∫
a
f
(
x
)
d
x
=
a
∫
f
(
x
)
d
x
(
a
konstan)
{\displaystyle \int af(x)\,dx=a\int f(x)\,dx\qquad {\mbox{(}}a{\mbox{ konstan)}}\,\!}
∫
[
f
(
x
)
+
g
(
x
)
]
d
x
=
∫
f
(
x
)
d
x
+
∫
g
(
x
)
d
x
{\displaystyle \int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx}
∫
[
f
(
x
)
−
g
(
x
)
]
d
x
=
∫
f
(
x
)
d
x
−
∫
g
(
x
)
d
x
{\displaystyle \int [f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx}
∫
f
(
x
)
g
(
x
)
d
x
=
f
(
x
)
∫
g
(
x
)
d
x
−
∫
[
f
′
(
x
)
(
∫
g
(
x
)
d
x
)
]
d
x
{\displaystyle \int f(x)g(x)\,dx=f(x)\int g(x)\,dx-\int \left[f'(x)\left(\int g(x)\,dx\right)\right]\,dx}
∫
[
f
(
x
)
]
n
f
′
(
x
)
d
x
=
[
f
(
x
)
]
n
+
1
n
+
1
+
C
(untuk
n
≠
−
1
)
{\displaystyle \int [f(x)]^{n}f'(x)\,dx={[f(x)]^{n+1} \over n+1}+C\qquad {\mbox{(untuk }}n\neq -1{\mbox{)}}\,\!}
∫
f
′
(
x
)
f
(
x
)
d
x
=
ln
|
f
(
x
)
|
+
C
{\displaystyle \int {f'(x) \over f(x)}\,dx=\ln {\left|f(x)\right|}+C}
∫
f
′
(
x
)
f
(
x
)
d
x
=
1
2
[
f
(
x
)
]
2
+
C
{\displaystyle \int {f'(x)f(x)}\,dx={1 \over 2}[f(x)]^{2}+C}
∫
f
(
x
)
d
x
=
F
(
x
)
+
C
{\displaystyle \int f(x)\,dx=F(x)+C}
∫
a
b
f
(
x
)
d
x
=
F
(
b
)
−
F
(
a
)
{\displaystyle \int _{a}^{b}f(x)\,dx=F(b)-F(a)}
∫
a
c
|
f
(
x
)
|
d
x
=
∫
a
b
f
(
x
)
d
x
+
∫
b
c
−
f
(
x
)
d
x
=
F
(
b
)
−
F
(
a
)
−
F
(
c
)
+
F
(
b
)
{\displaystyle \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)}
|
f
(
x
)
|
=
{
f
(
x
)
,
jika
f
(
x
)
≥
b
,
−
f
(
x
)
,
jika
f
(
x
)
<
b
.
{\displaystyle |f(x)|={\begin{cases}f(x),&{\mbox{jika }}f(x)\geq b,\\-f(x),&{\mbox{jika }}f(x)<b.\end{cases}}}
∫
d
x
=
x
+
C
{\displaystyle \int \,dx=x+C}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
C
jika
n
≠
−
1
{\displaystyle \int x^{n}\,dx={\frac {x^{n+1}}{n+1}}+C\qquad {\mbox{ jika }}n\neq -1}
∫
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
+
C
jika
n
≠
−
1
{\displaystyle \int (ax+b)^{n}\,dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\mbox{ jika }}n\neq -1}
∫
d
x
x
=
ln
|
x
|
+
C
{\displaystyle \int {dx \over x}=\ln {\left|x\right|}+C}
∫
d
x
a
2
−
x
2
=
arcsin
x
a
+
C
{\displaystyle \int {dx \over {\sqrt {a^{2}-x^{2}}}}=\arcsin {x \over a}+C}
∫
d
x
a
2
+
x
2
=
1
a
arctan
x
a
+
C
{\displaystyle \int {dx \over {a^{2}+x^{2}}}={1 \over a}\arctan {x \over a}+C}
∫
d
x
x
x
2
−
a
2
=
1
a
arcsec
x
a
+
C
{\displaystyle \int {dx \over x{\sqrt {x^{2}-a^{2}}}}={1 \over a}\operatorname {arcsec} {x \over a}+C}
Eksponen dan logaritma
∫
e
x
d
x
=
e
x
+
C
{\displaystyle \int e^{x}\,dx=e^{x}+C}
∫
a
x
d
x
=
a
x
ln
a
+
C
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln {a}}}+C}
∫
ln
x
d
x
=
x
ln
x
−
x
+
C
{\displaystyle \int \ln {x}\,dx=x\ln {x}-x+C}
∫
b
log
x
d
x
=
x
⋅
b
log
x
−
x
⋅
b
log
e
+
C
{\displaystyle \int \,^{b}\!\log {x}\,dx=x\cdot \,^{b}\!\log x-x\cdot \,^{b}\!\log e+C}
Trigonometri
∫
sin
x
d
x
=
−
cos
x
+
C
{\displaystyle \int \sin {x}\,dx=-\cos {x}+C}
∫
cos
x
d
x
=
sin
x
+
C
{\displaystyle \int \cos {x}\,dx=\sin {x}+C}
∫
tan
x
d
x
=
ln
|
sec
x
|
+
C
{\displaystyle \int \tan {x}\,dx=\ln {\left|\sec {x}\right|}+C}
∫
cot
x
d
x
=
−
ln
|
csc
x
|
+
C
{\displaystyle \int \cot {x}\,dx=-\ln {\left|\csc {x}\right|}+C}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
C
{\displaystyle \int \sec {x}\,dx=\ln {\left|\sec {x}+\tan {x}\right|}+C}
∫
csc
x
d
x
=
−
ln
|
csc
x
+
cot
x
|
+
C
{\displaystyle \int \csc {x}\,dx=-\ln {\left|\csc {x}+\cot {x}\right|}+C}
Hiperbolik
∫
sinh
x
d
x
=
cosh
x
+
C
{\displaystyle \int \sinh x\,dx=\cosh x+C}
∫
cosh
x
d
x
=
sinh
x
+
C
{\displaystyle \int \cosh x\,dx=\sinh x+C}
∫
tanh
x
d
x
=
ln
|
cosh
x
|
+
C
{\displaystyle \int \tanh x\,dx=\ln |\cosh x|+C}
∫
coth
x
d
x
=
ln
|
sinh
x
|
+
C
{\displaystyle \int \coth x\,dx=\ln |\sinh x|+C}
∫
sech
x
d
x
=
arctan
(
sinh
x
)
+
C
{\displaystyle \int {\mbox{sech}}\,x\,dx=\arctan(\sinh x)+C}
∫
csch
x
d
x
=
ln
|
tanh
x
2
|
+
C
{\displaystyle \int {\mbox{csch}}\,x\,dx=\ln \left|\tanh {x \over 2}\right|+C}
Berikut contoh penyelesaian cara biasa.
∫
x
3
−
c
o
s
3
x
+
e
5
x
+
2
−
1
2
x
+
5
d
x
{\displaystyle \int x^{3}-cos3x+e^{5x+2}-{\frac {1}{2x+5}}dx}
=
1
3
+
1
x
3
+
1
−
s
i
n
3
x
3
+
e
5
x
+
2
5
−
l
n
(
2
x
+
5
)
2
+
C
{\displaystyle ={\frac {1}{3+1}}x^{3+1}-{\frac {sin3x}{3}}+{\frac {e^{5x+2}}{5}}-{\frac {ln(2x+5)}{2}}+C}
=
x
4
4
−
s
i
n
3
x
3
+
e
5
x
+
2
5
−
l
n
(
2
x
+
5
)
2
+
C
{\displaystyle ={\frac {x^{4}}{4}}-{\frac {sin3x}{3}}+{\frac {e^{5x+2}}{5}}-{\frac {ln(2x+5)}{2}}+C}
Berikut contoh penyelesaian cara substitusi.
∫
ln
(
x
)
x
d
x
{\displaystyle \int {\frac {\ln(x)}{x}}\,dx}
t
=
ln
(
x
)
,
d
t
=
d
x
x
{\displaystyle t=\ln(x),\,dt={\frac {dx}{x}}}
Dengan menggunakan rumus di atas,
∫
ln
(
x
)
x
d
x
=
∫
t
d
t
=
1
2
t
2
+
C
=
1
2
ln
2
x
+
C
{\displaystyle {\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}}}
Cara 1
Rumus Integral parsial diselesaikan dengan rumus berikut.
∫
u
d
v
=
u
v
−
∫
v
d
u
{\displaystyle \int u\,dv=uv-\int v\,du}
Berikut contoh penyelesaian cara parsial dengan rumus.
∫
x
sin
(
x
)
d
x
{\displaystyle \int x\sin(x)\,dx}
u
=
x
,
d
u
=
1
d
x
,
d
v
=
sin
(
x
)
d
x
,
v
=
−
cos
(
x
)
{\displaystyle u=x,\,du=1\,dx,\,dv=\sin(x)\,dx,\,v=-\cos(x)}
Dengan menggunakan rumus di atas,
∫
x
sin
(
x
)
d
x
=
(
x
)
(
−
cos
(
x
)
)
−
∫
(
−
cos
(
x
)
)
(
1
d
x
)
=
−
x
cos
(
x
)
+
∫
cos
(
x
)
d
x
=
−
x
cos
(
x
)
+
sin
(
x
)
+
C
{\displaystyle {\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
∫
u
d
v
{\textstyle \int u\,dv}
Tanda
Turunan
Integral
+
u
{\displaystyle u}
d
v
{\displaystyle dv}
-
d
u
d
x
{\displaystyle {\frac {du}{dx}}}
v
{\displaystyle v}
+
d
2
u
d
x
2
{\displaystyle {\frac {d^{2}u}{dx^{2}}}}
∫
v
d
x
{\displaystyle \int v\,dx}
Berikut contoh penyelesaian cara parsial dengan tabel.
∫
x
sin
(
x
)
d
x
{\displaystyle \int x\sin(x)\,dx}
Tanda
Turunan
Integral
+
x
{\displaystyle x}
sin
(
x
)
{\displaystyle \sin(x)}
-
1
{\displaystyle 1}
−
cos
(
x
)
{\displaystyle -\cos(x)}
+
0
{\displaystyle 0}
−
sin
(
x
)
{\displaystyle -\sin(x)}
Dengan tabel di atas,
∫
x
sin
(
x
)
d
x
=
(
x
)
(
−
cos
(
x
)
)
−
(
1
)
(
−
sin
(
x
)
)
+
C
=
−
x
cos
(
x
)
+
sin
(
x
)
+
C
{\displaystyle {\begin{aligned}&\;\int x\sin(x)\,dx\\=&\;(x)(-\cos(x))-(1)(-\sin(x))+C\\=&\;-x\cos(x)+\sin(x)+C\end{aligned}}}
Berikut contoh penyelesaian cara parsial untuk persamaan pecahan (rasional).
∫
d
x
x
2
−
4
{\displaystyle \int {\frac {dx}{x^{2}-4}}}
Pertama, pisahkan pecahan tersebut.
=
1
x
2
−
4
=
1
(
x
+
2
)
(
x
−
2
)
=
A
x
+
2
+
B
x
−
2
=
A
(
x
−
2
)
+
B
(
x
+
2
)
x
2
−
4
=
(
A
+
B
)
x
−
2
(
A
−
B
)
x
2
−
4
{\displaystyle {\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
A
+
B
=
0
{\textstyle A+B=0}
A
−
B
=
−
1
2
{\textstyle A-B=-{\frac {1}{2}}}
A
=
−
1
4
{\textstyle A=-{\frac {1}{4}}}
B
=
1
4
{\textstyle B={\frac {1}{4}}}
∫
d
x
x
2
−
4
=
∫
−
1
4
(
x
+
2
)
+
1
4
(
x
−
2
)
d
x
=
1
4
∫
1
x
−
2
−
1
x
+
2
d
x
=
1
4
(
ln
|
x
−
2
|
−
ln
|
x
+
2
|
)
+
C
=
1
4
ln
|
x
−
2
x
+
2
|
+
C
{\displaystyle {\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
sunting
Bentuk
Trigonometri
a
2
−
b
2
x
2
{\displaystyle {\sqrt {a^{2}-b^{2}x^{2}}}}
x
=
a
b
sin
(
α
)
{\displaystyle x={\frac {a}{b}}\sin(\alpha )}
a
2
+
b
2
x
2
{\displaystyle {\sqrt {a^{2}+b^{2}x^{2}}}}
x
=
a
b
tan
(
α
)
{\displaystyle x={\frac {a}{b}}\tan(\alpha )}
b
2
x
2
−
a
2
{\displaystyle {\sqrt {b^{2}x^{2}-a^{2}}}}
x
=
a
b
sec
(
α
)
{\displaystyle x={\frac {a}{b}}\sec(\alpha )}
Berikut contoh penyelesaian cara substitusi trigonometri.
∫
d
x
x
2
x
2
+
4
{\displaystyle \int {\frac {dx}{x^{2}{\sqrt {x^{2}+4}}}}}
x
=
2
tan
(
A
)
,
d
x
=
2
sec
2
(
A
)
d
A
{\displaystyle x=2\tan(A),\,dx=2\sec ^{2}(A)\,dA}
Dengan substitusi di atas,
∫
d
x
x
2
x
2
+
4
=
∫
2
sec
2
(
A
)
d
A
(
2
tan
(
A
)
)
2
4
+
(
2
tan
(
A
)
)
2
=
∫
2
sec
2
(
A
)
d
A
4
tan
2
(
A
)
4
+
4
tan
2
(
A
)
=
∫
2
sec
2
(
A
)
d
A
4
tan
2
(
A
)
4
(
1
+
tan
2
(
A
)
)
=
∫
2
sec
2
(
A
)
d
A
4
tan
2
(
A
)
4
sec
2
(
A
)
=
∫
2
sec
2
(
A
)
d
A
(
4
tan
2
(
A
)
)
(
2
sec
(
A
)
)
=
∫
sec
(
A
)
d
A
4
tan
2
(
A
)
=
1
4
∫
sec
(
A
)
d
A
tan
2
(
A
)
=
1
4
∫
cos
(
A
)
sin
2
(
A
)
d
A
{\displaystyle {\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.
∫
cos
(
A
)
sin
2
(
A
)
d
A
{\displaystyle \int {\frac {\cos(A)}{\sin ^{2}(A)}}\,dA}
t
=
sin
(
A
)
,
d
t
=
cos
(
A
)
d
A
{\displaystyle t=\sin(A),\,dt=\cos(A)\,dA}
Dengan substitusi di atas,
1
4
∫
cos
(
A
)
sin
2
(
A
)
d
A
=
1
4
∫
d
t
t
2
=
1
4
(
−
1
t
)
+
C
=
−
1
4
t
+
C
=
−
1
4
sin
(
A
)
+
C
{\displaystyle {\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
sin
(
A
)
=
x
x
2
+
4
{\textstyle \sin(A)={\frac {x}{\sqrt {x^{2}+4}}}}
=
−
1
4
sin
(
A
)
+
C
=
−
x
2
+
4
4
x
+
C
{\displaystyle {\begin{aligned}=&\;-{\frac {1}{4\sin(A)}}+C\\=&\;-{\frac {\sqrt {x^{2}+4}}{4x}}+C\end{aligned}}}
Sumbu x
S
=
∫
x
1
x
2
1
+
(
f
′
(
x
)
)
2
d
x
{\displaystyle S=\int _{x_{1}}^{x_{2}}{\sqrt {1+(f'(x))^{2}}}\,dx}
Sumbu y
S
=
∫
y
1
y
2
1
+
(
f
′
(
y
)
)
2
d
y
{\displaystyle S=\int _{y_{1}}^{y_{2}}{\sqrt {1+(f'(y))^{2}}}\,dy}
Satu kurva
Sumbu x
L
=
∫
x
1
x
2
f
(
x
)
d
x
{\displaystyle L=\int _{x_{1}}^{x_{2}}f(x)\,dx}
Sumbu y
L
=
∫
y
1
y
2
f
(
y
)
d
y
{\displaystyle L=\int _{y_{1}}^{y_{2}}f(y)\,dy}
Dua kurva
Sumbu x
L
=
∫
x
1
x
2
(
f
(
x
2
)
−
f
(
x
1
)
)
d
x
{\displaystyle L=\int _{x_{1}}^{x_{2}}(f(x_{2})-f(x_{1}))\,dx}
Sumbu y
L
=
∫
y
1
y
2
(
f
(
y
2
)
−
f
(
y
1
)
)
d
y
{\displaystyle L=\int _{y_{1}}^{y_{2}}(f(y_{2})-f(y_{1}))\,dy}
atau juga
L
=
D
D
6
a
2
{\displaystyle L={\frac {D{\sqrt {D}}}{6a^{2}}}}
Sumbu x sebagai poros
L
p
=
2
π
∫
x
1
x
2
f
(
x
)
d
s
{\displaystyle L_{p}=2\pi \int _{x_{1}}^{x_{2}}f(x)\,ds}
dengan
d
s
=
1
+
(
f
′
(
x
)
)
2
d
x
{\displaystyle ds={\sqrt {1+(f'(x))^{2}}}\,dx}
Sumbu y sebagai poros
L
p
=
2
π
∫
y
1
y
2
f
(
y
)
d
s
{\displaystyle L_{p}=2\pi \int _{y_{1}}^{y_{2}}f(y)\,ds}
dengan
d
s
=
1
+
(
f
′
(
y
)
)
2
d
y
{\displaystyle ds={\sqrt {1+(f'(y))^{2}}}\,dy}
Satu kurva
Sumbu x sebagai poros
V
=
π
∫
x
1
x
2
(
f
(
x
)
)
2
d
x
{\displaystyle V=\pi \int _{x_{1}}^{x_{2}}(f(x))^{2}\,dx}
Sumbu y sebagai poros
V
=
π
∫
y
1
y
2
(
f
(
y
)
)
2
d
y
{\displaystyle V=\pi \int _{y_{1}}^{y_{2}}(f(y))^{2}\,dy}
Dua kurva
Sumbu x sebagai poros
V
=
π
∫
x
1
x
2
(
(
f
(
x
2
)
)
2
−
(
f
(
x
1
)
)
2
)
d
x
{\displaystyle V=\pi \int _{x_{1}}^{x_{2}}((f(x_{2}))^{2}-(f(x_{1}))^{2})\,dx}
Sumbu y sebagai poros
V
=
π
∫
y
1
y
2
(
(
f
(
y
2
)
)
2
−
(
f
(
y
1
)
)
2
)
d
y
{\displaystyle V=\pi \int _{y_{1}}^{y_{2}}((f(y_{2}))^{2}-(f(y_{1}))^{2})\,dy}
Jenis integral lipat yaitu integral lipat dua dan integral lipat tiga.
contoh Tentukan hasil dari:
∫
(
5
x
−
1
)
(
5
x
2
−
2
x
+
7
)
4
d
x
{\displaystyle \int (5x-1)(5x^{2}-2x+7)^{4}dx}
∫
x
2
c
o
s
3
x
d
x
{\displaystyle \int x^{2}cos3xdx}
∫
s
i
n
2
x
c
o
s
x
d
x
{\displaystyle \int sin^{2}xcosxdx}
∫
c
o
s
2
x
s
i
n
x
d
x
{\displaystyle \int cos^{2}xsinxdx}
∫
s
e
c
x
d
x
{\displaystyle \int secxdx}
∫
c
s
c
x
d
x
{\displaystyle \int cscxdx}
∫
1
2
x
2
−
5
x
+
3
d
x
{\displaystyle \int {\frac {1}{2x^{2}-5x+3}}dx}
∫
1
50
−
288
x
2
d
x
{\displaystyle \int {\frac {1}{\sqrt {50-288x^{2}}}}dx}
∫
1
12
+
75
x
2
d
x
{\displaystyle \int {\frac {1}{12+75x^{2}}}dx}
∫
1
11
x
605
x
2
−
245
d
x
{\displaystyle \int {\frac {1}{11x{\sqrt {605x^{2}-245}}}}dx}
jawaban Jawaban
misalkan
u
=
5
x
2
−
2
x
+
7
u
=
5
x
2
−
2
x
+
7
d
u
=
10
x
−
2
d
x
=
2
(
5
x
−
1
)
d
x
d
u
2
=
(
5
x
−
1
)
d
x
∫
(
5
x
−
1
)
(
5
x
2
−
2
x
+
7
)
4
d
x
=
∫
1
2
u
4
d
u
=
1
2
1
5
u
5
+
C
=
1
10
(
5
x
2
−
2
x
+
7
)
5
+
C
{\displaystyle {\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
gunakan integral parsial
∫
x
2
c
o
s
3
x
d
x
=
x
2
3
s
i
n
3
x
−
2
x
9
(
−
c
o
s
3
x
)
+
2
27
(
−
s
i
n
3
x
)
+
C
=
x
2
3
s
i
n
3
x
+
2
x
9
c
o
s
3
x
−
2
27
s
i
n
3
x
+
C
{\displaystyle {\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
∗
cara 1
misalkan
u
=
s
i
n
x
u
=
s
i
n
x
d
u
=
c
o
s
x
d
x
∫
s
i
n
2
x
c
o
s
x
d
x
=
∫
u
2
d
u
=
u
3
3
+
C
=
s
i
n
3
x
3
+
C
∗
cara 2
∫
s
i
n
2
x
c
o
s
x
d
x
=
∫
1
−
c
o
s
2
x
2
c
o
s
x
d
x
=
1
2
∫
(
1
−
c
o
s
2
x
)
c
o
s
x
d
x
=
1
2
∫
c
o
s
x
−
c
o
s
2
x
c
o
s
x
d
x
=
1
2
∫
c
o
s
x
−
1
2
(
c
o
s
3
x
+
c
o
x
x
)
d
x
=
1
2
∫
c
o
s
x
−
1
2
c
o
s
3
x
−
1
2
c
o
s
x
d
x
=
1
2
∫
1
2
c
o
s
x
−
1
2
c
o
s
3
x
d
x
=
1
2
(
1
2
s
i
n
x
−
1
6
s
i
n
3
x
)
+
C
=
s
i
n
x
4
−
s
i
n
3
x
12
+
C
∗
cara 3
∫
s
i
n
2
x
c
o
s
x
d
x
=
∫
s
i
n
x
s
i
n
x
c
o
s
x
d
x
=
1
2
∫
s
i
n
x
s
i
n
2
x
d
x
=
1
4
∫
−
(
c
o
s
3
x
−
c
o
s
(
−
x
)
)
d
x
=
1
4
∫
−
c
o
s
3
x
+
c
o
s
x
d
x
=
1
4
(
−
s
i
n
3
x
3
+
s
i
n
x
)
+
C
=
−
s
i
n
3
x
12
+
s
i
n
x
4
+
C
{\displaystyle {\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
∗
cara 1
misalkan
u
=
c
o
s
x
u
=
c
o
s
x
d
u
=
−
s
i
n
x
d
x
−
d
u
=
s
i
n
x
d
x
∫
c
o
s
2
x
s
i
n
x
d
x
=
−
∫
u
2
d
u
=
−
u
3
3
+
C
=
−
c
o
s
3
x
3
+
C
∗
cara 2
∫
c
o
s
2
x
s
i
n
x
d
x
=
∫
c
o
s
2
x
+
1
2
s
i
n
x
d
x
=
1
2
∫
(
c
o
s
2
x
+
1
)
s
i
n
x
d
x
=
1
2
∫
c
o
s
2
x
s
i
n
x
+
s
i
n
x
d
x
=
1
2
∫
1
2
(
s
i
n
3
x
−
s
i
n
x
)
+
s
i
n
x
d
x
=
1
2
∫
1
2
s
i
n
3
x
−
1
2
s
i
n
x
+
s
i
n
x
d
x
=
1
2
∫
1
2
s
i
n
3
x
+
1
2
s
i
n
x
d
x
=
1
2
(
−
1
6
c
o
s
3
x
−
1
2
c
o
s
x
)
+
C
=
−
c
o
s
3
x
12
−
c
o
s
x
4
+
C
∗
cara 3
∫
c
o
s
2
x
s
i
n
x
d
x
=
∫
c
o
s
x
c
o
s
x
s
i
n
x
d
x
=
1
2
∫
c
o
s
x
s
i
n
2
x
d
x
=
1
4
∫
(
s
i
n
3
x
−
s
i
n
(
−
x
)
)
d
x
=
1
4
∫
s
i
n
3
x
+
s
i
n
x
d
x
=
1
4
(
−
c
o
s
3
x
3
−
c
o
s
x
)
+
C
=
−
c
o
s
3
x
12
−
c
o
s
x
4
+
C
{\displaystyle {\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
∫
s
e
c
x
d
x
=
∫
s
e
c
x
s
e
c
x
+
t
a
n
x
s
e
c
x
+
t
a
n
x
d
x
=
∫
s
e
c
2
x
+
s
e
c
x
t
a
n
x
s
e
c
x
+
t
a
n
x
d
x
misalkan
u
=
s
e
c
x
+
t
a
n
x
u
=
s
e
c
x
+
t
a
n
x
d
u
=
s
e
c
x
t
a
n
x
+
s
e
c
2
x
d
x
=
∫
s
e
c
2
x
+
s
e
c
x
t
a
n
x
s
e
c
x
+
t
a
n
x
d
x
=
∫
1
u
d
u
=
l
n
u
+
C
=
l
n
|
s
e
c
x
+
t
a
n
x
|
+
C
{\displaystyle {\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
∫
c
s
c
x
d
x
=
∫
c
s
c
x
c
s
c
x
+
c
o
t
x
c
s
c
x
+
c
o
t
x
d
x
=
∫
c
s
c
2
x
+
c
s
c
x
c
o
t
x
c
s
c
x
+
c
o
t
x
d
x
misalkan
u
=
c
s
c
x
+
c
o
t
x
u
=
c
s
c
x
+
c
o
t
x
d
u
=
−
c
s
c
x
c
o
t
x
−
c
s
c
2
x
d
x
−
d
u
=
c
s
c
x
c
o
t
x
+
c
s
c
2
x
d
x
=
∫
c
s
c
2
x
+
c
s
c
x
c
o
t
x
c
s
c
x
+
c
o
t
x
d
x
=
−
∫
1
u
d
u
=
−
l
n
u
+
C
=
−
l
n
|
c
s
c
x
+
c
o
t
x
|
+
C
{\displaystyle {\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
∫
1
2
x
2
−
5
x
+
3
d
x
=
∫
1
(
2
x
−
3
)
(
x
−
1
)
d
x
=
∫
(
A
(
2
x
−
3
)
+
B
(
x
−
1
)
)
d
x
=
∫
(
A
(
x
−
1
)
+
B
(
2
x
−
3
)
(
2
x
−
3
)
(
x
−
1
)
)
d
x
=
∫
(
A
x
−
A
+
2
B
x
−
3
B
2
x
2
−
5
x
+
3
)
d
x
=
∫
(
(
A
+
2
B
)
x
+
(
−
A
−
3
B
)
2
x
2
−
5
x
+
3
)
d
x
cari nilai A dan B dari 0=A+2B dan 1=-A-3B adalah 2 dan -1
=
∫
(
A
(
2
x
−
3
)
+
B
(
x
−
1
)
)
d
x
=
∫
(
2
(
2
x
−
3
)
+
−
1
(
x
−
1
)
)
d
x
=
l
n
|
2
x
−
3
|
−
l
n
|
x
−
1
|
+
C
=
l
n
|
2
x
−
3
x
−
1
|
+
C
{\displaystyle {\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
∫
1
50
−
288
x
2
d
x
x
=
5
12
s
i
n
A
d
x
=
5
12
c
o
s
A
d
A
∫
1
50
−
288
x
2
d
x
=
∫
5
12
c
o
s
A
50
−
288
(
5
12
s
i
n
A
)
2
d
A
=
∫
5
12
c
o
s
A
50
−
288
⋅
25
144
s
i
n
2
A
d
A
=
∫
5
12
c
o
s
A
50
−
50
s
i
n
2
A
d
A
=
∫
5
12
c
o
s
A
50
(
1
−
s
i
n
2
A
)
d
A
=
∫
5
12
c
o
s
A
50
c
o
s
2
A
d
A
=
∫
5
12
c
o
s
A
5
c
o
s
A
2
d
A
=
∫
2
24
d
A
=
2
24
∫
d
A
=
A
+
C
A
=
a
r
c
s
i
n
12
x
5
=
2
24
a
r
c
s
i
n
12
x
5
+
C
{\displaystyle {\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
∫
1
12
+
75
x
2
d
x
x
=
2
5
t
a
n
A
d
x
=
2
5
s
e
c
2
A
d
A
∫
1
12
+
75
x
2
d
x
=
∫
2
5
s
e
c
2
A
12
+
75
(
2
5
t
a
n
A
)
2
d
A
=
∫
2
5
s
e
c
2
A
12
+
75
⋅
4
25
t
a
n
2
A
d
A
=
∫
2
5
s
e
c
2
A
12
+
12
t
a
n
2
A
d
A
=
∫
2
5
s
e
c
2
A
12
(
1
+
t
a
n
2
A
)
d
A
=
∫
2
5
s
e
c
2
A
12
s
e
c
2
A
d
A
=
∫
1
30
d
A
=
A
30
+
C
A
=
a
r
c
t
a
n
5
x
2
=
a
r
c
t
a
n
5
x
2
30
+
C
{\displaystyle {\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
∫
1
11
x
605
x
2
−
245
d
x
x
=
7
11
s
e
c
A
d
x
=
7
11
s
e
c
A
t
a
n
A
d
A
∫
1
11
x
605
x
2
−
245
d
x
=
∫
7
11
s
e
c
A
t
a
n
A
11
(
7
11
s
e
c
A
)
605
(
7
11
s
e
c
A
)
2
−
245
d
A
=
∫
s
e
c
A
t
a
n
A
11
s
e
c
A
605
⋅
49
121
s
e
c
2
A
−
245
d
A
=
∫
s
e
c
A
t
a
n
A
11
s
e
c
A
245
s
e
c
2
A
−
245
d
A
=
∫
s
e
c
A
t
a
n
A
11
s
e
c
A
245
(
s
e
c
2
A
−
1
)
d
A
=
∫
s
e
c
A
t
a
n
A
11
s
e
c
A
245
t
a
n
2
A
d
A
=
∫
s
e
c
A
t
a
n
A
77
5
s
e
c
A
t
a
n
A
d
A
=
∫
5
385
d
A
=
A
5
385
+
C
A
=
a
r
c
s
e
c
11
x
7
=
5
a
r
c
s
e
c
11
x
7
385
+
C
{\displaystyle {\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}}}
Tentukan hasil dari:
∫
1
5
2
x
−
3
d
x
{\displaystyle \int _{1}^{5}2x-3dx}
∫
0
4
|
x
−
2
|
d
x
{\displaystyle \int _{0}^{4}|x-2|dx}
∫
2
4
|
3
−
x
|
d
x
{\displaystyle \int _{2}^{4}|3-x|dx}
∫
1
5
x
2
−
6
x
+
8
d
x
{\displaystyle \int _{1}^{5}x^{2}-6x+8dx}
∫
−
3
5
|
x
2
−
2
x
−
8
|
d
x
{\displaystyle \int _{-3}^{5}|x^{2}-2x-8|dx}
jawaban Jawaban
∫
1
5
2
x
−
3
d
x
=
x
2
−
3
x
|
1
5
=
10
−
(
−
2
)
=
12
{\displaystyle {\begin{aligned}\int _{1}^{5}2x-3dx&=\left.x^{2}-3x\right|_{1}^{5}\\&=10-(-2)\\&=12\\\end{aligned}}}
Jawaban
∫
0
4
|
x
−
2
|
d
x
tentukan nilai harga nol
x
−
2
=
0
x
=
2
buatlah batas-batas wilayah yaitu kurang dari 2 bernilai - sedangkan minimal dari 2 bernilai + berarti
|
x
−
2
|
=
{
x
−
2
,
jika
x
≥
2
,
−
(
x
−
2
)
,
jika
x
<
2.
∫
0
4
|
x
−
2
|
d
x
=
∫
2
4
x
−
2
d
x
+
∫
0
2
−
(
x
−
2
)
d
x
=
(
x
2
2
−
2
x
)
|
2
4
+
(
−
x
2
2
+
2
x
)
|
0
2
=
0
−
(
−
2
)
+
2
−
0
=
4
{\displaystyle {\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
∫
2
4
|
3
−
x
|
d
x
tentukan nilai syarat-syarat
3
−
x
≥
0
harga nol
x
=
3
buatlah batas-batas wilayah yaitu lebih dari 3 bernilai - sedangkan maksimal dari 3 bernilai + berarti
|
3
−
x
|
=
{
3
−
x
,
jika
x
≤
3
,
−
(
3
−
x
)
,
jika
x
>
3.
∫
2
4
|
3
−
x
|
d
x
=
∫
2
3
3
−
x
d
x
+
∫
3
4
−
(
3
−
x
)
d
x
=
(
3
x
−
x
2
2
)
|
2
3
+
(
−
3
x
+
x
2
2
)
|
3
4
=
4.5
−
4
−
4
−
(
−
4.5
)
=
1
{\displaystyle {\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
∫
1
5
x
2
−
6
x
+
8
d
x
=
x
3
3
−
3
x
2
+
8
x
|
1
5
=
20
3
−
16
3
=
4
3
=
1
1
3
{\displaystyle {\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
∫
−
3
5
|
x
2
−
2
x
−
8
|
d
x
tentukan nilai syarat-syarat
x
2
−
2
x
−
8
≥
0
harga nol
(
x
+
2
)
(
x
−
4
)
=
0
x
=
−
2
atau
x
=
4
buatlah batas-batas wilayah yaitu antara -2 dan 4 bernilai - sedangkan maksimal dari -2 dan minimal dari 4 bernilai + berarti
|
x
2
−
2
x
−
8
|
=
{
x
2
−
2
x
−
8
,
jika
x
≤
−
2
,
−
(
x
2
−
2
x
−
8
)
,
jika
−
2
<
x
<
4
,
x
2
−
2
x
−
8
,
jika
x
≥
4
∫
−
3
5
|
x
2
−
2
x
−
8
|
d
x
=
∫
−
3
−
2
x
2
−
2
x
−
8
d
x
+
∫
−
2
4
−
(
x
2
−
2
x
−
8
)
d
x
+
∫
4
5
x
2
−
2
x
−
8
d
x
=
(
x
3
3
−
x
2
−
8
x
)
|
−
3
−
2
+
(
−
x
3
3
+
x
2
+
8
x
)
|
−
2
4
+
(
x
3
3
−
x
2
−
8
x
)
|
4
5
=
28
3
−
6
+
80
3
−
(
−
28
3
)
−
70
3
−
(
−
80
3
)
=
128
3
=
42
2
3
{\displaystyle {\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}}}
Tentukan luas (tak tentu) dengan persamaan garis
y
=
x
{\displaystyle y=x}
y dengan cara integral! Jawaban
L
=
∫
x
d
x
L
=
1
2
x
2
L
=
1
2
x
y
(
y
=
x
)
{\displaystyle {\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
{\displaystyle y=x^{2}}
y dengan cara integral! Jawaban
L
=
∫
x
2
d
x
L
=
1
3
x
3
L
=
1
3
x
y
(
y
=
x
2
)
{\displaystyle {\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
=
x
{\displaystyle y={\sqrt {x}}}
y dengan cara integral! Jawaban
L
=
∫
x
d
x
L
=
2
3
x
3
2
L
=
2
3
x
y
(
y
=
x
)
{\displaystyle {\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
{\displaystyle L=s^{2}}
Jawaban
dengan posisi
y
=
s
dan titik (''s'', ''s''),
L
=
∫
0
s
s
d
x
L
=
s
x
|
0
s
L
=
s
s
−
0
L
=
s
2
{\displaystyle {\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}}}
Buktikan luas persegi panjang
L
=
p
l
{\displaystyle L=pl}
Jawaban
Dengan posisi
y
=
l
dan titik (''p'', ''l''),
L
=
∫
0
p
l
d
x
L
=
l
x
|
0
p
L
=
p
l
−
0
L
=
p
l
{\displaystyle {\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
=
a
t
2
{\displaystyle L={\frac {at}{2}}}
Jawaban
dengan posisi
y
=
−
t
x
a
+
t
dan titik (''a'', ''t''),
L
=
∫
0
a
(
−
t
x
a
+
t
)
d
x
L
=
−
t
x
2
2
a
+
t
x
|
0
a
L
=
−
t
a
2
2
a
+
t
a
−
0
+
0
L
=
−
t
a
2
+
t
a
L
=
a
t
2
{\displaystyle {\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}}}
Buktikan volume tabung
V
=
π
r
2
t
{\displaystyle V=\pi r^{2}t}
Jawaban
Dengan posisi
y
=
r
dan titik (''t'', ''r''),
V
=
π
∫
0
t
r
2
d
x
V
=
π
r
2
x
|
0
t
V
=
π
r
2
t
−
0
V
=
π
r
2
t
{\displaystyle {\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}}}
Buktikan volume kerucut
V
=
π
r
2
t
3
{\displaystyle V={\frac {\pi r^{2}t}{3}}}
Jawaban
dengan posisi
y
=
r
x
t
dan titik (''t'', ''r''),
V
=
π
∫
0
t
(
r
x
t
)
2
d
x
V
=
π
r
2
x
3
3
t
2
|
0
t
V
=
π
r
2
t
3
3
t
2
−
0
V
=
π
r
2
t
3
{\displaystyle {\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}}}
Buktikan volume bola
V
=
4
π
r
3
3
{\displaystyle V={\frac {4\pi r^{3}}{3}}}
Jawaban
dengan posisi
y
=
r
2
−
x
2
serta titik (-''r'', 0) dan (''r'', 0),
V
=
π
∫
−
r
r
(
r
2
−
x
2
)
2
d
x
V
=
π
∫
−
r
r
r
2
−
x
2
d
x
V
=
π
r
2
x
−
x
3
3
|
−
r
r
V
=
π
(
r
3
−
r
3
3
−
(
−
r
3
+
r
3
3
)
)
V
=
4
π
r
3
3
{\displaystyle {\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}}}
Buktikan luas permukaan bola
L
=
4
π
r
2
{\displaystyle L=4\pi r^{2}}
Jawaban
dengan posisi
y
=
r
2
−
x
2
serta titik (-''r'', 0) dan (''r'', 0), Kita tahu bahwa turunannya adalah
y
′
=
−
x
r
2
−
x
2
selanjutnya
d
s
=
1
+
(
−
x
r
2
−
x
2
)
2
d
x
d
s
=
1
+
x
2
r
2
−
x
2
d
x
d
s
=
r
2
r
2
−
x
2
d
x
d
s
=
r
r
2
−
x
2
d
x
sehingga
L
=
2
π
∫
−
r
r
r
2
−
x
2
⋅
r
r
2
−
x
2
d
x
L
=
2
π
∫
−
r
r
r
d
x
L
=
2
π
r
x
|
−
r
r
L
=
2
π
(
r
r
−
r
(
−
r
)
)
L
=
2
π
(
r
2
+
r
2
)
L
=
4
π
r
2
{\displaystyle {\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}}}
Buktikan keliling lingkaran
K
=
2
π
r
{\displaystyle K=2\pi r}
Jawaban
dengan posisi
y
=
r
2
−
x
2
serta titik (-''r'', 0) dan (''r'', 0)
kita tahu bahwa turunannya adalah
y
′
=
−
x
r
2
−
x
2
sehingga
K
=
4
∫
0
r
1
+
(
−
x
r
2
−
x
2
)
2
d
x
K
=
4
∫
0
r
1
+
(
x
2
r
2
−
x
2
)
d
x
K
=
4
∫
0
r
r
2
r
2
−
x
2
d
x
K
=
4
r
∫
0
r
1
r
2
−
x
2
d
x
K
=
4
r
∫
0
r
1
r
2
−
x
2
d
x
K
=
4
r
arcsin
(
x
r
)
|
0
r
K
=
4
r
(
arcsin
(
r
r
)
−
arcsin
(
0
r
)
)
K
=
4
r
(
arcsin
(
1
)
−
arcsin
(
0
)
)
)
K
=
4
r
(
π
2
)
K
=
2
π
r
{\displaystyle {\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}}}
Buktikan luas lingkaran
L
=
π
r
2
{\displaystyle L=\pi r^{2}}
Jawaban
Dengan posisi
y
=
r
2
−
x
2
serta titik (-''r'', 0) dan (''r'', 0), dibuat trigonometri dan turunannya terlebih dahulu.
sin
(
θ
)
=
x
r
x
=
r
sin
(
θ
)
d
x
=
r
cos
(
θ
)
d
θ
dengan turunan di atas
L
=
4
∫
0
r
r
2
−
x
2
d
x
L
=
4
∫
0
r
r
2
−
(
r
sin
(
θ
)
)
2
d
x
L
=
4
∫
0
r
r
2
−
r
2
sin
2
(
θ
)
d
x
L
=
4
∫
0
r
r
2
(
1
−
sin
2
(
θ
)
)
d
x
L
=
4
∫
0
r
r
cos
(
θ
)
d
x
L
=
4
∫
0
r
r
cos
(
θ
)
(
r
cos
(
θ
)
d
θ
)
L
=
4
∫
0
r
r
2
cos
2
(
θ
)
d
θ
L
=
4
r
2
∫
0
r
(
1
+
cos
(
2
θ
)
2
)
d
θ
L
=
2
r
2
∫
0
r
(
1
+
cos
(
2
θ
)
)
d
θ
L
=
2
r
2
(
θ
+
1
2
sin
(
2
θ
)
)
|
0
r
L
=
2
r
2
(
θ
+
sin
(
θ
)
cos
(
θ
)
)
|
0
r
L
=
2
r
2
(
arcsin
(
x
r
)
+
(
x
r
)
(
r
2
−
x
2
r
)
)
|
0
r
L
=
2
r
2
(
arcsin
(
r
r
)
+
(
r
r
)
(
r
2
−
r
2
r
)
)
−
(
arcsin
(
0
r
)
+
(
0
r
)
(
r
2
−
0
2
r
)
)
L
=
2
r
2
(
arcsin
(
1
)
+
0
−
(
arcsin
(
0
)
+
0
)
)
L
=
2
r
2
(
π
2
)
L
=
π
r
2
{\displaystyle {\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}}}
Buktikan luas elips
L
=
π
a
b
{\displaystyle L=\pi ab}
Jawaban
Dengan posisi
y
=
b
a
2
−
x
2
a
serta (-''a'', 0) dan (''a'', 0),
L
=
4
∫
0
r
b
a
2
−
x
2
a
d
x
L
=
4
b
a
∫
0
a
a
2
−
x
2
d
x
dengan anggapan bahwa lingkaran mempunyai memotong titik (-''a'', 0) dan (''a'', 0) serta pusatnya setitik dengan pusat elips,
L
elips
L
ling
=
4
b
a
∫
0
a
a
2
−
x
2
d
x
4
∫
0
a
a
2
−
x
2
d
x
L
elips
L
ling
=
b
a
L
elips
=
b
a
L
ling
L
elips
=
b
a
π
a
2
L
elips
=
π
a
b
{\displaystyle {\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}}}
Berapa luas daerah yang dibatasi y=x2 -2x dan y=4x+7! Jawaban
cari titik potong dari kedua persamaan tersebut
y
1
=
y
2
x
−
2
x
=
4
x
+
7
x
2
−
6
x
−
7
=
0
(
x
+
1
)
(
x
−
7
)
=
0
x
=
−
1
atau
x
=
7
y
=
4
(
−
1
)
+
7
=
3
y
=
4
(
7
)
+
7
=
35
jadi titik potong (-1,3) dan (7,35) kemudian buatlah gambar kedua persamaan tersebut
L
=
∫
−
1
7
4
x
+
7
−
(
x
2
−
2
x
)
d
x
=
∫
−
1
7
−
x
2
+
6
x
+
7
d
x
=
−
x
3
3
+
3
x
2
+
7
x
|
−
1
7
=
−
7
3
3
+
3
(
7
)
2
+
7
(
7
)
−
(
−
(
−
1
)
3
3
+
3
(
−
1
)
2
+
7
(
−
1
)
)
=
245
3
−
(
−
13
3
)
=
258
3
=
86
{\displaystyle {\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}}}
Berapa volume benda putar yang dibatasi y=6-1,5x, y=x-4 dan x=0 mengelilingi sumbu x sejauh 360°! Jawaban
cari titik potong dari kedua persamaan tersebut
y
1
=
y
2
6
−
1
,
5
x
=
x
−
4
2
,
5
x
=
10
x
=
4
y
=
4
−
4
=
0
jadi titik potong (4,0) kemudian buatlah gambar kedua persamaan tersebut
untuk x=0
y
=
6
−
1
,
5
(
0
)
=
6
y
=
0
−
4
=
−
4
L
=
π
∫
0
4
(
(
6
−
1
,
5
x
)
2
−
(
x
−
4
)
2
)
d
x
=
π
∫
0
4
(
36
−
18
x
+
2
,
25
x
2
−
(
x
2
−
8
x
+
16
)
)
d
x
=
π
∫
0
4
(
1
,
25
x
2
−
10
x
+
20
)
d
x
=
π
(
1
,
25
x
3
3
−
5
x
2
+
20
x
)
|
0
4
=
π
(
1
,
25
(
4
)
3
3
−
5
(
4
)
2
+
20
(
4
)
−
(
1
,
25
(
0
)
3
3
−
5
(
0
)
2
+
20
(
0
)
)
)
=
π
(
30
3
−
0
)
=
10
π
{\displaystyle {\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}}}
![{\displaystyle \int [f(x)+g(x)]\,dx=\int f(x)\,dx+\int g(x)\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5dc2c6e2acadf432d22ca42bc6a21af25e48e64d)
![{\displaystyle \int [f(x)-g(x)]\,dx=\int f(x)\,dx-\int g(x)\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5c5f0ce4c12f5715703d07f46dbd9e067acdd3f)
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![{\displaystyle \int [f(x)]^{n}f'(x)\,dx={[f(x)]^{n+1} \over n+1}+C\qquad {\mbox{(untuk }}n\neq -1{\mbox{)}}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fee3d8370de7075ba5cd461320474837c6282f46)
![{\displaystyle \int {f'(x)f(x)}\,dx={1 \over 2}[f(x)]^{2}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/55ddd18a718860218bb3d10a0ccc72af8e420c58)
