Soal-Soal Matematika/Integral

Penulisan turunan

Turunan pertama dari f(x) adalah f'(x). Turunan kedua dan ketiga adalah f''(x) dan f'''(x). Jika turunan keempat dan seterusnya boleh ditulis ' atau angka yakni f''''(x) atau f⁴(x).

Kaidah umum

1. $\int af(x)dx=a\int f(x)dx\text{(}a\text{ konstan)}$
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. $\int f(x)g(x)dx=f(x)\int g(x)dx-\int [f^{\prime }(x)(\int g(x)dx)]dx$
5. $\int [f(x){]}^n{f}^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}+C\text{(untuk }n\ne -1\text{)}$
6. $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$
7. $\int f^{\prime }(x)f(x)dx=\frac{1}{2}[f(x){]}^2+C$
8. $\int f(x)dx=F(x)+C$
9. ${\int }_a^{b}f(x)dx=F(b)-F(a)$
10. $\int f(x)\circ g(x)dx$ = f’(g(x)) g’(x)
11. $\int f(x)\circ f^{-1}(x)dx$ = f’(f^-1(x)) f^-1’(x) = 1
12. $\int f^{-1}(x)dx=\frac{1}{f^{\prime }(f^{\prime }(x))}$

rumus sederhana

$\int dx=x+C$

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\text{ jika }n\ne -1$

$\int (ax+b{)}^{n}dx=\frac{(ax+b{)}^{n+1}}{a(n+1)}+C\text{ jika }n\ne -1$

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

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

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

$\int \frac{dx}{x\sqrt{b^2{x}^2-a^2}}=\frac{1}{a}\mathrm{arcsec}\frac{bx}{a}+C$

Eksponen dan logaritma

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

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

$\int \ln xdx=x\cdot \ln x-x+C$

$\int {}^b\log xdx=x\cdot {}^b\log x-x\cdot {}^b\log e+C$

Trigonometri

$\int \sin xdx=-\cos x+C$

$\int \cos xdx=\sin x+C$

$\int \tan xdx=-\ln |\cos x|+C$

$\int \cot xdx=\ln |\sin x|+C$

$\int \sec xdx=\ln |\sec x+\tan x|+C$

$\int \csc xdx=-\ln |\csc x+\cot x|+C$

Hiperbolik

$\int \sinh xdx=\cosh x+C$

$\int \cosh xdx=\sinh x+C$

$\int \tanh xdx=\ln |\cosh x|+C$

$\int \coth xdx=\ln |\sinh x|+C$

$\int \text{sech}xdx=\arctan (\sinh x)+C$

$\int \text{csch}xdx=\ln |\tanh \frac{x}{2}|+C$

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

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

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

$=\int x^5-20x^4+150x^3-500x^2+625xdx$

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

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

integral substitusi

Berikut contoh penyelesaian cara substitusi.

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

$u=\ln xdu=\frac{dx}{x}$

Dengan menggunakan rumus di atas,

Jawaban

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

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

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

Dengan menggunakan rumus di atas,

Jawaban

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

integral parsial

Cara 1

Rumus

Integral parsial diselesaikan dengan rumus berikut.

$\int udv=uv-\int vdu$

Berikut contoh penyelesaian cara parsial dengan rumus.

1. $\int x\sin xdx$

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

Dengan menggunakan rumus di atas,

Jawaban

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

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

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

Dengan menggunakan rumus di atas,

Jawaban

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

3. $\int x\ln xdx$

$u=\ln x,du=\frac{1}{x}dx,dv=xdx,v=\frac{x^2}{2}$

Dengan menggunakan rumus di atas,

Jawaban

$$\begin{array}{c}\int x\ln xdx\\ (\ln x)(\frac{x^2}{2})-\int (\frac{x^2}{2})(\frac{1}{x}dx)\\ \frac{x^2}{2}\ln x-\int \frac{x^2}{2}(\frac{1}{x})dx\\ \frac{x^2}{2}\ln x-\int \frac{x}{2}dx\\ \frac{x^2}{2}\ln x-\frac{x^2}{4}+C\end{array}$$

4. $\int e^x\sin xdx$

ini terjadi dua kali integral parsial

$u=\sin x,du=\cos xdx,dv=e^{x}dx,v=e^x$

Dengan menggunakan rumus di atas,

Jawaban

$$\begin{array}{c}\int e^x\sin xdx\\ (\sin x)(e^x)-\int (e^x)(\cos xdx)\\ e^x\sin x-\int e^x\cos xdx\\ u=\cos x,du=-\sin xdx,dv=e^{x}dx,v=e^x\\ (\cos x)(e^x)-\int (e^x)(-\sin xdx)\\ e^x\cos x+\int e^x\sin xdx\\ e^x\sin x-(e^x\cos x+\int e^x\sin xdx)\\ e^x\sin x-e^x\cos x-\int e^x\sin xdx\\ \text{misalkan }I=\int e^x\sin xdx\\ I& =e^x\sin x-e^x\cos x-I\\ 2I& =e^x\sin x-e^x\cos x\\ I& =\frac{e^x\sin x-e^x\cos x}{2}\\ & =\frac{e^x\sin x}{2}-\frac{e^x\cos x}{2}+C\end{array}$$

Cara 2

Tabel

Untuk $\int udv$, berlaku ketentuan sebagai berikut.

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

Berikut contoh penyelesaian cara parsial dengan tabel.

1. $\int x\sin xdx$

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

Dengan tabel di atas,

Jawaban

$$\begin{array}{c}\int x\sin xdx\\ (x)(-\cos x)-(1)(-\sin x)+C\\ -x\cos x+\sin x+C\end{array}$$

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

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

3. $\int x\ln xdx$

Tanda	Turunan	Integral
+	$\ln x$	$x$
-	$\frac{1}{x}$	$\frac{x^2}{2}$

Dengan tabel di atas,

Jawaban

$$\begin{array}{c}\int x\ln xdx\\ (\ln x)(\frac{x^2}{2})-\int (\frac{1}{x})(\frac{x^2}{2})dx\\ \frac{x^2}{2}\ln x-\int \frac{x}{2}dx\\ \frac{x^2}{2}\ln x-\frac{x^2}{4}+C\end{array}$$

4. $\int e^x\sin xdx$

ini terjadi dua kali integral parsial

Tanda	Turunan	Integral
+	$\sin x$	$e^x$
-	$\cos x$	$e^x$
+	$-\sin x$	$e^x$

Dengan tabel di atas,

Jawaban

$$\begin{array}{c}\int e^x\sin xdx\\ (\sin x)(e^x)-(\cos x)(e^x)+\int (-\sin x)(e^x)dx\\ e^x\sin x-e^x\cos x-\int e^x\sin x)dx\\ \text{misalkan }I=\int e^x\sin xdx\\ I& =e^x\sin x-e^x\cos x-I\\ 2I& =e^x\sin x-e^x\cos x\\ I& =\frac{e^x\sin x-e^x\cos x}{2}\\ & =\frac{e^x\sin x}{2}-\frac{e^x\cos x}{2}+C\end{array}$$

integral pecahan parsial

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

1. $\int \frac{1}{x^2-4}dx$

Pertama, pisahkan pecahan tersebut.

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

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

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

2. $\int \frac{x^2+5x+4}{(x-3{)}^3}dx$

cara 1

Pertama, pisahkan pecahan tersebut.

$\begin{array}{cc}=& \frac{x^2+5x+4}{(x-3{)}^3}\\ =& \frac{A}{x-3}+\frac{B}{(x-3{)}^2}+\frac{C}{(x-3{)}^3}\\ & \text{untuk mencari C yaitu }(3{)}^2+5(3)+4=28\\ =& \frac{A}{x-3}+\frac{B}{(x-3{)}^2}+\frac{28}{(x-3{)}^3}\\ =& \frac{A(x-3{)}^2+B(x-3)}{(x-3{)}^3}+\frac{28}{(x-3{)}^3}\\ =& \frac{A(x^2-6x+9)+B(x-3)}{(x-3{)}^3}+\frac{28}{(x-3{)}^3}\\ =& \frac{A{x}^2-6Ax+9A+Bx-3B}{(x-3{)}^3}+\frac{28}{(x-3{)}^3}\\ =& \frac{A{x}^2+(-6A+B)x+(9-3B)}{(x-3{)}^3}+\frac{28}{(x-3{)}^3}\end{array}$

Kita tahu bahwa $A=1$ dan $-6A+B=5$ dapat diselesaikan, yaitu $A=1$ dan Templat:Nowrap

$\begin{array}{cc}& \int \frac{x^2+5x+4}{(x-3{)}^3}dx\\ =& \int (\frac{1}{x-3}+\frac{11}{(x-3{)}^2}+\frac{28}{(x-3{)}^3})dx\\ =& \ln |x-3|+(-11(x-3{)}^{-1})+(-14(x-3{)}^{-2})+C\\ =& \ln |x-3|-\frac{11}{x-3}-\frac{14}{(x-3{)}^2}+C\end{array}$

cara 2

Pertama, ubah u=x-3 menjadi x=u+3 dan du=dx tersebut.

$\begin{array}{cc}=& \frac{x^2+5x+4}{(x-3{)}^3}\\ =& \frac{(u+3{)}^2+5(u+3)+4}{u^3}\\ =& \frac{u^2+11u+28}{u^3}\\ & \int \frac{x^2+5x+4}{(x-3{)}^3}dx\\ =& \int \frac{u^2+11u+28}{u^3}du\\ =& \int (\frac{u^2}{u^3}+\frac{11u}{u^3}+\frac{28}{u^3})du\\ =& \int (\frac{1}{u}+\frac{11}{u^2}+\frac{28}{u^3})du\\ =& \ln u+(-11u^{-1})+(-14u^{-2})+C\\ =& \ln |x-3|+(-11(x-3{)}^{-1})+(-14(x-3{)}^{-2})+C\\ =& \ln |x-3|-\frac{11}{x-3}-\frac{14}{(x-3{)}^2}+C\end{array}$

3. $\int \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}dx$

Pertama, penyelesaian dengan persamaan polinomial tersebut.

$\begin{array}{cc}=& \frac{x^4+2x^3-13x^2-7x+21}{x^2-4x+3}\\ =& x^2+6x+8+\frac{7x-3}{x^2-4x+3}\\ =& \frac{1}{3}{x}^3+3x^2+8x+\int \frac{7x-3}{x^2-4x+3}dx\end{array}$

Kedua, pisahkan pecahan tersebut.

$\begin{array}{cc}=& \frac{7x-3}{x^2-4x+3}\\ =& \frac{7x-3}{(x-1)(x-3)}\\ =& \frac{A}{x-1}+\frac{B}{x-3}\\ =& \frac{A(x-3)+B(x-1)}{x^2-4x+3}\\ =& \frac{(A+B)x+(-3A-B)}{x^2-4}\end{array}$

Kita tahu bahwa $A+B=7$ dan $-3A-B=-3$ dapat diselesaikan, yaitu $A=-2$ dan Templat:Nowrap

$\begin{array}{cc}& \frac{1}{3}{x}^3+3x^2+8x+\int \frac{7x-3}{x^2-4x+3}dx\\ =& \frac{1}{3}{x}^3+3x^2+8x+\int (-\frac{2}{(x-1)}+\frac{9}{x-3})dx\\ =& \frac{1}{3}{x}^3+3x^2+8x-2\ln |x-1|+9\ln |x-3|dx+C\end{array}$

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.

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

$x=2\tan A,dx=2{\sec }^{2}AdA$

Dengan substitusi di atas,

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

Substitusi berikut dapat dibuat.

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

$t=\sin A,dt=\cos AdA$

Dengan substitusi di atas,

$\begin{array}{cc}& \frac{1}{4}\int \frac{\cos A}{{\sin }^{2}A}dA\\ =& \frac{1}{4}\int \frac{dt}{t^2}\\ =& \frac{1}{4}(-\frac{1}{t})+C\\ =& -\frac{1}{4t}+C\\ =& -\frac{1}{4\sin A}+C\end{array}$

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

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

integral mutlak

$\int |f(x)|dA$

buatlah $f(x)\ge 0$ 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 $|f(x)|=\{\begin{array}{cc}f(x),& \text{jika }f(x)\ge b,\\ -f(x),& \text{jika }f(x)<b.\end{array}$. 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

${\int }_{-a}^{a}f(x)dx={\int }_0^a(f(x)+f(-x))dx$

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

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

${\int }_a^{c}f(x)dx={\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx$

${\int }_a^{b}f(x\pm k)dx={\int }_{a\pm k}^{b\pm k}f(x)dx$

${\int }_a^{b}f(x\pm k)dx={\int }_{a\cdot k}^{b\cdot k}f(x)dx$

${\int }_a^{b}f(k\cdot x)dx=\frac{1}{k}{\int }_{a\pm k}^{b\pm k}f(x)dx$

${\int }_a^{b}f(k\cdot x)dx=\frac{1}{k}{\int }_{a\cdot k}^{b\cdot k}f(x)dx$

integral terbalik

${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$

Jenis integral lainnya

panjang busur

Sumbu x

$S={\int }_{x_1}^{x_2}\sqrt{1+(f^{\prime }(x){)}^2}dx$

Sumbu y

$S={\int }_{y_1}^{y_2}\sqrt{1+(f^{\prime }(y){)}^2}dy$

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

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

Sumbu y sebagai poros

$L_p=2\pi {\int }_{y_1}^{y_2}f(y)ds$

dengan

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

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:

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

jawaban

Jawaban

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

Jawaban

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

Jawaban

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

Jawaban

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

Jawaban

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

Jawaban

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

Jawaban

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

Jawaban

$$\begin{array}{c}\int \frac{1}{\sqrt{50-288x^2}}dx\\ x& =\frac{5}{12}sinA\\ dx& =\frac{5}{12}cosAdA\\ \int \frac{1}{\sqrt{50-288x^2}}dx& =\int \frac{\frac{5}{12}cosA}{\sqrt{50-288(\frac{5}{12}sinA{)}^2}}dA\\ & =\int \frac{\frac{5}{12}cosA}{\sqrt{50-\frac{288\cdot 25}{144}si{n}^{2}A}}dA\\ & =\int \frac{\frac{5}{12}cosA}{\sqrt{50-50si{n}^{2}A}}dA\\ & =\int \frac{\frac{5}{12}cosA}{\sqrt{50(1-si{n}^{2}A)}}dA\\ & =\int \frac{\frac{5}{12}cosA}{\sqrt{50co{s}^{2}A}}dA\\ & =\int \frac{\frac{5}{12}cosA}{5cosA\sqrt{2}}dA\\ & =\int \frac{\sqrt{2}}{24}dA\\ & =\frac{\sqrt{2}}{24}\int dA\\ & =A+C\\ A& =arcsin\frac{12x}{5}\\ & =\frac{\sqrt{2}}{24}arcsin\frac{12x}{5}+C\end{array}$$

Jawaban

$$\begin{array}{c}\int \frac{1}{12+75x^2}dx\\ x& =\frac{2}{5}tanA\\ dx& =\frac{2}{5}se{c}^{2}AdA\\ \int \frac{1}{12+75x^2}dx& =\int \frac{\frac{2}{5}se{c}^{2}A}{12+75(\frac{2}{5}tanA{)}^2}dA\\ & =\int \frac{\frac{2}{5}se{c}^{2}A}{12+\frac{75\cdot 4}{25}ta{n}^{2}A}dA\\ & =\int \frac{\frac{2}{5}se{c}^{2}A}{12+12ta{n}^{2}A}dA\\ & =\int \frac{\frac{2}{5}se{c}^{2}A}{12(1+ta{n}^{2}A)}dA\\ & =\int \frac{\frac{2}{5}se{c}^{2}A}{12se{c}^{2}A}dA\\ & =\int \frac{1}{30}dA\\ & =\frac{A}{30}+C\\ A& =arctan\frac{5x}{2}\\ & =\frac{arctan\frac{5x}{2}}{30}+C\end{array}$$

Jawaban

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

Jawaban

$$\begin{array}{c}\int (1+si{n}^{2}x+si{n}^{4}x+si{n}^{6}x+\dots )dx\\ a=1,r=si{n}^{2}x\\ S_{\mathrm{\infty }}& =\frac{a}{1-r}\\ & =\frac{1}{1-si{n}^{2}x}\\ & =\frac{1}{co{s}^{2}x}\\ & =se{c}^{2}x\\ \int se{c}^{2}xdx& =tanx+C\end{array}$$

Jawaban

$$\begin{array}{c}\int (1+co{s}^{2}x+co{s}^{4}x+co{s}^{6}x+\dots )dx\\ a=1,r=co{s}^{2}x\\ S_{\mathrm{\infty }}& =\frac{a}{1-r}\\ & =\frac{1}{1-co{s}^{2}x}\\ & =\frac{1}{si{n}^{2}x}\\ & =cs{c}^{2}x\\ \int cs{c}^{2}xdx& =-cotx+C\end{array}$$

Jawaban

$$\begin{array}{c}\int (sinx+si{n}^{3}x+si{n}^{5}x+si{n}^{7}x+\dots )dx\\ a=sinx,r=si{n}^{2}x\\ S_{\mathrm{\infty }}& =\frac{a}{1-r}\\ & =\frac{sinx}{1-si{n}^{2}x}\\ & =\frac{sinx}{co{s}^{2}x}\\ & =secx\cdot tanx\\ \int secx\cdot tanxdx& =secx+C\end{array}$$

Jawaban

$$\begin{array}{c}\int (cosx+co{s}^{3}x+co{s}^{5}x+co{s}^{7}x+\dots )dx\\ a=cosx,r=co{s}^{2}x\\ S_{\mathrm{\infty }}& =\frac{a}{1-r}\\ & =\frac{cosx}{1-co{s}^{2}x}\\ & =\frac{cosx}{si{n}^{2}x}\\ & =cscx\cdot cotx\\ \int cscx\cdot cotxdx& =-cscx+C\end{array}$$

Jawaban

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

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

$$\begin{array}{cc}{\int }_1^{5}2x-3dx& ={x^2-3x|}_1^5\\ & =10-(-2)\\ & =12\end{array}$$

Jawaban

$$\begin{array}{c}{\int }_0^4|x-2|dx\\ \text{ tentukan nilai harga nol }\\ x-2& =0\\ x& =2\\ \text{ buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti }\\ |x-2|=\{\begin{array}{cc}x-2,& \text{jika }x\ge 2,\\ -(x-2),& \text{jika }x<2.\end{array}\\ {\int }_0^4|x-2|dx& ={\int }_2^{4}x-2dx+{\int }_0^2-(x-2)dx\\ & ={(\frac{x^2}{2}-2x)|}_2^4+{(\frac{-x^2}{2}+2x)|}_0^2\\ & =0-(-2)+2-0\\ & =4\end{array}$$

Jawaban

$$\begin{array}{c}{\int }_2^4|3-x|dx\\ \text{ tentukan nilai syarat-syarat }\\ 3-x& \ge 0\\ \text{ harga nol }\\ x& =3\\ \text{ buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti }\\ |3-x|=\{\begin{array}{cc}3-x,& \text{jika }x\le 3,\\ -(3-x),& \text{jika }x>3.\end{array}\\ {\int }_2^4|3-x|dx& ={\int }_2^{3}3-xdx+{\int }_3^4-(3-x)dx\\ & ={(3x-\frac{x^2}{2})|}_2^3+{(-3x+\frac{x^2}{2})|}_3^4\\ & =4.5-4-4-(-4.5)\\ & =1\end{array}$$

Jawaban

$$\begin{array}{cc}{\int }_1^5{x}^2-6x+8dx& ={\frac{x^3}{3}-3x^2+8x|}_1^5\\ & =\frac{20}{3}-\frac{16}{3}\\ & =\frac{4}{3}\\ & =1\frac{1}{3}\end{array}$$

Jawaban

$$\begin{array}{c}{\int }_{-3}^5|x^2-2x-8|dx\\ \text{ tentukan nilai syarat-syarat }\\ x^2-2x-8& \ge 0\\ \text{ harga nol }\\ (x+2)(x-4)& =0\\ x=-2\text{ atau }x=4\\ \text{ buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti }\\ |x^2-2x-8|=\{\begin{array}{cc}{x}^2-2x-8,& \text{jika }x\le -2,\\ -(x^2-2x-8),& \text{jika }-2<x<4,\\ x^2-2x-8,& \text{jika }x\ge 4\end{array}\\ {\int }_{-3}^5|x^2-2x-8|dx& ={\int }_{-3}^{-2}{x}^2-2x-8dx+{\int }_{-2}^4-(x^2-2x-8)dx+{\int }_4^5{x}^2-2x-8dx\\ & ={(\frac{x^3}{3}-x^2-8x)|}_{-3}^{-2}+{(\frac{-x^3}{3}+x^2+8x)|}_{-2}^4+{(\frac{x^3}{3}-x^2-8x)|}_4^5\\ & =\frac{28}{3}-6+\frac{80}{3}-(-\frac{28}{3})-\frac{70}{3}-(-\frac{80}{3})\\ & =\frac{128}{3}\\ & =42\frac{2}{3}\end{array}$$

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

Jawaban

$$\begin{array}{cc}L& =\int xdx\\ L& =\frac{1}{2}{x}^2\\ L& =\frac{1}{2}xy(y=x)\end{array}$$

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

Jawaban

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

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

Jawaban

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

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

Jawaban

$$\begin{array}{c}\text{dengan posisi }y=s\text{ dan titik (''s'', ''s''), }\\ L& ={\int }_0^{s}sdx\\ L& =sx{|}_0^s\\ L& =ss-0\\ L& =s^2\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{Dengan posisi }y=l\text{dan titik (''p'', ''l''), }\\ L& ={\int }_0^{p}ldx\\ L& =lx{|}_0^p\\ L& =pl-0\\ L& =pl\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{dengan posisi }y=\frac{-tx}{a}+t\text{ dan titik (''a'', ''t''), }\\ L& ={\int }_0^a(\frac{-tx}{a}+t)dx\\ L& ={\frac{-t{x}^2}{2a}+tx|}_0^a\\ L& =\frac{-t{a}^2}{2a}+ta-0+0\\ L& =\frac{-ta}{2}+ta\\ L& =\frac{at}{2}\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{ Dengan posisi }y=r\text{ dan titik (''t'', ''r''), }\\ V& =\pi {\int }_0^t{r}^{2}dx\\ V& =\pi {r^{2}x|}_0^t\\ V& =\pi r^{2}t-0\\ V& =\pi r^{2}t\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{dengan posisi }y=\frac{rx}{t}\text{ dan titik (''t'', ''r''), }\\ V& =\pi {\int }_0^t{\left(\frac{rx}{t}\right)}^{2}dx\\ V& =\pi {\frac{r^2{x}^3}{3t^2}|}_0^t\\ V& =\pi \frac{r^2{t}^3}{3t^2}-0\\ V& =\frac{\pi r^{2}t}{3}\end{array}$$

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

Jawaban

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

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

Jawaban

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

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

Jawaban

$$\begin{array}{c}\text{dengan posisi }y=\sqrt{r^2-x^2}\text{ serta titik (-''r'', 0) dan (''r'', 0) }\\ \text{kita tahu bahwa turunannya adalah }\\ y^{\prime }& =\frac{-x}{\sqrt{r^2-x^2}}\\ \text{sehingga }\\ K& =4{\int }_0^r\sqrt{1+(\frac{-x}{\sqrt{r^2-x^2}}{)}^2}dx\\ K& =4{\int }_0^r\sqrt{1+(\frac{x^2}{r^2-x^2})}dx\\ K& =4{\int }_0^r\sqrt{\frac{r^2}{r^2-x^2}}dx\\ K& =4r{\int }_0^r\sqrt{\frac{1}{r^2-x^2}}dx\\ K& =4r{\int }_0^r\frac{1}{\sqrt{r^2-x^2}}dx\\ K& =4r{\arcsin \left(\frac{x}{r}\right)|}_0^r\\ K& =4r(\arcsin \left(\frac{r}{r}\right)-\arcsin \left(\frac{0}{r}\right))\\ K& =4r(\arcsin \left(1\right)-\arcsin \left(0\right)))\\ K& =4r\left(\frac{\pi }{2}\right)\\ K& =2\pi r\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{Dengan posisi }y=\sqrt{r^2-x^2}\text{ serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu. }\\ \sin (\theta )& =\frac{x}{r}\\ x& =r\sin (\theta )\\ dx& =r\cos (\theta )d\theta \\ \text{dengan turunan di atas }\\ L& =4{\int }_0^r\sqrt{r^2-x^2}dx\\ L& =4{\int }_0^r\sqrt{r^2-(r\sin (\theta ){)}^2}dx\\ L& =4{\int }_0^r\sqrt{r^2-r^2{\sin }^2(\theta )}dx\\ L& =4{\int }_0^r\sqrt{r^2(1-{\sin }^2(\theta ))}dx\\ L& =4{\int }_0^{r}r\cos (\theta )dx\\ L& =4{\int }_0^{r}r\cos (\theta )(r\cos (\theta )d\theta )\\ L& =4{\int }_0^r{r}^2{\cos }^2(\theta )d\theta \\ L& =4r^2{\int }_0^r\left(\frac{1+\cos (2\theta )}{2}\right)d\theta \\ L& =2r^2{\int }_0^r(1+\cos (2\theta ))d\theta \\ L& =2r^2(\theta +\frac{1}{2}\sin (2\theta )){|}_0^r\\ L& =2r^2(\theta +\sin (\theta )\cos (\theta )){|}_0^r\\ L& =2r^2(\arcsin \left(\frac{x}{r}\right)+\left(\frac{x}{r}\right)\left(\frac{r^2-x^2}{r}\right)){|}_0^r\\ L& =2r^2(\arcsin \left(\frac{r}{r}\right)+\left(\frac{r}{r}\right)\left(\frac{r^2-r^2}{r}\right))-(\arcsin \left(\frac{0}{r}\right)+\left(\frac{0}{r}\right)\left(\frac{r^2-0^2}{r}\right))\\ L& =2r^2(\arcsin (1)+0-(\arcsin (0)+0))\\ L& =2r^2\left(\frac{\pi }{2}\right)\\ L& =\pi r^2\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{Dengan posisi }y=\frac{b\sqrt{a^2-x^2}}{a}\text{ serta (-''a'', 0) dan (''a'', 0), }\\ L& =4{\int }_0^r\frac{b\sqrt{a^2-x^2}}{a}dx\\ L& =\frac{4b}{a}{\int }_0^a\sqrt{a^2-x^2}dx\\ \text{dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips, }\\ \frac{L_{\text{elips}}}{L_{\text{ling}}}& =\frac{\frac{4b}{a}{\int }_0^a\sqrt{a^2-x^2}dx}{4{\int }_0^a\sqrt{a^2-x^2}dx}\\ \frac{L_{\text{elips}}}{L_{\text{ling}}}& =\frac{b}{a}\\ L_{\text{elips}}& =\frac{b}{a}{L}_{\text{ling}}\\ L_{\text{elips}}& =\frac{b}{a}\pi a^2\\ L_{\text{elips}}& =\pi ab\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{ cari titik potong dari kedua persamaan tersebut }\\ y_1& =y_2\\ x^{-}2x& =4x+7\\ x^2-6x-7& =0\\ (x+1)(x-7)& =0\\ x=-1& \text{ atau }x=7\\ y& =4(-1)+7=3\\ y& =4(7)+7=35\\ \text{jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut }\\ L& ={\int }_{-1}^{7}4x+7-(x^2-2x)dx\\ & ={\int }_{-1}^7-x^2+6x+7dx\\ & =-\frac{x^3}{3}+3x^2+7x{|}_{-1}^7\\ & =-\frac{7^3}{3}+3(7{)}^2+7(7)-(-\frac{(-1{)}^3}{3}+3(-1{)}^2+7(-1))\\ & =\frac{245}{3}-(-\frac{13}{3})\\ & =\frac{258}{3}\\ & =86\end{array}$$

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

Jawaban

$$\begin{array}{c}\text{ cari titik potong dari kedua persamaan tersebut }\\ y_1& =y_2\\ 6-1,5x& =x-4\\ 2,5x& =10\\ x& =4\\ y& =4-4=0\\ \text{jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut }\\ \text{untuk x=0 }\\ y& =6-1,5(0)=6\\ y& =0-4=-4\\ L& =\pi {\int }_0^4((6-1,5x{)}^2-(x-4{)}^2)dx\\ & =\pi {\int }_0^4(36-18x+2,25x^2-(x^2-8x+16))dx\\ & =\pi {\int }_0^4(1,25x^2-10x+20)dx\\ & =\pi (\frac{1,25x^3}{3}-5x^2+20x){|}_0^4\\ & =\pi (\frac{1,25(4{)}^3}{3}-5(4{)}^2+20(4)-(\frac{1,25(0{)}^3}{3}-5(0{)}^2+20(0)))\\ & =\pi (\frac{30}{3}-0)\\ & =10\pi \end{array}$$