∫
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)}
∫
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
=
1
a
arctan
x
a
+
C
{\displaystyle \int {dx \over {a^{2}+x^{2}}}={1 \over a}\arctan {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
Gagal mengurai (SVG (MathML dapat diaktifkan melalui plugin peramban): Respons tak sah ("Math extension cannot connect to Restbase.") dari peladen "http://localhost:6011/id.wikibooks.org/v1/":): {\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 luas (tak tentu) dengan persamaan garis
y
=
x
{\displaystyle y=x}
y dengan cara integral!
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!
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!
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}}
Dengan posisi
y
=
s
{\displaystyle y=s}
s , s ),
L
=
∫
0
s
s
d
x
L
=
s
x
|
0
s
L
=
s
s
−
0
L
=
s
2
{\displaystyle {\begin{aligned}L&=\int _{0}^{s}s\,dx\\L&=sx|_{0}^{s}\\L&=ss-0\\L&=s^{2}\end{aligned}}}
Buktikan luas persegi panjang
L
=
p
l
{\displaystyle L=pl}
Dengan posisi
y
=
l
{\displaystyle y=l}
p , l ),
L
=
∫
0
p
l
d
x
L
=
l
x
|
0
p
L
=
p
l
−
0
L
=
p
l
{\displaystyle {\begin{aligned}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}}}
Dengan posisi
y
=
−
t
x
a
+
t
{\displaystyle y={\frac {-tx}{a}}+t}
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}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}
Dengan posisi
y
=
r
{\displaystyle y=r}
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}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}}}
Dengan posisi
y
=
r
x
t
{\displaystyle y={\frac {rx}{t}}}
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}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}}}
Dengan posisi
y
=
r
2
−
x
2
{\displaystyle y={\sqrt {r^{2}-x^{2}}}}
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}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}}
Dengan posisi
y
=
r
2
−
x
2
{\displaystyle y={\sqrt {r^{2}-x^{2}}}}
r , 0) dan (r , 0),
Kita tahu bahwa turunannya adalah
y
′
=
−
x
r
2
−
x
2
{\displaystyle y'={\frac {-x}{\sqrt {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
{\displaystyle {\begin{aligned}ds&={\sqrt {1+({\frac {-x}{\sqrt {r^{2}-x^{2}}}})^{2}}}\,dx\\ds&={\sqrt {1+{\frac {x^{2}}{r^{2}-x^{2}}}}}\,dx\\ds&={\sqrt {\frac {r^{2}}{r^{2}-x^{2}}}}\,dx\\ds&={\frac {r}{\sqrt {r^{2}-x^{2}}}}\,dx\\\end{aligned}}}
sehingga
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}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}
Dengan posisi
y
=
r
2
−
x
2
{\displaystyle y={\sqrt {r^{2}-x^{2}}}}
r , 0) dan (r , 0),
Kita tahu bahwa turunannya adalah
y
′
=
−
x
r
2
−
x
2
{\displaystyle y'={\frac {-x}{\sqrt {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}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}}
Dengan posisi
y
=
r
2
−
x
2
{\displaystyle y={\sqrt {r^{2}-x^{2}}}}
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
{\displaystyle {\begin{aligned}\sin(\theta )&={\frac {x}{r}}\\x&=r\sin(\theta )\\dx&=r\cos(\theta )\,d\theta \end{aligned}}}
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}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}
Dengan posisi
y
=
b
a
2
−
x
2
a
{\displaystyle y={\frac {b{\sqrt {a^{2}-x^{2}}}}{a}}}
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
{\displaystyle {\begin{aligned}L&=4\int _{0}^{r}{\frac {b{\sqrt {a^{2}-x^{2}}}}{a}}\,dx\\L&={\frac {4b}{a}}\int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx\end{aligned}}}
Dengan anggapan bahwa lingkaran mempunyai memotong titik (-a , 0) dan (a , 0) serta pusatnya setitik dengan pusat elips,
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}{\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}}}
![{\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)
