最后更新: 2024-10-04 11:13 查看: 367 次反馈刷题

定积分的分部法

## 定积分的分部法 
设函数 $u(x) 、 v(x)$ 在区间 $(a, b)$ 上具有连续导数 $u^{\prime}(x) 、 v^{\prime}(x)$, 由于 $(u v)^{\prime}=u^{\prime} v+u v^{\prime}$, 即 $u v^{\prime}=(u v)^{\prime}-u^{\prime} v$, 因此

$$
\int_a^b u v^{\prime} \mathrm{d} x=\int_a^b(u v)^{\prime} \mathrm{d} x-\int_a^b u^{\prime} v \mathrm{~d} x,
$$

即

$$
\int_a^b u v^{\prime} \mathrm{d} x=\left.u v\right|_a ^b-\int_a^b u^{\prime} v \mathrm{~d} x
$$

或

$$ \boxed {
\int_a^b u \mathrm{~d} v=\left.u v\right|_a ^b-\int_a^b v \mathrm{~d} u .
}
$$

称上式为定积分的分部积分公式.

`例` 求 $\int_0^1 x \ln (1+x) \mathrm{d} x$.
解
$$
 \begin{aligned}
\int_0^1 x \ln (1+x) \mathrm{d} x & =\frac{1}{2} \int_0^1 \ln (1+x) \mathrm{d} x^2 \\
& =\left.\frac{1}{2} x^2 \ln (1+x)\right|_0 ^1-\frac{1}{2} \int_0^1 \frac{x^2}{1+x} \mathrm{~d} x \\
& =\frac{1}{2} \ln 2-\frac{1}{2} \int_0^1\left(x-1+\frac{1}{1+x}\right) \mathrm{d} x \\
& =\frac{1}{2} \ln 2-\left.\frac{1}{2}\left[\frac{1}{2} x^2-x+\ln (1+x)\right]\right|_0 ^1 \\
& =\frac{1}{2} \ln 2-\frac{1}{2}\left(\frac{1}{2}-1+\ln 2\right) \\
& =\frac{1}{4} .
\end{aligned}
$$

`例` 计算 $\int_0^1 \arctan x \mathrm{~d} x$.
解 令 $u=\arctan x, \mathrm{~d} v=\mathrm{d} x$, 则 $\mathrm{d} u=\frac{\mathrm{d} x}{1+x^2}, v=x$,

$$
\begin{aligned}
\int_0^1 \arctan x \mathrm{~d} x & =[x \arctan x]_0^1-\int_0^1 \frac{x \mathrm{~d} x}{1+x^2} \\
& =1 \cdot \frac{\pi}{4}-\frac{1}{2} \int_0^1 \frac{1}{1+x^2} \mathrm{~d}\left(1+x^2\right) \\
& =\frac{\pi}{4}-\left.\frac{1}{2} \ln \left(1+x^2\right)\right|_0 ^1 \\
& =\frac{\pi}{4}+\frac{\ln 2}{2} .
\end{aligned}
$$

`例` 求 $\int_{\frac{1}{e}}^e|\ln x| \mathrm{d} x$.

$$
\text { 解 } \begin{aligned}
\int_{\frac{1}{\mathrm{e}}}^{\mathrm{e}}|\ln x| \mathrm{d} x & =-\int_{\frac{1}{\mathrm{e}}}^1 \ln x \mathrm{~d} x+\int_1^{\mathrm{e}} \ln x \mathrm{~d} x \\
& =-\left.(x \ln x-x)\right|_{\frac{1}{\mathrm{e}}} ^1+\left.(x \ln x-x)\right|_1 ^{\mathrm{e}} \\
& =1-\frac{2}{\mathrm{e}}+1=2\left(1-\frac{1}{\mathrm{e}}\right) .
\end{aligned}
$$

`例`求 $\int_0^{\frac{\pi}{2}} x^2 \sin x \mathrm{~d} x$.
解 由分部积分公式得

$$
\int_0^{\frac{\pi}{2}} x^2 \sin x \mathrm{~d} x=\int_0^{\frac{\pi}{2}} x^2 \mathrm{~d}(-\cos x)=\left.x^2(-\cos x)\right|_0 ^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} \cos x \mathrm{~d}\left(x^2\right)=2 \int_0^{\frac{\pi}{2}} x \cos x \mathrm{~d} x ;
$$

再用一次分部积分公式得

$$
\int_0^{\frac{\pi}{2}} x \cos x \mathrm{~d} x=\int_0^{\frac{\pi}{2}} x \mathrm{~d}(\sin x)=\left.x \sin x\right|_0 ^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \sin x \mathrm{~d} x=\frac{\pi}{2}+\left.\cos x\right|_0 ^{\frac{\pi}{2}}=\frac{\pi}{2}-1 \text {; }
$$

从而

$$
\int_0^{\frac{\pi}{2}} x^2 \sin x \mathrm{~d} x=2 \int_0^{\frac{\pi}{2}} x \cos x \mathrm{~d} x=\pi-2
$$

`例` 求 $\int_0^{\frac{\pi}{2}} \mathrm{e}^x \sin x \mathrm{~d} x$.
解 $\int_0^{\frac{\pi}{2}} \mathrm{e}^x \sin x \mathrm{~d} x=\mathrm{e}^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \mathrm{e}^x \cos x \mathrm{~d} x=\mathrm{e}^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \cos x \mathrm{de}^x$

$$
\begin{aligned}
& =\mathrm{e}^{\frac{\pi}{2}}-\left.\mathrm{e}^x \cos x\right|_0 ^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}} \mathrm{e}^x \mathrm{~d} \cos x \\
& =\mathrm{e}^{\frac{\pi}{2}}+1-\int_0^{\frac{\pi}{2}} \mathrm{e}^x \sin x \mathrm{~d} x
\end{aligned}
$$

即

$$
\int_0^{\frac{\pi}{2}} \mathrm{e}^x \sin x \mathrm{~d} x=\mathrm{e}^{\frac{\pi}{2}}+1-\int_0^{\frac{\pi}{2}} \mathrm{e}^x \sin x \mathrm{~d} x,
$$

故 $\quad 2 \int_0^{\frac{\pi}{2}} \mathrm{e}^x \sin x \mathrm{~d} x=\mathrm{e}^{\frac{\pi}{2}}+1$,
从而 $\quad \int_0^{\frac{\pi}{2}} \mathrm{e}^x \sin x \mathrm{~d} x=\frac{1}{2}\left(\mathrm{e}^{\frac{\pi}{2}}+1\right)$.

`例` 计算 $\int_{\frac{1}{2}}^1 \mathrm{e}^{-\sqrt{2 x-1}} \mathrm{~d} x$.
解 令 $t=\sqrt{2 x-1}$, 则 $t \mathrm{~d} t=\mathrm{d} x$, 当 $x=\frac{1}{2}$ 时, $t=0$; 当 $x=1$ 时, $t=1$; 于是有 $\int_{1 / 2}^1 \mathrm{e}^{-\sqrt{2 x-1}} \mathrm{~d} x=\int_0^1 t \mathrm{e}^{-t} \mathrm{~d} t$
再使用分部积分法, 令 $u=t, \mathrm{~d} v=\mathrm{e}^{-t} \mathrm{~d} t$, 则 $\mathrm{d} u=\mathrm{d} t, v=-\mathrm{e}^{-t}$.

$$
\text { 从而 } \int_0^1 t \mathrm{e}^{-t} \mathrm{~d} t=-\left.t \mathrm{e}^{-t}\right|_0 ^1+\int_0^1 \mathrm{e}^{-t} \mathrm{~d} t=-\frac{1}{\mathrm{e}}-\left.\left(\mathrm{e}^{-t}\right)\right|_0 ^1=1-\frac{2}{\mathrm{e}} \text {. }
$$

`例` 导出 $I_n=\int_0^{\frac{\pi}{2}} \sin ^n x \mathrm{~d} x\left(n\right.$ 为非负整数)的递推公式, 并计算 $\int_0^\pi \sin ^5 \frac{\pi}{2} \mathrm{~d} x$.
解 易见 $I_0=\int_0^{\frac{\pi}{2}} \mathrm{~d} x=\frac{\pi}{2}, I_1=\int_0^{\frac{\pi}{2}} \sin x \mathrm{~d} x=1$, 当 $n \geq 2$ 时

$$
\begin{aligned}
I_n=\int_0^{\frac{\pi}{2}} \sin ^n x \mathrm{~d} x & =-\int_0^{\frac{\pi}{2}} \sin ^{n-1} x \mathrm{~d} \cos x \\
& =(n-1) \int_0^{\frac{\pi}{2}} \sin ^{n-2} x\left(1-\sin ^2 x\right) \mathrm{d} x \\
& =(n-1) \int_0^{\frac{\pi}{2}} \sin ^{n-2} x \mathrm{~d} x-(n-1) \int_0^{\frac{\pi}{2}} \sin ^n x \mathrm{~d} x \\
& =(n-1) I_{n-2}-(n-1) I_n
\end{aligned}
$$

从而得到递推公式 $I_n=\frac{n-1}{n} I_{n-2}$. 反复用此公式直到下标为 0 或 1 , 得

$$
I_n= \begin{cases}\frac{2 m-1}{2 m} \cdot \frac{2 m-3}{2 m-2} \cdots \cdots \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}, & n=2 m \\ \frac{2 m}{2 m+1} \cdot \frac{2 m-2}{2 m-1} \cdots \cdots \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3}, & n=2 m+1\end{cases}
$$

其中 $m$ 为自然数.
令 $\frac{x}{2}=t$, 则 $\mathrm{d} x=2 \mathrm{~d} t$, 于是,

$$
\int_0^\pi \sin ^5 \frac{x}{2} \mathrm{~d} x=2 \int_0^{\frac{\pi}{2}} \sin ^5 t \mathrm{~d} t=2 \cdot \frac{4}{5} \cdot \frac{2}{3}=\frac{16}{15} .

开VIP会员

非会员每天6篇，会员每天16篇，VIP会员无限制访问

题库训练自我测评投稿

上一篇：定积分的换元法

下一篇：平面图形的面积-直角坐标系法

本文对您是否有用？有用 (0) 无用 (0)