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

$\int \,dx=x+C$

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

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

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

$\int {dx \over {\sqrt {a^{2}-b^{2}x^{2}}}}={\frac {1}{b}}\arcsin \,{bx \over a}+C$

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

$\int {dx \over x{\sqrt {b^{2}x^{2}-a^{2}}}}={1 \over a}\operatorname {arcsec} \,{bx \over a}+C$

Eksponen dan logaritma

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

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

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

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

Trigonometri

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

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

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

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

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

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

Hiperbolik

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

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

$\int \tanh \,x\,dx=\ln {\left|\cosh \,x\right|}+C$

$\int \coth \,x\,dx=\ln {\left|\sinh \,x\right|}+C$

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

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

Jenis integral

integral biasa

Berikut contoh penyelesaian cara biasa.

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

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

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

2. $\int x(x-5)^{4}\,dx$

$=\int x(x^{4}-4x^{3}(5)+6x^{2}(25)-4x(125)+625)\,dx$

$=\int x(x^{4}-20x^{3}+150x^{2}-500x+625)\,dx$

$=\int x^{5}-20x^{4}+150x^{3}-500x^{2}+625x\,dx$

$={\frac {x^{6}}{6}}-4x^{5}+{\frac {150x^{4}}{4}}-{\frac {500x^{3}}{3}}+{\frac {625x^{2}}{2}}+C$

$={\frac {x^{6}}{6}}-4x^{5}+{\frac {75x^{4}}{2}}-{\frac {500x^{3}}{3}}+{\frac {625x^{2}}{2}}+C$

integral substitusi

Berikut contoh penyelesaian cara substitusi.

1. $\int {\frac {\ln \,x}{x}}\,dx$

$u=\ln \,x\,du={\frac {dx}{x}}$

Dengan menggunakan rumus di atas,

Jawaban

$${\begin{aligned}\int {\frac {\ln(x)}{x}}dx\\\int udu\\{\frac {1}{2}}u^{2}+C\\{\frac {1}{2}}ln^{2}x+C\\\end{aligned}}$$

2. $\int x(x-5)^{4}\,dx$

$u=x-5,\,du=dx,\,x=u+5$

Dengan menggunakan rumus di atas,

Jawaban

$${\begin{aligned}\int x(x-5)^{4}\,dx\\\int (u+5)u^{4}\,du\\\int (u^{5}+5u^{4})\,du\\{\frac {1}{6}}u^{6}+u^{5}+C\\{\frac {1}{6}}(x-5)^{6}+(x-5)^{5}+C\\{\frac {1}{6}}(x^{6}-6x^{5}(5)+15x^{4}(25)-20x^{3}(125)+15x^{2}(625)-6x(3125)+15625)+(x^{5}-5x^{4}(5)+10x^{3}(25)-10x^{2}(125)+5x(625)+3125)+C\\{\frac {1}{6}}x^{6}-5x^{5}+{\frac {125x^{4}}{2}}-{\frac {1250x^{3}}{3}}+{\frac {3125x^{2}}{2}}-3125x+{\frac {15625}{6}}+x^{5}-25x^{4}+250x^{3}-1250x^{2}+3125x+3125+C\\{\frac {1}{6}}x^{6}-4x^{5}+{\frac {75x^{4}}{2}}-{\frac {500x^{3}}{3}}+{\frac {625x^{2}}{2}}+C\\\end{aligned}}$$

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.

1. $\int x\sin \,x\,dx$

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

Dengan menggunakan rumus di atas,

Jawaban

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

2. $\int x(x-5)^{4}\,dx$

$u=x,\,du=1\,dx,\,dv=(x-5)^{4}\,dx,\,v={\frac {1}{5}}(x-5)^{5}$

Dengan menggunakan rumus di atas,

Jawaban

$${\begin{aligned}\int x(x-5)^{4}\,dx\\(x)({\frac {1}{5}}(x-5)^{5})-\int ({\frac {1}{5}}(x-5)^{5})(1\,dx)\\{\frac {x}{5}}(x-5)^{5}-\int {\frac {1}{5}}(x-5)^{5}\,dx\\{\frac {x}{5}}(x-5)^{5}-{\frac {1}{30}}(x-5)^{6}\,dx\\\end{aligned}}$$

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.

1. $\int x\sin \,x\,dx$

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

Dengan tabel di atas,

Jawaban

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

2. $\int x(x-5)^{4}\,dx$

Tanda	Turunan	Integral
+	$x$	$(x-5)^{4}$
-	$1$	${\frac {1}{5}}(x-5)^{5}$
+	$0$	${\frac {1}{30}}(x-5)^{6}$

Dengan tabel di atas,

Jawaban

$${\begin{aligned}\int x(x-5)^{4}\,dx\\(x)({\frac {1}{5}}(x-5)^{5})-(1)({\frac {1}{30}}(x-5)^{6})+C\\{\frac {x}{5}}(x-5)^{5}-{\frac {1}{30}}(x-5)^{6}+C\end{aligned}}$$

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	Substitusi 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.

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

integral mutlak

$\int |f(x)|\,dA$

buatlah $f(x)\geq 0$ jika hasil x adalah lebih dari 0 maka f(x) sedangkan kurang dari 0 maka -f(x)

$\int _{a}^{c}|f(x)|\,dx=\int _{a}^{b}f(x)\,dx+\int _{b}^{c}-f(x)\,dx=F(b)-F(a)-F(c)+F(b)$ jika $|f(x)|={\begin{cases}f(x),&{\mbox{jika }}f(x)\geq b,\\-f(x),&{\mbox{jika }}f(x)<b.\end{cases}}$. sebelum membuat f(x) dan -f(x) menentukan nilai x sebagai f(x) ≥ 0 untuk batas-batas wilayah yang menempatkan hasil positif (f(x)) dan negatif (-(f(x)) serta hasil integral mutlak tidak mungkin negatif.

integral fungsi ganjil dan integral fungsi genap

$\int _{-a}^{a}f(x)\,dx$

dengan mengingat fungsi ganjil dan fungsi genap yaitu f(-x)=-f(x) untuk fungsi ganjil dan f(-x)=f(x) untuk fungsi genap maka berlaku untuk integral:

integral fungsi ganjil

$\int _{-a}^{a}f(x)\,dx$ = 0

integral fungsi genap

$\int _{-a}^{a}f(x)\,dx$ = $2\int _{0}^{a}f(x)\,dx$

integral notasi

$\int _{a}^{b}f(x)\,dx=\int _{a-p}^{b-p}f(x)\,dx=\int _{a+p}^{b+p}f(x)\,dx$

$\int _{a}^{c}f(x)\,dx=\int _{a}^{b}f(x)\,dx+\int _{b}^{c}f(x)\,dx$

integral terbalik

$\int _{a}^{b}f(x)\,dx=-\int _{b}^{a}f(x)\,dx$

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

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

atau juga $V=|{\frac {D^{2}{\sqrt {D}}}{30a^{3}}}|$

Integral lipat

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

contoh

1. Tentukan hasil dari:

• $\int (5x-1)(5x^{2}-2x+7)^{4}dx$
• $\int x^{2}cos3xdx$
• $\int sin^{2}xcosxdx$
• $\int cos^{2}xsinxdx$
• $\int secxdx$
• $\int cscxdx$
• $\int {\frac {1}{2x^{2}-5x+3}}dx$
• $\int {\frac {1}{\sqrt {50-288x^{2}}}}dx$
• $\int {\frac {1}{12+75x^{2}}}dx$
• $\int {\frac {1}{11x{\sqrt {605x^{2}-245}}}}dx$
• $\int (1+sin^{2}x+sin^{4}x+sin^{6}x+\dots )dx$
• $\int (1+cos^{2}x+cos^{4}x+cos^{6}x+\dots )dx$
• $\int (sinx+sin^{3}x+sin^{5}x+sin^{7}x+\dots )dx$
• $\int (cosx+cos^{3}x+cos^{5}x+cos^{7}x+\dots )dx$
• $\int {\sqrt {\frac {x}{\sqrt {\frac {x}{\sqrt {\frac {x}{\dots }}}}}}}dx$

jawaban

Jawaban

$${\begin{aligned}{\text{misalkan }}u&=5x^{2}-2x+7\\u&=5x^{2}-2x+7\\du&=10x-2dx\\&=2(5x-1)dx\\{\frac {du}{2}}&=(5x-1)dx\\\int (5x-1)(5x^{2}-2x+7)^{4}dx&=\int {\frac {1}{2}}u^{4}du\\&={\frac {1}{2}}{\frac {1}{5}}u^{5}+C\\&={\frac {1}{10}}(5x^{2}-2x+7)^{5}+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}{\text{gunakan integral parsial }}\\\int x^{2}cos3xdx&={\frac {x^{2}}{3}}sin3x-{\frac {2x}{9}}(-cos3x)+{\frac {2}{27}}(-sin3x)+C\\&={\frac {x^{2}}{3}}sin3x+{\frac {2x}{9}}cos3x-{\frac {2}{27}}sin3x+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}*{\text{cara 1}}\\{\text{misalkan }}u&=sinx\\u&=sinx\\du&=cosxdx\\\int sin^{2}xcosxdx&=\int u^{2}du\\&={\frac {u^{3}}{3}}+C\\&={\frac {sin^{3}x}{3}}+C\\*{\text{cara 2}}\\\int sin^{2}xcosxdx&=\int {\frac {1-cos2x}{2}}cosxdx\\&={\frac {1}{2}}\int (1-cos2x)cosxdx\\&={\frac {1}{2}}\int cosx-cos2xcosxdx\\&={\frac {1}{2}}\int cosx-{\frac {1}{2}}(cos3x+coxx)dx\\&={\frac {1}{2}}\int cosx-{\frac {1}{2}}cos3x-{\frac {1}{2}}cosxdx\\&={\frac {1}{2}}\int {\frac {1}{2}}cosx-{\frac {1}{2}}cos3xdx\\&={\frac {1}{2}}({\frac {1}{2}}sinx-{\frac {1}{6}}sin3x)+C\\&={\frac {sinx}{4}}-{\frac {sin3x}{12}}+C\\*{\text{cara 3}}\\\int sin^{2}xcosxdx&=\int sinxsinxcosxdx\\&={\frac {1}{2}}\int sinxsin2xdx\\&={\frac {1}{4}}\int -(cos3x-cos(-x))dx\\&={\frac {1}{4}}\int -cos3x+cosxdx\\&={\frac {1}{4}}(-{\frac {sin3x}{3}}+sinx)+C\\&=-{\frac {sin3x}{12}}+{\frac {sinx}{4}}+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}*{\text{cara 1}}\\{\text{misalkan }}u&=cosx\\u&=cosx\\du&=-sinxdx\\-du&=sinxdx\\\int cos^{2}xsinxdx&=-\int u^{2}du\\&=-{\frac {u^{3}}{3}}+C\\&=-{\frac {cos^{3}x}{3}}+C\\*{\text{cara 2}}\\\int cos^{2}xsinxdx&=\int {\frac {cos2x+1}{2}}sinxdx\\&={\frac {1}{2}}\int (cos2x+1)sinxdx\\&={\frac {1}{2}}\int cos2xsinx+sinxdx\\&={\frac {1}{2}}\int {\frac {1}{2}}(sin3x-sinx)+sinxdx\\&={\frac {1}{2}}\int {\frac {1}{2}}sin3x-{\frac {1}{2}}sinx+sinxdx\\&={\frac {1}{2}}\int {\frac {1}{2}}sin3x+{\frac {1}{2}}sinxdx\\&={\frac {1}{2}}(-{\frac {1}{6}}cos3x-{\frac {1}{2}}cosx)+C\\&=-{\frac {cos3x}{12}}-{\frac {cosx}{4}}+C\\*{\text{cara 3}}\\\int cos^{2}xsinxdx&=\int cosxcosxsinxdx\\&={\frac {1}{2}}\int cosxsin2xdx\\&={\frac {1}{4}}\int (sin3x-sin(-x))dx\\&={\frac {1}{4}}\int sin3x+sinxdx\\&={\frac {1}{4}}(-{\frac {cos3x}{3}}-cosx)+C\\&=-{\frac {cos3x}{12}}-{\frac {cosx}{4}}+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int secxdx&=\int secx{\frac {secx+tanx}{secx+tanx}}dx\\&=\int {\frac {sec^{2}x+secxtanx}{secx+tanx}}dx\\{\text{misalkan }}u&=secx+tanx\\u&=secx+tanx\\du&=secxtanx+sec^{2}xdx\\&=\int {\frac {sec^{2}x+secxtanx}{secx+tanx}}dx\\&=\int {\frac {1}{u}}du\\&=lnu+C\\&=ln|secx+tanx|+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int cscxdx&=\int cscx{\frac {cscx+cotx}{cscx+cotx}}dx\\&=\int {\frac {csc^{2}x+cscxcotx}{cscx+cotx}}dx\\{\text{misalkan }}u&=cscx+cotx\\u&=cscx+cotx\\du&=-cscxcotx-csc^{2}xdx\\-du&=cscxcotx+csc^{2}xdx\\&=\int {\frac {csc^{2}x+cscxcotx}{cscx+cotx}}dx\\&=-\int {\frac {1}{u}}du\\&=-lnu+C\\&=-ln|cscx+cotx|+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int {\frac {1}{2x^{2}-5x+3}}dx&=\int {\frac {1}{(2x-3)(x-1)}}dx\\&=\int ({\frac {A}{(2x-3)}}+{\frac {B}{(x-1)}})dx\\&=\int ({\frac {A(x-1)+B(2x-3)}{(2x-3)(x-1)}})dx\\&=\int ({\frac {Ax-A+2Bx-3B}{2x^{2}-5x+3}})dx\\&=\int ({\frac {(A+2B)x+(-A-3B)}{2x^{2}-5x+3}})dx\\{\text{cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1}}\\&=\int ({\frac {A}{(2x-3)}}+{\frac {B}{(x-1)}})dx\\&=\int ({\frac {2}{(2x-3)}}+{\frac {-1}{(x-1)}})dx\\&=ln|2x-3|-ln|x-1|+C\\&=ln|{\frac {2x-3}{x-1}}|+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int {\frac {1}{\sqrt {50-288x^{2}}}}dx\\x&={\frac {5}{12}}sinA\\dx&={\frac {5}{12}}cosAdA\\\int {\frac {1}{\sqrt {50-288x^{2}}}}dx&=\int {\frac {{\frac {5}{12}}cosA}{\sqrt {50-288({\frac {5}{12}}sinA)^{2}}}}dA\\&=\int {\frac {{\frac {5}{12}}cosA}{\sqrt {50-{\frac {288\cdot 25}{144}}sin^{2}A}}}dA\\&=\int {\frac {{\frac {5}{12}}cosA}{\sqrt {50-50sin^{2}A}}}dA\\&=\int {\frac {{\frac {5}{12}}cosA}{\sqrt {50(1-sin^{2}A)}}}dA\\&=\int {\frac {{\frac {5}{12}}cosA}{\sqrt {50cos^{2}A}}}dA\\&=\int {\frac {{\frac {5}{12}}cosA}{5cosA{\sqrt {2}}}}dA\\&=\int {\frac {\sqrt {2}}{24}}dA\\&={\frac {\sqrt {2}}{24}}\int dA\\&=A+C\\A&=arcsin{\frac {12x}{5}}\\&={\frac {\sqrt {2}}{24}}arcsin{\frac {12x}{5}}+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int {\frac {1}{12+75x^{2}}}dx\\x&={\frac {2}{5}}tanA\\dx&={\frac {2}{5}}sec^{2}AdA\\\int {\frac {1}{12+75x^{2}}}dx&=\int {\frac {{\frac {2}{5}}sec^{2}A}{12+75({\frac {2}{5}}tanA)^{2}}}dA\\&=\int {\frac {{\frac {2}{5}}sec^{2}A}{12+{\frac {75\cdot 4}{25}}tan^{2}A}}dA\\&=\int {\frac {{\frac {2}{5}}sec^{2}A}{12+12tan^{2}A}}dA\\&=\int {\frac {{\frac {2}{5}}sec^{2}A}{12(1+tan^{2}A)}}dA\\&=\int {\frac {{\frac {2}{5}}sec^{2}A}{12sec^{2}A}}dA\\&=\int {\frac {1}{30}}dA\\&={\frac {A}{30}}+C\\A&=arctan{\frac {5x}{2}}\\&={\frac {arctan{\frac {5x}{2}}}{30}}+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int {\frac {1}{11x{\sqrt {605x^{2}-245}}}}dx\\x&={\frac {7}{11}}secA\\dx&={\frac {7}{11}}secAtanAdA\\\int {\frac {1}{11x{\sqrt {605x^{2}-245}}}}dx&=\int {\frac {{\frac {7}{11}}secAtanA}{11({\frac {7}{11}}secA){\sqrt {605({\frac {7}{11}}secA)^{2}-245}}}}dA\\&=\int {\frac {secAtanA}{11secA{\sqrt {{\frac {605\cdot 49}{121}}sec^{2}A-245}}}}dA\\&=\int {\frac {secAtanA}{11secA{\sqrt {245sec^{2}A-245}}}}dA\\&=\int {\frac {secAtanA}{11secA{\sqrt {245(sec^{2}A-1)}}}}dA\\&=\int {\frac {secAtanA}{11secA{\sqrt {245tan^{2}A}}}}dA\\&=\int {\frac {secAtanA}{77{\sqrt {5}}secAtanA}}dA\\&=\int {\frac {\sqrt {5}}{385}}dA\\&={\frac {A{\sqrt {5}}}{385}}+C\\A&=arcsec{\frac {11x}{7}}\\&={\frac {{\sqrt {5}}arcsec{\frac {11x}{7}}}{385}}+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int (1+sin^{2}x+sin^{4}x+sin^{6}x+\dots )dx\\a=1,r=sin^{2}x\\S_{\infty }&={\frac {a}{1-r}}\\&={\frac {1}{1-sin^{2}x}}\\&={\frac {1}{cos^{2}x}}\\&=sec^{2}x\\\int sec^{2}xdx&=tanx+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int (1+cos^{2}x+cos^{4}x+cos^{6}x+\dots )dx\\a=1,r=cos^{2}x\\S_{\infty }&={\frac {a}{1-r}}\\&={\frac {1}{1-cos^{2}x}}\\&={\frac {1}{sin^{2}x}}\\&=csc^{2}x\\\int csc^{2}xdx&=-cotx+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int (sinx+sin^{3}x+sin^{5}x+sin^{7}x+\dots )dx\\a=sinx,r=sin^{2}x\\S_{\infty }&={\frac {a}{1-r}}\\&={\frac {sinx}{1-sin^{2}x}}\\&={\frac {sinx}{cos^{2}x}}\\&=secx\cdot tanx\\\int secx\cdot tanxdx&=secx+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int (cosx+cos^{3}x+cos^{5}x+cos^{7}x+\dots )dx\\a=cosx,r=cos^{2}x\\S_{\infty }&={\frac {a}{1-r}}\\&={\frac {cosx}{1-cos^{2}x}}\\&={\frac {cosx}{sin^{2}x}}\\&=cscx\cdot cotx\\\int cscx\cdot cotxdx&=-cscx+C\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int {\sqrt {\frac {x}{\sqrt {\frac {x}{\sqrt {\frac {x}{\dots }}}}}}}dx\\{\text{misalkan }}y={\sqrt {\frac {x}{\sqrt {\frac {x}{\sqrt {\frac {x}{\dots }}}}}}}\\y&={\sqrt {\frac {x}{\sqrt {\frac {x}{\sqrt {\frac {x}{\dots }}}}}}}\\y^{2}&={\frac {x}{\sqrt {\frac {x}{\sqrt {\frac {x}{\dots }}}}}}\\&={\frac {x}{y}}\\y^{3}&=x\\3y^{2}dy&=dx\\\int y\cdot 3y^{2}dy\\\int 3y^{3}dy\\{\frac {3y^{4}}{4}}+C\\\end{aligned}}$$

1. Tentukan hasil dari:

• $\int _{1}^{5}2x-3dx$
• $\int _{0}^{4}|x-2|dx$
• $\int _{2}^{4}|3-x|dx$
• $\int _{1}^{5}x^{2}-6x+8dx$
• $\int _{-3}^{5}|x^{2}-2x-8|dx$

jawaban

Jawaban

$${\begin{aligned}\int _{1}^{5}2x-3dx&=\left.x^{2}-3x\right|_{1}^{5}\\&=10-(-2)\\&=12\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int _{0}^{4}|x-2|dx\\{\text{ tentukan nilai harga nol }}\\x-2&=0\\x&=2\\{\text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti }}\\|x-2|={\begin{cases}x-2,&{\mbox{jika }}x\geq 2,\\-(x-2),&{\mbox{jika }}x<2.\end{cases}}\\\int _{0}^{4}|x-2|dx&=\int _{2}^{4}x-2dx+\int _{0}^{2}-(x-2)dx\\&=\left.({\frac {x^{2}}{2}}-2x)\right|_{2}^{4}+\left.({\frac {-x^{2}}{2}}+2x)\right|_{0}^{2}\\&=0-(-2)+2-0\\&=4\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int _{2}^{4}|3-x|dx\\{\text{ tentukan nilai syarat-syarat }}\\3-x&\geq 0\\{\text{ harga nol }}\\x&=3\\{\text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti }}\\|3-x|={\begin{cases}3-x,&{\mbox{jika }}x\leq 3,\\-(3-x),&{\mbox{jika }}x>3.\end{cases}}\\\int _{2}^{4}|3-x|dx&=\int _{2}^{3}3-xdx+\int _{3}^{4}-(3-x)dx\\&=\left.(3x-{\frac {x^{2}}{2}})\right|_{2}^{3}+\left.(-3x+{\frac {x^{2}}{2}})\right|_{3}^{4}\\&=4.5-4-4-(-4.5)\\&=1\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int _{1}^{5}x^{2}-6x+8dx&=\left.{\frac {x^{3}}{3}}-3x^{2}+8x\right|_{1}^{5}\\&={\frac {20}{3}}-{\frac {16}{3}}\\&={\frac {4}{3}}\\&=1{\frac {1}{3}}\\\end{aligned}}$$

Jawaban

$${\begin{aligned}\int _{-3}^{5}|x^{2}-2x-8|dx\\{\text{ tentukan nilai syarat-syarat }}\\x^{2}-2x-8&\geq 0\\{\text{ harga nol }}\\(x+2)(x-4)&=0\\x=-2{\text{ atau }}x=4\\{\text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti }}\\|x^{2}-2x-8|={\begin{cases}x^{2}-2x-8,&{\mbox{jika }}x\leq -2,\\-(x^{2}-2x-8),&{\mbox{jika }}-2<x<4,\\x^{2}-2x-8,&{\mbox{jika }}x\geq 4\end{cases}}\\\int _{-3}^{5}|x^{2}-2x-8|dx&=\int _{-3}^{-2}x^{2}-2x-8dx+\int _{-2}^{4}-(x^{2}-2x-8)dx+\int _{4}^{5}x^{2}-2x-8dx\\&=\left.({\frac {x^{3}}{3}}-x^{2}-8x)\right|_{-3}^{-2}+\left.({\frac {-x^{3}}{3}}+x^{2}+8x)\right|_{-2}^{4}+\left.({\frac {x^{3}}{3}}-x^{2}-8x)\right|_{4}^{5}\\&={\frac {28}{3}}-6+{\frac {80}{3}}-(-{\frac {28}{3}})-{\frac {70}{3}}-(-{\frac {80}{3}})\\&={\frac {128}{3}}\\&=42{\frac {2}{3}}\\\end{aligned}}$$

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

Jawaban

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

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

Jawaban

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

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

Jawaban

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

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

Jawaban

$${\begin{aligned}{\text{dengan posisi }}y=s{\text{ dan titik (''s'', ''s''), }}\\L&=\int _{0}^{s}s\,dx\\L&=sx|_{0}^{s}\\L&=ss-0\\L&=s^{2}\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{Dengan posisi }}y=l{\text{dan titik (''p'', ''l''), }}\\L&=\int _{0}^{p}l\,dx\\L&=lx|_{0}^{p}\\L&=pl-0\\L&=pl\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{dengan posisi }}y={\frac {-tx}{a}}+t{\text{ dan titik (''a'', ''t''), }}\\L&=\int _{0}^{a}\left({\frac {-tx}{a}}+t\right)\,dx\\L&=\left.{\frac {-tx^{2}}{2a}}+tx\right|_{0}^{a}\\L&={\frac {-ta^{2}}{2a}}+ta-0+0\\L&={\frac {-ta}{2}}+ta\\L&={\frac {at}{2}}\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{ Dengan posisi }}y=r{\text{ dan titik (''t'', ''r''), }}\\V&=\pi \int _{0}^{t}r^{2}\,dx\\V&=\pi \left.r^{2}x\right|_{0}^{t}\\V&=\pi r^{2}t-0\\V&=\pi r^{2}t\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{dengan posisi }}y={\frac {rx}{t}}{\text{ dan titik (''t'', ''r''), }}\\V&=\pi \int _{0}^{t}\left({\frac {rx}{t}}\right)^{2}\,dx\\V&=\pi \left.{\frac {r^{2}x^{3}}{3t^{2}}}\right|_{0}^{t}\\V&=\pi {\frac {r^{2}t^{3}}{3t^{2}}}-0\\V&={\frac {\pi r^{2}t}{3}}\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{dengan posisi }}y={\sqrt {r^{2}-x^{2}}}{\text{ serta titik (-''r'', 0) dan (''r'', 0), }}\\V&=\pi \int _{-r}^{r}\left({\sqrt {r^{2}-x^{2}}}\right)^{2}\,dx\\V&=\pi \int _{-r}^{r}r^{2}-x^{2}dx\\V&=\pi \left.r^{2}x-{\frac {x^{3}}{3}}\right|_{-r}^{r}\\V&=\pi \left(r^{3}-{\frac {r^{3}}{3}}-\left(-r^{3}+{\frac {r^{3}}{3}}\right)\right)\\V&={\frac {4\pi r^{3}}{3}}\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{dengan posisi }}y={\sqrt {r^{2}-x^{2}}}{\text{ serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah }}\\y'&={\frac {-x}{\sqrt {r^{2}-x^{2}}}}\\{\text{selanjutnya }}\\ds&={\sqrt {1+({\frac {-x}{\sqrt {r^{2}-x^{2}}}})^{2}}}\,dx\\ds&={\sqrt {1+{\frac {x^{2}}{r^{2}-x^{2}}}}}\,dx\\ds&={\sqrt {\frac {r^{2}}{r^{2}-x^{2}}}}\,dx\\ds&={\frac {r}{\sqrt {r^{2}-x^{2}}}}\,dx\\{\text{sehingga }}\\L&=2\pi \int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\cdot {\frac {r}{\sqrt {r^{2}-x^{2}}}}\,dx\\L&=2\pi \int _{-r}^{r}r\,dx\\L&=2\pi rx|_{-r}^{r}\\L&=2\pi (rr-r(-r))\\L&=2\pi (r^{2}+r^{2})\\L&=4\pi r^{2}\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{dengan posisi }}y={\sqrt {r^{2}-x^{2}}}{\text{ serta titik (-''r'', 0) dan (''r'', 0) }}\\{\text{kita tahu bahwa turunannya adalah }}\\y'&={\frac {-x}{\sqrt {r^{2}-x^{2}}}}\\{\text{sehingga }}\\K&=4\int _{0}^{r}{\sqrt {1+({\frac {-x}{\sqrt {r^{2}-x^{2}}}})^{2}}}\,dx\\K&=4\int _{0}^{r}{\sqrt {1+({\frac {x^{2}}{r^{2}-x^{2}}})}}\,dx\\K&=4\int _{0}^{r}{\sqrt {\frac {r^{2}}{r^{2}-x^{2}}}}\,dx\\K&=4r\int _{0}^{r}{\sqrt {\frac {1}{r^{2}-x^{2}}}}\,dx\\K&=4r\int _{0}^{r}{\frac {1}{\sqrt {r^{2}-x^{2}}}}\,dx\\K&=4r\left.\arcsin \left({\frac {x}{r}}\right)\right|_{0}^{r}\\K&=4r\left(\arcsin \left({\frac {r}{r}}\right)-\arcsin \left({\frac {0}{r}}\right)\right)\\K&=4r\left(\arcsin \left(1\right)-\arcsin \left(0\right)\right))\\K&=4r\left({\frac {\pi }{2}}\right)\\K&=2\pi r\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{Dengan posisi }}y={\sqrt {r^{2}-x^{2}}}{\text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. }}\\\sin(\theta )&={\frac {x}{r}}\\x&=r\sin(\theta )\\dx&=r\cos(\theta )\,d\theta \\{\text{dengan turunan di atas }}\\L&=4\int _{0}^{r}{\sqrt {r^{2}-x^{2}}}\,dx\\L&=4\int _{0}^{r}{\sqrt {r^{2}-(r\sin(\theta ))^{2}}}\,dx\\L&=4\int _{0}^{r}{\sqrt {r^{2}-r^{2}\sin ^{2}(\theta )}}\,dx\\L&=4\int _{0}^{r}{\sqrt {r^{2}(1-\sin ^{2}(\theta ))}}\,dx\\L&=4\int _{0}^{r}r\cos(\theta )\,dx\\L&=4\int _{0}^{r}r\cos(\theta )(r\cos(\theta )\,d\theta )\\L&=4\int _{0}^{r}r^{2}\cos ^{2}(\theta )\,d\theta \\L&=4r^{2}\int _{0}^{r}\left({\frac {1+\cos(2\theta )}{2}}\right)\,d\theta \\L&=2r^{2}\int _{0}^{r}(1+\cos(2\theta ))\,d\theta \\L&=2r^{2}\left(\theta +{\frac {1}{2}}\sin(2\theta )\right)|_{0}^{r}\\L&=2r^{2}(\theta +\sin(\theta )\cos(\theta ))|_{0}^{r}\\L&=2r^{2}\left(\arcsin \left({\frac {x}{r}}\right)+\left({\frac {x}{r}}\right)\left({\frac {r^{2}-x^{2}}{r}}\right)\right)|_{0}^{r}\\L&=2r^{2}\left(\arcsin \left({\frac {r}{r}}\right)+\left({\frac {r}{r}}\right)\left({\frac {r^{2}-r^{2}}{r}}\right)\right)-\left(\arcsin \left({\frac {0}{r}}\right)+\left({\frac {0}{r}}\right)\left({\frac {r^{2}-0^{2}}{r}}\right)\right)\\L&=2r^{2}(\arcsin(1)+0-(\arcsin(0)+0))\\L&=2r^{2}\left({\frac {\pi }{2}}\right)\\L&=\pi r^{2}\end{aligned}}$$

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

Jawaban

$${\begin{aligned}{\text{Dengan posisi }}y={\frac {b{\sqrt {a^{2}-x^{2}}}}{a}}{\text{ serta (-''a'', 0) dan (''a'', 0), }}\\L&=4\int _{0}^{r}{\frac {b{\sqrt {a^{2}-x^{2}}}}{a}}\,dx\\L&={\frac {4b}{a}}\int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx\\{\text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, }}\\{\frac {L_{\text{elips}}}{L_{\text{ling}}}}&={\frac {{\frac {4b}{a}}\int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx}{4\int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx}}\\{\frac {L_{\text{elips}}}{L_{\text{ling}}}}&={\frac {b}{a}}\\L_{\text{elips}}&={\frac {b}{a}}L_{\text{ling}}\\L_{\text{elips}}&={\frac {b}{a}}\pi a^{2}\\L_{\text{elips}}&=\pi ab\end{aligned}}$$

1. Berapa luas daerah yang dibatasi y=x²-2x dan y=4x+7!

Jawaban

$${\begin{aligned}{\text{ cari titik potong dari kedua persamaan tersebut }}\\y_{1}&=y_{2}\\x^{-}2x&=4x+7\\x^{2}-6x-7&=0\\(x+1)(x-7)&=0\\x=-1&{\text{ atau }}x=7\\y&=4(-1)+7=3\\y&=4(7)+7=35\\{\text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut }}\\L&=\int _{-1}^{7}4x+7-(x^{2}-2x)\,dx\\&=\int _{-1}^{7}-x^{2}+6x+7\,dx\\&=-{\frac {x^{3}}{3}}+3x^{2}+7x|_{-1}^{7}\\&=-{\frac {7^{3}}{3}}+3(7)^{2}+7(7)-(-{\frac {(-1)^{3}}{3}}+3(-1)^{2}+7(-1))\\&={\frac {245}{3}}-(-{\frac {13}{3}})\\&={\frac {258}{3}}\\&=86\\\end{aligned}}$$

1. Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°!

Jawaban

$${\begin{aligned}{\text{ cari titik potong dari kedua persamaan tersebut }}\\y_{1}&=y_{2}\\6-1,5x&=x-4\\2,5x&=10\\x&=4\\y&=4-4=0\\{\text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut }}\\{\text{untuk x=0 }}\\y&=6-1,5(0)=6\\y&=0-4=-4\\L&=\pi \int _{0}^{4}((6-1,5x)^{2}-(x-4)^{2})\,dx\\&=\pi \int _{0}^{4}(36-18x+2,25x^{2}-(x^{2}-8x+16))\,dx\\&=\pi \int _{0}^{4}(1,25x^{2}-10x+20)\,dx\\&=\pi ({\frac {1,25x^{3}}{3}}-5x^{2}+20x)|_{0}^{4}\\&=\pi ({\frac {1,25(4)^{3}}{3}}-5(4)^{2}+20(4)-({\frac {1,25(0)^{3}}{3}}-5(0)^{2}+20(0)))\\&=\pi ({\frac {30}{3}}-0)\\&=10\pi \\\end{aligned}}$$