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微积分学
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不定积分
∫
3
x
2
d
x
{\displaystyle \int {\frac {3x}{2}}dx}
解答：欲找到函数
F
{\displaystyle F}
，使
F
′
(
x
)
=
3
x
2
{\displaystyle F'(x)={\frac {3x}{2}}}
由于
d
d
x
x
2
=
2
x
{\displaystyle {\frac {d}{dx}}x^{2}=2x}
因此须找到常数
a
{\displaystyle a}
，使
d
d
x
a
x
2
=
2
a
x
=
3
x
2
{\displaystyle {\frac {d}{dx}}ax^{2}=2ax={\frac {3x}{2}}}
解出
a
{\displaystyle a}
，得
2
a
x
=
3
x
2
⟹
a
=
3
4
{\displaystyle 2ax={\frac {3x}{2}}\implies a={\frac {3}{4}}}
故
∫
3
x
2
=
3
4
x
2
+
C
{\displaystyle \int {\frac {3x}{2}}={\frac {3}{4}}x^{2}+C}
求导验证：
d
d
x
(
3
4
x
2
+
C
)
=
3
4
(
2
x
)
=
3
x
2
{\displaystyle {\frac {d}{dx}}\left({\frac {3}{4}}x^{2}+C\right)={\frac {3}{4}}(2x)={\frac {3x}{2}}}
求
f
(
x
)
=
2
x
4
{\displaystyle f(x)=2x^{4}}
的原函数
解答：由于
d
d
x
x
5
=
5
x
4
{\displaystyle {\frac {d}{dx}}x^{5}=5x^{4}}
须找到常数
a
{\displaystyle a}
，使
d
d
x
a
x
5
=
5
a
x
4
=
2
x
4
{\displaystyle {\frac {d}{dx}}ax^{5}=5ax^{4}=2x^{4}}
解出
a
{\displaystyle a}
，得
5
a
x
4
=
2
x
4
⟹
a
=
2
5
{\displaystyle 5ax^{4}=2x^{4}\implies a={\frac {2}{5}}}
故原函数为
2
5
x
5
+
C
{\displaystyle \mathbf {{\frac {2}{5}}x^{5}+C} }
求导验证：
d
d
x
∫
2
x
4
d
x
=
d
d
x
(
2
5
x
5
+
C
)
=
2
5
(
5
x
4
)
=
2
x
4
{\displaystyle {\frac {d}{dx}}\int 2x^{4}dx={\frac {d}{dx}}({\frac {2}{5}}x^{5}+C)={\frac {2}{5}}(5x^{4})=2x^{4}}
∫
(
7
x
2
+
3
cos
x
−
e
x
)
d
x
{\displaystyle \int (7x^{2}+3\cos x-e^{x})dx}
解答：
∫
(
7
x
2
+
3
cos
x
−
e
x
)
d
x
=
7
∫
x
2
d
x
+
3
∫
cos
x
d
x
−
∫
e
x
d
x
=
7
(
x
3
3
)
+
3
sin
x
−
e
x
+
C
=
7
3
x
3
+
3
sin
x
−
e
x
+
C
{\displaystyle {\begin{aligned}\int (7x^{2}+3\cos x-e^{x})dx&=7\int x^{2}dx+3\int \cos xdx-\int e^{x}dx\\&=7({\frac {x^{3}}{3}})+3\sin x-e^{x}+C\\&=\mathbf {{\frac {7}{3}}x^{3}+3\sin x-e^{x}+C} \end{aligned}}}
∫
(
2
5
x
+
sin
x
)
d
x
{\displaystyle \int ({\frac {2}{5x}}+\sin x)dx}
解答：
∫
(
2
5
x
+
sin
x
)
d
x
=
2
5
∫
d
x
x
+
∫
sin
x
d
x
=
2
5
ln
|
x
|
−
cos
x
+
C
{\displaystyle {\begin{aligned}\int ({\frac {2}{5x}}+\sin x)dx&={\frac {2}{5}}\int {\frac {dx}{x}}+\int \sin xdx\\&=\mathbf {{\frac {2}{5}}\ln |x|-\cos x+C} \end{aligned}}}
∫
x
sin
2
x
2
d
x
{\displaystyle \int x\sin 2x^{2}dx}
解答：令
u
=
2
x
2
{\displaystyle u=2x^{2}}
，
d
u
=
4
x
d
x
{\displaystyle du=4xdx}
，
d
x
=
d
u
4
x
{\displaystyle dx={\frac {du}{4x}}}
∫
x
sin
2
x
2
d
x
=
∫
x
sin
u
d
u
4
x
=
1
4
∫
sin
u
d
u
=
−
cos
u
4
+
C
=
−
cos
2
x
2
4
+
C
{\displaystyle {\begin{aligned}\int x\sin 2x^{2}dx&=\int x\sin u{\frac {du}{4x}}\\&={\frac {1}{4}}\int \sin udu\\&=-{\frac {\cos u}{4}}+C\\&=-\mathbf {{\frac {\cos 2x^{2}}{4}}+C} \end{aligned}}}
∫
−
3
e
sin
x
cos
x
d
x
{\displaystyle \int -3e^{\sin x}\cos xdx}
解答：令
u
=
sin
x
{\displaystyle u=\sin x}
，
d
u
=
cos
x
d
x
{\displaystyle du=\cos xdx}
，则
d
x
=
d
u
cos
x
{\displaystyle dx={\frac {du}{\cos x}}}
∫
−
3
e
sin
x
cos
x
d
x
=
−
3
∫
e
u
cos
x
d
u
cos
x
=
−
3
∫
e
u
d
u
=
−
3
e
u
+
C
=
−
3
e
sin
x
+
C
{\displaystyle {\begin{aligned}\int -3e^{\sin x}\cos xdx&=-3\int e^{u}\cos x{\frac {du}{\cos x}}\\&=-3\int e^{u}du\\&=-3e^{u}+C\\&=\mathbf {-3e^{\sin x}+C} \end{aligned}}}
∫
2
x
−
5
x
3
d
x
{\displaystyle \int {\frac {2x-5}{x^{3}}}dx}
解答：令
u
=
2
(
x
−
5
)
{\displaystyle u=2(x-5)}
，
v
=
∫
d
x
x
3
=
−
1
2
x
2
{\displaystyle v=\int {\frac {dx}{x^{3}}}=-{\frac {1}{2x^{2}}}}
∫
2
x
−
5
x
3
d
x
=
∫
u
d
v
=
u
v
−
∫
v
d
u
=
(
2
x
−
5
)
(
−
1
2
x
2
)
−
∫
(
−
1
2
x
2
)
2
d
x
=
5
−
2
x
2
x
2
+
∫
d
x
x
2
=
5
−
2
x
2
x
2
−
1
x
=
5
−
2
x
2
x
2
−
2
x
2
x
2
=
5
−
4
x
2
x
2
{\displaystyle {\begin{aligned}\int {\frac {2x-5}{x^{3}}}dx&=\int udv\\&=uv-\int vdu\\&=(2x-5)(-{\frac {1}{2x^{2}}})-\int (-{\frac {1}{2x^{2}}})2dx\\&={\frac {5-2x}{2x^{2}}}+\int {\frac {dx}{x^{2}}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {1}{x}}\\&={\frac {5-2x}{2x^{2}}}-{\frac {2x}{2x^{2}}}\\&=\mathbf {\frac {5-4x}{2x^{2}}} \end{aligned}}}
∫
(
2
x
−
1
)
e
−
3
x
+
1
d
x
{\displaystyle \int (2x-1)e^{-3x+1}dx}
解答：令
u
=
2
x
−
1
{\displaystyle u=2x-1}
，
d
v
=
e
−
3
x
+
1
d
x
{\displaystyle dv=e^{-3x+1}dx}
则
d
u
=
2
d
x
{\displaystyle du=2dx}
，
v
=
∫
e
−
3
x
+
1
d
x
{\displaystyle v=\int e^{-3x+1}dx}
欲求
v
{\displaystyle v}
，令
w
=
−
3
x
+
1
{\displaystyle w=-3x+1}
，
d
w
=
−
3
d
x
{\displaystyle dw=-3dx}
，
d
x
=
−
d
w
3
{\displaystyle dx={\frac {-dw}{3}}}
，则
v
=
∫
e
−
3
x
+
1
d
x
=
∫
e
w
(
−
1
3
)
d
w
=
−
e
w
3
=
−
e
−
3
x
+
1
3
{\displaystyle v=\int e^{-3x+1}dx=\int e^{w}({\frac {-1}{3}})dw={\frac {-e^{w}}{3}}={\frac {-e^{-3x+1}}{3}}}
，故
∫
(
2
x
−
1
)
e
−
3
x
+
1
d
x
=
∫
u
d
v
=
u
v
−
∫
v
d
u
=
(
2
x
−
1
)
−
e
−
3
x
+
1
3
−
∫
−
e
−
3
x
+
1
3
(
2
)
d
x
=
(
1
−
2
x
)
e
−
3
x
+
1
3
+
2
3
∫
e
−
3
x
+
1
d
x
=
(
1
−
2
x
)
e
−
3
x
+
1
3
+
2
3
∫
−
e
w
3
d
w
=
3
(
1
−
2
x
)
e
−
3
x
+
1
9
−
2
9
e
w
=
(
3
−
6
x
)
e
−
3
x
+
1
9
−
2
9
e
−
3
x
+
1
=
(
1
−
6
x
)
e
−
3
x
+
1
9
{\displaystyle {\begin{aligned}\int (2x-1)e^{-3x+1}dx&=\int udv\\&=uv-\int vdu\\&=(2x-1){\frac {-e^{-3x+1}}{3}}-\int {\frac {-e^{-3x+1}}{3}}(2)dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int e^{-3x+1}dx\\&={\frac {(1-2x)e^{-3x+1}}{3}}+{\frac {2}{3}}\int {\frac {-e^{w}}{3}}dw\\&={\frac {3(1-2x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{w}\\&={\frac {(3-6x)e^{-3x+1}}{9}}-{\frac {2}{9}}e^{-3x+1}\\&=\mathbf {\frac {(1-6x)e^{-3x+1}}{9}} \end{aligned}}}
章节导航
:
目录
·
预备知识
·
极限
·
导数
·
积分
·
极坐标方程与参数方程
·
数列和级数
·
多元函数微积分
·
扩展知识
·
附录
