定积分 2
练习
被积函数中含有变限积分函数的定积分计算
例题 1
设 \(f(x)=\displaystyle\int_{0}^{x} \dfrac{\sin t}{\pi - t} \mathrm{d}t\)，计算 \(\displaystyle\int_{0}^{\pi} f(x) \mathrm{d}x\)
Success
利用分部积分法：
\[
\begin{aligned}
\displaystyle\int_{0}^{\pi} f(x) \mathrm{d}x &= x f(x) \bigg|_{0}^{\pi} - \displaystyle\int_{0}^{\pi} x \mathrm{d}f(x) \\
&= \pi f(\pi) - \displaystyle\int_{0}^{\pi} x \cdot \dfrac{\sin x}{\pi - x} \mathrm{d}x \\
&= \pi \displaystyle\int_{0}^{\pi} \dfrac{\sin t}{\pi - t} \mathrm{d}t - \displaystyle\int_{0}^{\pi} \dfrac{x \sin x}{\pi - x} \mathrm{d}x \\
&= \displaystyle\int_{0}^{\pi} \dfrac{(\pi - x) \sin x}{\pi - x} \mathrm{d}x = \displaystyle\int_{0}^{\pi} \sin x \mathrm{d}x = 2
\end{aligned}
\]
类题 1
设 \(f(x)=\displaystyle\int_{1}^{x} \dfrac{\ln(1+t)}{t} \mathrm{d}t\)，计算 \(\displaystyle\int_{0}^{1} \dfrac{f(x)}{\sqrt{x}} \mathrm{d}x\)
Success
\[
\begin{aligned}
\displaystyle\int_{0}^{1} \dfrac{f(x)}{\sqrt{x}} \mathrm{d}x &= 2\displaystyle\int_{0}^{1} f(x) \mathrm{d}\sqrt{x} = 2\sqrt{x} f(x) \bigg|_{0}^{1} - 2\displaystyle\int_{0}^{1} \sqrt{x} \mathrm{d}f(x) \\
&= -2 \displaystyle\int_{0}^{1} \sqrt{x} \dfrac{\ln(1+x)}{x} \mathrm{d}x \xrightarrow{x \to x^2} -4 \displaystyle\int_{0}^{1} \ln(1+x^2) \mathrm{d}x \\
&= -4 \ln 2 + 8 \displaystyle\int_{0}^{1} \dfrac{x^2}{1+x^2} \mathrm{d}x = 8 - 2\pi - 4\ln 2
\end{aligned}
\]
类题 2
设 \(f''(x)=\arctan(x-1)^2\)，且 \(f(0)=0\)，求 \(I=\displaystyle\int_{0}^{1} f(x) \mathrm{d}x\)
Success
\[
\begin{aligned}
\displaystyle\int_{0}^{1} f(x) \mathrm{d}x &= xf(x)\bigg|_0^1 - \displaystyle\int_{0}^{1} x \mathrm{d}f(x) = f(1) - \displaystyle\int_{0}^{1} x f'(x) \mathrm{d}x \\
&= \displaystyle\int_{0}^{1} \arctan(x-1)^2 \mathrm{d}x - \displaystyle\int_{0}^{1} x \arctan(x-1)^2 \mathrm{d}x \\
&\xrightarrow{(x-1)^2 \to x} \dfrac{1}{2} \displaystyle\int_{0}^{1} \arctan x \mathrm{d}x \\
&= \dfrac{\pi}{8} - \dfrac{1}{2} \displaystyle\int_{0}^{1} \dfrac{x}{1+x^2} \mathrm{d}x = \dfrac{\pi}{8} - \dfrac{1}{4} \ln 2
\end{aligned}
\]
类题 3
已知函数 \(f(x)\) 在 \(\left[0, \dfrac{3}{2}\pi\right]\) 连续，在 \(\left(0, \dfrac{3}{2}\pi\right)\) 内是 \(\dfrac{\cos x}{2x - 3\pi}\) 的一个原函数，\(f(0)=0\)。
(1) 求 \(f(x)\) 在 \(\left[0, \dfrac{3}{2}\pi\right]\) 上的平均值。
(2) 证明 \(f(x)\) 在 \(\left(0, \dfrac{3}{2}\pi\right)\) 内有唯一零点。
Success
(1)
\(f(x)\) 在 \(\left[0, \dfrac{3}{2}\pi\right]\) 上的平均值为 \(\overline{f(x)} = \dfrac{1}{\frac{3\pi}{2}} \displaystyle\int_{0}^{\frac{3\pi}{2}} f(x) \mathrm{d}x\)。
利用分部积分法计算定积分：
\[
\begin{aligned}
\displaystyle\int_{0}^{\frac{3\pi}{2}} f(x) \mathrm{d}x &= x f(x) \bigg|_{0}^{\frac{3\pi}{2}} - \displaystyle\int_{0}^{\frac{3\pi}{2}} x \mathrm{d}f(x) \\
&= \dfrac{3\pi}{2} f\left(\dfrac{3\pi}{2}\right) - 0 - \displaystyle\int_{0}^{\frac{3\pi}{2}} x \cdot \dfrac{\cos x}{2x - 3\pi} \mathrm{d}x \\
&= \dfrac{3\pi}{2} \displaystyle\int_{0}^{\frac{3\pi}{2}} \dfrac{\cos t}{2t - 3\pi} \mathrm{d}t - \displaystyle\int_{0}^{\frac{3\pi}{2}} \dfrac{x \cos x}{2x - 3\pi} \mathrm{d}x \\
&\xlongequal{\text{前项 } t \text{ 换为 } x} \displaystyle\int_{0}^{\frac{3\pi}{2}} \dfrac{\frac{3\pi}{2} \cos x}{2x - 3\pi} \mathrm{d}x - \displaystyle\int_{0}^{\frac{3\pi}{2}} \dfrac{x \cos x}{2x - 3\pi} \mathrm{d}x \\
&= \displaystyle\int_{0}^{\frac{3\pi}{2}} \dfrac{\left(\frac{3\pi}{2} - x\right)\cos x}{2x - 3\pi} \mathrm{d}x = \displaystyle\int_{0}^{\frac{3\pi}{2}} \dfrac{-\frac{1}{2}(2x - 3\pi)\cos x}{2x - 3\pi} \mathrm{d}x \\
&= -\dfrac{1}{2} \displaystyle\int_{0}^{\frac{3\pi}{2}} \cos x \mathrm{d}x
\end{aligned}
\]
于是：
\[
\overline{f(x)} = \dfrac{2}{3\pi} \cdot \left( -\dfrac{1}{2} \displaystyle\int_{0}^{\frac{3\pi}{2}} \cos x \mathrm{d}x \right) = -\dfrac{1}{3\pi} \sin x \bigg|_{0}^{\frac{3\pi}{2}} = \dfrac{1}{3\pi}
\]
(2)
令 \(f'(x) = \dfrac{\cos x}{2x - 3\pi} = 0\)，解得 \(x=\dfrac{\pi}{2}\)。
于是 \(f(x)\) 在 \(\left(0, \dfrac{\pi}{2}\right)\) 内单调递减，在 \(\left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right)\) 内单调递增。
由于 \(f(0) = 0\implies f\left(\dfrac{\pi}{2}\right)<f(0) = 0\)
若 \(f\left(\dfrac{3\pi}{2}\right)<0\)，则与平均值大于 \(0\) 矛盾
故 \(f(x)\) 在 \((0,\dfrac{3\pi}{2})\) 有唯一零点
例题 2
设 \(f(x)\) 为非负连续函数，满足 \(f(x)\displaystyle\int_{0}^{x} f(x-t) \mathrm{d}t = \sin^4 x\)，求 \(\displaystyle\int_{0}^{\frac{\pi}{2}} f(x) \mathrm{d}x\)
Success
\[
f(x)\displaystyle\int_{0}^{x} f(x-t) \mathrm{d}t \xlongequal{t \to x-t} f(x)\displaystyle\int_{0}^{x} f(t) \mathrm{d}t = \dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d}x} \left( \displaystyle\int_{0}^{x} f(t) \mathrm{d}t \right)^2 = \cos^4 x
\]
于是
\[
\left( \displaystyle\int_{0}^{x} f(t) \mathrm{d}t \right)^2 = 2 \displaystyle\int_{0}^{x} \cos^4 t \mathrm{d}t
\]
令 \(x = \dfrac{\pi}{2}\)，得到 \(\displaystyle\int_{0}^{\frac{\pi}{2}} f(t) \mathrm{d}t = \sqrt{\dfrac{3\pi}{8}}\)。（非负连续函数积分值为正）
类题
设 \(f(x)\) 为连续函数，且 \(x > -1\) 时， \(f(x) \left[ \displaystyle\int_{0}^{x} f(t) \mathrm{d}t + 1 \right] = \dfrac{x e^x}{2(1+x)^2}\)，求 \(f(x)\)
Success
\[
\dfrac{1}{2} \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \displaystyle\int_{0}^{x} f(t) \mathrm{d}t + 1 \right]^2 = \dfrac{x e^x}{2(1+x)^2}
\]
于是
\[
\left( \displaystyle\int_{0}^{x} f(t) \mathrm{d}t + 1 \right)^2 - 1 = 2\displaystyle\int_{0}^{x} \dfrac{t e^t}{2(1+t)^2} \mathrm{d}t = \dfrac{e^x}{x+1} - 1
\]
\[
\displaystyle\int_{0}^{x} f(t) \mathrm{d}t = -1 \pm \sqrt{\dfrac{e^x}{x+1}}
\]
求导：\(f(x) = \pm \dfrac{x}{2x+2} \sqrt{\dfrac{e^x}{x+1}}\)
被积函数中含有导函数的定积分计算
例题
设 \(\displaystyle\int_{0}^{2} f(x) \mathrm{d}x = 4\)， \(f(2)=1\)， \(f'(2)=0\)，求 \(\displaystyle\int_{0}^{1} x^2 f''(2x) \mathrm{d}x\)
Success
\[
\begin{aligned}
\displaystyle\int_{0}^{1} x^2 f''(2x) \mathrm{d}x &\xlongequal{2x \to x} \dfrac{1}{8} \displaystyle\int_{0}^{2} x^2 f''(x) \mathrm{d}x = \dfrac{1}{8} \displaystyle\int_{0}^{2} x^2 \mathrm{d}f'(x) \\
&= -\dfrac{1}{4} \displaystyle\int_{0}^{2} x f'(x) \mathrm{d}x = -\dfrac{1}{4} \displaystyle\int_{0}^{2} x \mathrm{d}f(x) = \dfrac{1}{2}
\end{aligned}
\]
利用“定积分的结果是一个数字”来求解某些待定函数的问题
例题 1
设 \(f(x)=x^2 - x \displaystyle\int_{0}^{2} f(x) \mathrm{d}x + 2\displaystyle\int_{0}^{1} f(x) \mathrm{d}x\)，求 \(f(x)\)
Success
设 \(\displaystyle\int_{0}^{2} f(x) \mathrm{d}x = a\), \(\displaystyle\int_{0}^{1} f(x) \mathrm{d}x = b\)，则 \(f(x) = x^2 - ax + 2b\)。
代入积分计算：
\[
a = \displaystyle\int_{0}^{2} (x^2 - ax + 2b) \mathrm{d}x = \dfrac{8}{3} - 2a + 4b
\]
\[
b = \displaystyle\int_{0}^{1} (x^2 - ax + 2b) \mathrm{d}x = \dfrac{1}{3} - \dfrac{a}{2} + 2b
\]
联立解得 \(a = \dfrac{4}{3}, b = \dfrac{1}{3}\)，所以 \(f(x) = x^2 - \dfrac{4}{3}x + \dfrac{2}{3}\)。
例题 2
设 \(f(x)\) 为连续函数， \(f(x) = \dfrac{x}{1+\cos^2 x} + \displaystyle\int_{-\pi}^{\pi} f(x) \sin x \mathrm{d}x\)，求 \(f(x)\)
Success
设 \(\displaystyle\int_{-\pi}^{\pi} f(x) \sin x \mathrm{d}x = a\)，则 \(f(x) = \dfrac{x}{1+\cos^2 x} + a\)。
\[
a = \displaystyle\int_{-\pi}^{\pi} \left( \dfrac{x}{1+\cos^2 x} + a \right) \sin x \mathrm{d}x = \displaystyle\int_{-\pi}^{\pi} \dfrac{x \sin x}{1+\cos^2 x} \mathrm{d}x + a \displaystyle\int_{-\pi}^{\pi} \sin x \mathrm{d}x
\]
第二项为 0，第一项利用偶函数性质：
\[
a = 2 \displaystyle\int_{0}^{\pi} \dfrac{x \sin x}{1+\cos^2 x} \mathrm{d}x = \dfrac{\pi^2}{2}
\]
因此 \(f(x) = \dfrac{x}{1+\cos^2 x} + \dfrac{\pi^2}{2}\)。
例题 3
连续函数 \(f\) 和 \(g\) 满足 \(f(x)=3x^2 + 1 + \displaystyle\int_{0}^{1} g(x) \mathrm{d}x\), \(g(x)=-x + 6x^2 \displaystyle\int_{0}^{1} f(x) \mathrm{d}x\)，求 \(f(x), g(x)\)
Success
设 \(\displaystyle\int_{0}^{1} f(x) \mathrm{d}x = a\), \(\displaystyle\int_{0}^{1} g(x) \mathrm{d}x = b\)。
则 \(f(x) = 3x^2 + 1 + b\), \(g(x) = -x + 6ax^2\)。
积分得方程组：
\[
a = \displaystyle\int_{0}^{1} (3x^2 + 1 + b) \mathrm{d}x = 2 + b
\]
\[
b = \displaystyle\int_{0}^{1} (-x + 6ax^2) \mathrm{d}x = -\dfrac{1}{2} + 2a
\]
解得 \(a = -\dfrac{3}{2}, b = -\dfrac{7}{2}\)。
因此 \(f(x) = 3x^2 - \dfrac{5}{2}, g(x) = -9x^2 - x\)。
利用分部积分，导出积分的递推公式
例题 1
华里士公式 (Wallis formula)：
\(I_n = \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^n x \mathrm{d}x = \displaystyle\int_{0}^{\frac{\pi}{2}} \cos^n x \mathrm{d}x = \dfrac{(n-1)!!}{n!!} I^*\)
其中 \(I^* = \dfrac{\pi}{2}\) (n为偶数)，\(I^* = 1\) (n为奇数)。
例题 2
求积分 \(J_n = \displaystyle\int_{0}^{1} x \ln^n x \mathrm{d}x\) 的递推关系，并计算 \(J_n\)
Success
\[
J_n = \displaystyle\int_{0}^{1} \ln^n x \mathrm{d}\left(\dfrac{x^2}{2}\right) = \dfrac{1}{2} x^2 \ln^n x \bigg|_{0}^{1} - \dfrac{n}{2} \displaystyle\int_{0}^{1} x \ln^{n-1} x \mathrm{d}x = -\dfrac{n}{2} J_{n-1} \quad (n \ge 1)
\]
依次相乘得到 \(J_n = \left(-\dfrac{1}{2}\right)^n n! J_0\)。
而 \(J_0 = \displaystyle\int_{0}^{1} x \mathrm{d}x = \dfrac{1}{2}\)，因此 \(J_n = (-1)^n \dfrac{n!}{2^{n+1}}\)。
例题 3
设 \(J_n = \displaystyle\int_{0}^{\frac{\pi}{4}} \tan^n x \mathrm{d}x\)，判断 \(J_n\) 的单调性，并利用分部积分导出 \(J_n\) 和 \(J_{n+2}\) 的递推关系，并计算 \(\lim\limits_{n \to \infty} n J_n\)
Success
在积分区间 \(\left(0, \dfrac{\pi}{4}\right)\) 内，\(\tan x \in (0, 1)\)，于是 \(\tan^n x\) 为正且单调递减，故 \(J_n\) 单调递减。
\[
J_{n+2} = \displaystyle\int_{0}^{\frac{\pi}{4}} \tan^n x (\sec^2 x - 1) \mathrm{d}x = \displaystyle\int_{0}^{\frac{\pi}{4}} \tan^n x \mathrm{d} \tan x - J_n = \dfrac{1}{n+1} - J_n
\]
即递推关系为 \(J_n + J_{n+2} = \dfrac{1}{n+1}\)。
\(J_n\) 单调递减且有下界 0，故极限存在。在递推关系中令 \(n \to \infty\)，得 \(\lim\limits_{n \to \infty} J_n = 0\)
\[
\lim\limits_{n \to \infty} n J_n = \dfrac{1}{2} \lim_{n \to \infty} [n J_n + (n+2) J_{n+2}] = \dfrac{1}{2} \lim_{n \to \infty} \dfrac{n}{n+1} = \dfrac{1}{2}
\]
例题 4
设 \(a_n = \displaystyle\int_{0}^{1} x^n \sqrt{1-x^2} \mathrm{d}x\), \(b_n = \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^n x \cdot \cos^n x \mathrm{d}x\)，求 \(\lim_{n \to \infty} \dfrac{b_n}{a_n}\)
Success
\[
a_n \xlongequal{x \to \sin x} \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^n x \cos^2 x \mathrm{d}x = \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^n x \mathrm{d}x - \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^{n+2} x \mathrm{d}x
\]
\[
b_n = \dfrac{1}{2^n} \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^n(2x) \mathrm{d}x = \dfrac{1}{2^n} \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^n x \mathrm{d}x
\]
通过比较 Wallis 公式的比值极限，最终可得 \(\lim_{n \to \infty} \dfrac{b_n}{a_n} = 0\)。
例题 5
证明：\(J_n = \displaystyle\int_{0}^{\frac{\pi}{2}} \cos^n x \cdot \sin nx \mathrm{d}x = \dfrac{1}{2^{n+1}} \left( 2 + \dfrac{2^2}{2} + \dfrac{2^3}{3} + \dots + \dfrac{2^n}{n} \right)\)
Success
考虑到 \(\cos^n x = \left( \dfrac{e^{ix} + e^{-ix}}{2} \right)^n\)，展开整理后可得：
\[
J_n = \dfrac{1}{2^n} \sum_{k=0}^{n} C_n^k \displaystyle\int_{0}^{\frac{\pi}{2}} \sin nx \cos(n-2k)x \mathrm{d}x
\]
通过积化和差及区间再现等技巧计算每一项积分，最终求和得到结论。
分段函数的定积分
例题 1
设 \(f(x) = \displaystyle\int_{0}^{1} |t^2 - x^2| \mathrm{d}t (x>0)\)，求 \(f'(x)\) 并求 \(f(x)\) 的最小值。
Success
\[
f(x) = \begin{cases}
\displaystyle\int_{0}^{x} (x^2 - t^2) \mathrm{d}t + \displaystyle\int_{x}^{1} (t^2 - x^2) \mathrm{d}t = \dfrac{4}{3}x^3 - x^2 + \dfrac{1}{3}, & (0 < x < 1) \\
\displaystyle\int_{0}^{1} (x^2 - t^2) \mathrm{d}t = x^2 - \dfrac{1}{3}, & (x \ge 1)
\end{cases}
\]
求导得：
\[
f'(x) = \begin{cases}
4x^2 - 2x, & (0 < x < 1) \\
2x, & (x \ge 1)
\end{cases}
\]
令 \(f'(x)=0\) 得 \(x=\dfrac{1}{2}\)。\(f(x)\) 在 \(\left(0, \dfrac{1}{2}\right)\) 减，在 \(\left(\dfrac{1}{2}, +\infty\right)\) 增。
最小值 \(f\left(\dfrac{1}{2}\right) = \dfrac{1}{4}\)。
例题 2
设 \(f(x) = \begin{cases} 2x + \dfrac{3}{2}x^2, & 0 \le x < 1 \\ \dfrac{1+x}{x(1+xe^x)}, & 1 \le x \le 2 \end{cases}\)，求函数 \(F(x) = \displaystyle\int_{0}^{x} f(t) \mathrm{d}t\) 的表达式。
Success
当 \(0 \le x < 1\) 时， \(F(x) = x^2 + \dfrac{1}{2}x^3\)。
当 \(1 \le x \le 2\) 时，
\[
F(x) = \dfrac{3}{2} + \displaystyle\int_{1}^{x} \dfrac{1+t}{t(1+te^t)} \mathrm{d}t = \dfrac{3}{2} + \displaystyle\int_{1}^{x} \dfrac{\mathrm{d}(te^t)}{te^t(1+te^t)} = \dfrac{1}{2} + \ln(1+e) + \ln\left(\dfrac{xe^x}{1+xe^x}\right)
\]
例题 3
已知 \(x \ge 0\) 时， \(f(x) = \begin{cases} \sin x, & 0 \le x < \dfrac{\pi}{2} \\ 0, & x \ge \dfrac{\pi}{2} \end{cases}\)，且 \(g(x) = \begin{cases} 1, & x \ge 0 \\ 0, & x < 0 \end{cases}\)，求 \(F(x) = \displaystyle\int_{0}^{x} f(t)g(x-t) \mathrm{d}t (x \ge 0)\)
Success
当 \(0 \le x < \dfrac{\pi}{2}\) 时， \(F(x) = \displaystyle\int_{0}^{x} (x-t) \sin t \mathrm{d}t = x - \sin x\)。
当 \(x \ge \dfrac{\pi}{2}\) 时， \(F(x) = \displaystyle\int_{0}^{\frac{\pi}{2}} (x-t) \sin t \mathrm{d}t = x - 1\)。