac {-ta}{2}}+ta\\L&={\frac {at}{2}}\end{aligned}}$

• Buktikan volume tabung $V=\pi r^{2}t$ dengan cara integral!

Dengan posisi $y=r$ dan titik (t, r),

${\begin{aligned}V&=\pi \int _{0}^{t}r^{2}\,dx\\V&=\pi \left.r^{2}x\right|_{0}^{t}\\V&=\pi r^{2}t-0\\V&=\pi r^{2}t\end{aligned}}$

• Buktikan volume kerucut $V={\frac {\pi r^{2}t}{3}}$ dengan cara integral!

Dengan posisi $y={\frac {rx}{t}}$ dan titik (t, r),

${\begin{aligned}V&=\pi \int _{0}^{t}\left({\frac {rx}{t}}\right)^{2}\,dx\\V&=\pi \left.{\frac {r^{2}x^{3}}{3t^{2}}}\right|_{0}^{t}\\V&=\pi {\frac {r^{2}t^{3}}{3t^{2}}}-0\\V&={\frac {\pi r^{2}t}{3}}\end{aligned}}$

• Buktikan volume bola $V={\frac {4\pi r^{3}}{3}}$ dengan cara integral!

Dengan posisi $y={\sqrt {r^{2}-x^{2}}}$ serta titik (-r, 0) dan (r, 0),

${\begin{aligned}V&=\pi \int _{-r}^{r}\left({\sqrt {r^{2}-x^{2}}}\right)^{2}\,dx\\V&=\pi \int _{-r}^{r}r^{2}-x^{2}dx\\V&=\pi \left.r^{2}x-{\frac {x^{3}}{3}}\right|_{-r}^{r}\\V&=\pi \left(r^{3}-{\frac {r^{3}}{3}}-\left(-r^{3}+{\frac {r^{3}}{3}}\right)\right)\\V&={\frac {4\pi r^{3}}{3}}\end{aligned}}$

• Buktikan luas permukaan bola $L=4\pi r^{2}$ dengan cara integral!

Dengan posisi $y={\sqrt {r^{2}-x^{2}}}$ serta titik (-r, 0) dan (r, 0),

Kita tahu bahwa turunannya adalah

$y'={\frac {-x}{\sqrt {r^{2}-x^{2}}}}$

selanjutnya

${\begin{aligned}ds&={\sqrt {1+({\frac {-x}{\sqrt {r^{2}-x^{2}}}})^{2}}}\,dx\\ds&={\sqrt {1+{\frac {x^{2}}{r^{2}-x^{2}}}}}\,dx\\ds&={\sqrt {\frac {r^{2}}{r^{2}-x^{2}}}}\,dx\\ds&={\frac {r}{\sqrt {r^{2}-x^{2}}}}\,dx\\\end{aligned}}$

sehingga

${\begin{aligned}L&=2\pi \int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\cdot {\frac {r}{\sqrt {r^{2}-x^{2}}}}\,dx\\L&=2\pi \int _{-r}^{r}r\,dx\\L&=2\pi rx_{-r}^{r}\\L&=2\pi (rr-r(-r))\\L&=2\pi (r^{2}+r^{2})\\L&=4\pi r^{2}\end{aligned}}$

• Buktikan keliling lingkaran $K=2\pi r$ dengan cara integral!

Dengan posisi $y={\sqrt {r^{2}-x^{2}}}$ serta titik (-r, 0) dan (r, 0),

Kita tahu bahwa turunannya adalah

$y'={\frac {-x}{\sqrt {r^{2}-x^{2}}}}$

sehingga

${\begin{aligned}K&=4\int _{0}^{r}{\sqrt {1+({\frac {-x}{\sqrt {r^{2}-x^{2}}}})^{2}}}\,dx\\K&=4\int _{0}^{r}{\sqrt {1+({\frac {x^{2}}{r^{2}-x^{2}}})}}\,dx\\K&=4\int _{0}^{r}{\sqrt {\frac {r^{2}}{r^{2}-x^{2}}}}\,dx\\K&=4r\int _{0}^{r}{\sqrt {\frac {1}{r^{2}-x^{2}}}}\,dx\\K&=4r\int _{0}^{r}{\frac {1}{\sqrt {r^{2}-x^{2}}}}\,dx\\K&=4r\;\left.\arcsin \left({\frac {x}{r}}\right)\right|_{0}^{r}\\K&=4r\left(\arcsin \left({\frac {r}{r}}\right)-\arcsin \left({\frac {0}{r}}\right)\right)\\K&=4r\left(\arcsin \left(1\right)-\arcsin \left(0\right)\right))\\K&=4r\left({\frac {\pi }{2}}\right)\\K&=2\pi r\end{aligned}}$

• Buktikan luas lingkaran $L=\pi r^{2}$ dengan cara integral!

Dengan posisi $y={\sqrt {r^{2}-x^{2}}}$ serta titik (-r, 0) dan (r, 0), dibuat trigonometri dan turunannya terlebih dahulu.

${\begin{aligned}\sin(\theta )&={\frac {x}{r}}\\x&=r\sin(\theta )\\dx&=r\cos(\theta )\,d\theta \end{aligned}}$

Dengan turunan di atas,

${\begin{aligned}L&=4\int _{0}^{r}{\sqrt {r^{2}-x^{2}}}\,dx\\L&=4\int _{0}^{r}{\sqrt {r^{2}-(r\sin(\theta ))^{2}}}\,dx\\L&=4\int _{0}^{r}{\sqrt {r^{2}-r^{2}\sin ^{2}(\theta )}}\,dx\\L&=4\int _{0}^{r}{\sqrt {r^{2}(1-\sin ^{2}(\theta ))}}\,dx\\L&=4\int _{0}^{r}r\cos(\theta )\,dx\\L&=4\int _{0}^{r}r\cos(\theta )(r\cos(\theta )\,d\theta )\\L&=4\int _{0}^{r}r^{2}\cos ^{2}(\theta )\,d\theta \\L&=4r^{2}\int _{0}^{r}\left({\frac {1+\cos(2\theta )}{2}}\right)\,d\theta \\L&=2r^{2}\int _{0}^{r}(1+\cos(2\theta ))\,d\theta \\L&=2r^{2}\left(\theta +{\frac {1}{2}}\sin(2\theta )\right)_{0}^{r}\\L&=2r^{2}(\theta +\sin(\theta )\cos(\theta ))_{0}^{r}\\L&=2r^{2}\left(\arcsin \left({\frac {x}{r}}\right)+\left({\frac {x}{r}}\right)\left({\frac {r^{2}-x^{2}}{r}}\right)\right)_{0}^{r}\\L&=2r^{2}\left(\arcsin \left({\frac {r}{r}}\right)+\left({\frac {r}{r}}\right)\left({\frac {r^{2}-r^{2}}{r}}\right)\right)-\left(\arcsin \left({\frac {0}{r}}\right)+\left({\frac {0}{r}}\right)\left({\frac {r^{2}-0^{2}}{r}}\right)\right)\\L&=2r^{2}(\arcsin(1)+0-(\arcsin(0)+0))\\L&=2r^{2}\left({\frac {\pi }{2}}\right)\\L&=\pi r^{2}\end{aligned}}$

• Buktikan luas elips $L=\pi ab$ dengan cara integral!

Dengan posisi $y={\frac {b{\sqrt {a^{2}-x^{2}}}}{a}}$ serta (-a, 0) dan (a, 0),

${\begin{aligned}L&=4\int _{0}^{r}{\frac {b{\sqrt {a^{2}-x^{2}}}}{a}}\,dx\\L&={\frac {4b}{a}}\int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx\end{aligned}}$

Dengan anggapan bahwa lingkaran mempunyai memotong titik (-a, 0) dan (a, 0) serta pusatnya setitik dengan pusat elips,

${\begin{aligned}{\frac {L_{\text{elips}}}{L_{\text{ling}}}}&={\frac {{\frac {4b}{a}}\int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx}{4\int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx}}\\{\frac {L_{\text{elips}}}{L_{\text{ling}}}}&={\frac {b}{a}}\\L_{\text{elips}}&={\frac {b}{a}}L_{\text{ling}}\\L_{\text{elips}}&={\frac {b}{a}}\pi a^{2}\\L_{\text{elips}}&=\pi ab\end{aligned}}$