高数强化NO15|定积分的计算|分段函数的积分|对称区间上的积分

定积分的计算

基本考查

$$\begin{array}{rl}& \text{解：}\frac{{\int }_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{x^2}{\sqrt{1-x^2}}dx}{\frac{\sqrt{3}}{2}-\frac{1}{2}}\stackrel{x=\sin t}{=}\frac{2}{\sqrt{3}-1}{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{{\sin }^{2}t}{\cos t}\cdot \cos tdt\\ & =(\sqrt{3}+1){\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1-\cos 2t}{2}dt=(\sqrt{3}+1)\left(\frac{1}{2}\cdot \frac{\pi }{6}-\frac{\sin 2t}{4}{|}_{\frac{\pi }{6}}^{\frac{\pi }{3}}\right)\\ & =\frac{\pi }{12}(\sqrt{3}+1)\end{array}$$
$$\boxed{【小结】}\text{积分平均值简称为平均值。}$$

$$\begin{array}{rl}& \text{解：}f\left(\frac{1}{x}\right)={\int }_1^{\frac{1}{x}}\frac{\ln t}{1+t}dt\stackrel{u=\frac{1}{t}}{=}{\int }_1^x\frac{\ln \frac{1}{u}}{1+\frac{1}{u}}d\left(\frac{1}{u}\right)={\int }_1^x\frac{-\ln u}{1+\frac{1}{u}}\left(-\frac{1}{u^2}\right)du\\ & ={\int }_1^x\frac{\ln u}{u^2+u}du\\ & \\ & f(x)+f\left(\frac{1}{x}\right)={\int }_1^x\left(\frac{\ln t}{1+t}+\frac{\ln t}{t^2+t}\right)dt={\int }_1^x\frac{\ln t}{t}dt=\frac{1}{2}{\ln }^{2}x\end{array}$$

$$\begin{array}{rl}& \text{解：设}{\int }_0^{1}f(x)dx=A\text{，两边同时在}(0,1)\text{上积分.}\\ & {\int }_0^{1}f(x)dx={\int }_0^1\frac{1}{1+x^2}dx+{\int }_0^{1}A{x}^{3}dx\\ & A={\int }_0^1\frac{1}{1+x^2}dx+A{\int }_0^1{x}^{3}dx\\ & A=\arctan x{|}_0^1+\frac{1}{4}A\\ & \frac{3}{4}A=\frac{\pi }{4}⟹A=\frac{\pi }{3}\end{array}$$
$$\boxed{【小结】}\text{函数的解析式中如果出现定积分，则先将该定积分令成}A\text{，再代入计算。}$$

分段函数的积分

1. 显式分段函数
   

$$\begin{array}{rl}& \text{解：法1：}F(x)={\int }_1^{x}f(t)dt=\{\begin{array}{ll}{\int }_1^{x}tdt,& 0\le x<1\\ {\int }_1^{x}1dt,& 1\le x\le 2\end{array}=\{\begin{array}{ll}\frac{x^3}{3}-\frac{1}{3},& 0\le x<1\\ x-1,& 1\le x\le 2\end{array}\end{array}$$
$$\begin{array}{rl}& \text{法2：}F(1)={\int }_1^{1}f(t)dt=0,\text{排除}A；\\ & F(x)\text{连续，排除}C.\text{故选}(D)\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{方法一直接分段计算在概率论中经常会考查，数一、三同学必须要掌握，但}\\ & \text{在高数中往往以选择题出现，学生面对选择题时应该要掌握方法二。}\\ & 2.\text{方法二是用两点进行判断：①变限积分一定能找到一个零点；②变限积分一定连续。}\end{array}$$
2. 隐式分段函数
$$\begin{array}{rl}& (\text{①符号}\text{sgn}(x)②\text{绝对值. ③取整. ④最大值 ⑤最小值})\\ & \text{思路：化为显式.}\end{array}$$

$$\begin{array}{rl}& \text{解：当}-1\le x\le 0\text{时，}-1\le t\le x\le 0.\\ & {\int }_{-1}^x(1-|t|)dt={\int }_{-1}^x(1+t)dt={\left[t+\frac{t^2}{2}\right]}_{-1}^x=\left(x+\frac{x^2}{2}\right)-\left(-1+\frac{1}{2}\right)=\frac{1}{2}{x}^2+x+\frac{1}{2}=\frac{1}{2}(x+1{)}^2\\ & \\ & \text{当}x>0\text{时}\\ & {\int }_{-1}^x(1-|t|)dt={\int }_{-1}^0(1+t)dt+{\int }_0^x(1-t)dt=\frac{1}{2}+\left(x-\frac{x^2}{2}\right)\\ & \\ & \text{故}f(x)=\{\begin{array}{ll}\frac{1}{2}+x-\frac{x^2}{2},& x>0\\ \frac{1}{2}(x+1{)}^2,& -1\le x\le 0\end{array}\end{array}$$
$$\boxed{【小结】}\text{被积函数中自变量要根据上下限范围讨论，把对应区间的解析式写出来，再求积分。}$$

对称区间上的积分

$$\begin{array}{rl}& \text{解：}{\int }_{-\pi }^{\pi }{\sin }^{3}xdx=0\\ & \\ & {\int }_{-\pi }^{\pi }\sqrt{{\pi }^2-x^2}dx=\frac{\pi }{2}\cdot {\pi }^2\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}1.\text{见到对称区间，优先考虑利用被积函数的奇偶性进行化简，即}\\ & {\int }_{-a}^{a}f(x)dx=\{\begin{array}{ll}0,& f(x)\text{为奇函数}\\ 2{\int }_0^{a}f(x)dx,& f(x)\text{为偶函数}\end{array}。\\ & 2.\text{有些特殊的积分可以考虑通过定义（算图像面积）解决。}\end{array}$$

$$\begin{array}{rl}& (a+b+c{)}^2=a^2+b^2+c^2+2ab+2ac+2bc\\ & \\ & {\int }_0^{\pi }x\sin xdx=\frac{\pi }{2}{\int }_0^{\pi }\sin xdx=\pi \\ & \\ & {\int }_0^{\pi }{\cos }^{2}xdx=2{\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx=2\cdot \frac{\pi }{4}=\frac{\pi }{2}\\ & \\ & \text{解：}{\int }_{-\pi }^{\pi }\left(x^2+a^2{\cos }^{2}x+b^2{\sin }^{2}x-2ax\cos x-2bx\sin x+2ab\sin x\cos x\right)dx\\ & =2{\int }_0^{\pi }{x}^{2}dx+2a^2{\int }_0^{\pi }{\cos }^{2}xdx+2b^2{\int }_0^{\pi }{\sin }^{2}xdx-4b{\int }_0^{\pi }x\sin xdx\\ & =\frac{2}{3}{\pi }^3+2a^2\cdot \frac{\pi }{2}+2b^2\cdot \frac{\pi }{2}-4b\cdot \pi =\frac{2}{3}{\pi }^3+\pi a^2+\pi (b^2-4b)\\ & \\ & \text{当}a=0,b=2\text{时，取最小. }{a}_1\cos x+b_1\sin x=2\sin x,\text{选}(A)\end{array}$$
$$\begin{array}{rl}& \text{三角函数定积分.}\\ & \\ & ①{\int }_0^{\frac{\pi }{2}}\sin tdt=1,{\int }_0^{\pi }\sin tdt=2,{\int }_{\frac{\pi }{2}}^{\pi }\sin tdt=1,{\int }_0^{2\pi }\sin tdt=0\\ & {\int }_0^{\frac{\pi }{2}}{\sin }^{2}tdt=\frac{\pi }{4},{\int }_0^{\pi }{\sin }^{2}tdt=2{\int }_0^{\frac{\pi }{2}}{\sin }^{2}tdt=\frac{\pi }{2}\\ & {\int }_0^{\frac{3\pi }{2}}{\sin }^{2}tdt=3{\int }_0^{\frac{\pi }{2}}{\sin }^{2}tdt=\frac{3\pi }{4},{\int }_0^{2\pi }{\sin }^{2}tdt=4{\int }_0^{\frac{\pi }{2}}{\sin }^{2}tdt=\pi \\ & {\int }_0^{\frac{\pi }{2}}{\sin }^{3}tdt=\frac{2}{3},{\int }_0^{\pi }{\sin }^{3}tdt=2{\int }_0^{\frac{\pi }{2}}{\sin }^{3}tdt=\frac{4}{3}\\ & {\int }_0^{\frac{3\pi }{2}}{\sin }^{3}tdt=\frac{2}{3},{\int }_0^{2\pi }{\sin }^{3}tdt=0\\ & \\ & ②{\int }_0^{\frac{\pi }{2}}\cos tdt=1,{\int }_0^{\pi }\cos tdt=0,{\int }_{\frac{\pi }{2}}^{\pi }\cos tdt=-1\\ & {\int }_0^{2\pi }\cos tdt=0\end{array}$$
$$\begin{array}{rl}& ③{\int }_0^{\pi }xf(\sin x)dx\stackrel{u=\pi -x}{=}{\int }_{\pi }^0(\pi -u)f(\sin (\pi -u))d(\pi -u)\\ & ={\int }_0^{\pi }(\pi -u)f(\sin u)du\\ & \\ & 2{\int }_0^{\pi }xf(\sin x)dx={\int }_0^{\pi }\pi f(\sin u)du\\ & {\int }_0^{\pi }xf(\sin x)dx=\frac{\pi }{2}{\int }_0^{\pi }f(\sin u)du=\pi {\int }_0^{\frac{\pi }{2}}f(\sin u)du\end{array}$$
$$\begin{array}{rl}& {\int }_0^{2\pi }xf(\cos x)dx\stackrel{u=2\pi -x}{=}{\int }_{2\pi }^0(2\pi -u)f(\cos (2\pi -u))d(2\pi -u)\\ & ={\int }_0^{2\pi }(2\pi -u)f(\cos u)du\\ & \\ & {\int }_0^{2\pi }xf(\cos x)dx=\frac{1}{2}{\int }_0^{2\pi }2\pi f(\cos u)du=\pi {\int }_0^{2\pi }f(\cos u)du\\ & \\ & {\int }_0^{\pi }x{\sin }^{4}xdx=\frac{\pi }{2}{\int }_0^{\pi }{\sin }^{4}xdx=\frac{\pi }{2}\cdot 2{\int }_0^{\frac{\pi }{2}}{\sin }^{4}xdx\\ & =\pi \cdot \frac{3!!}{4!!}\cdot \frac{\pi }{2}=\frac{3}{16}{\pi }^2\\ & \\ & {\int }_0^{2\pi }x{\cos }^{4}xdx=\pi {\int }_0^{2\pi }{\cos }^{4}xdx=\pi \cdot 4{\int }_0^{\frac{\pi }{2}}{\cos }^{4}xdx\\ & =4\pi \cdot \frac{3!!}{4!!}\cdot \frac{\pi }{2}=4\pi \cdot \frac{3}{8}\cdot \frac{\pi }{2}=\frac{3}{4}{\pi }^2\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{关于三角函数的常用公式：}\\ & (1)\text{华里士公式：}{\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx={\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx=\{\begin{array}{ll}\frac{n-1}{n}\frac{n-3}{n-2}\cdots \frac{1}{2}\frac{\pi }{2},& n=2k\\ \frac{n-1}{n}\frac{n-3}{n-2}\cdots \frac{2}{3},& n=2k+1\end{array}；\\ & \\ & (2){\int }_0^{\pi }{\sin }^{n}xdx=2{\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx，{\int }_0^{\pi }{\cos }^{n}xdx=\{\begin{array}{ll}2{\int }_0^{\frac{\pi }{2}}{\cos }^{n}xdx,& n=2k\\ 0,& n=2k+1\end{array}\\ & {\int }_0^{2\pi }{\sin }^{n}xdx={\int }_0^{2\pi }{\cos }^{n}xdx=\{\begin{array}{ll}4{\int }_0^{\frac{\pi }{2}}{\sin }^{n}xdx,& n=2k\\ 0,& n=2k+1\end{array}\\ & \\ & (3){\int }_0^{\pi }xf(\sin x)dx=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx=\pi {\int }_0^{\frac{\pi }{2}}f(\sin x)dx\\ & {\int }_0^{2\pi }xf(\cos x)dx=\pi {\int }_0^{2\pi }f(\cos x)dx。\end{array}$$

$$\begin{array}{rl}& (1){\int }_0^{2\pi }(1+\sin x{)}^{3}dx\\ & ={\int }_0^{2\pi }\left(1+3\sin x+3{\sin }^{2}x+{\sin }^{3}x\right)dx\\ & ={\int }_0^{2\pi }\left(1+3{\sin }^{2}x\right)dx\\ & =2\pi +3\cdot 4{\int }_0^{\frac{\pi }{2}}{\sin }^{2}xdx\\ & =2\pi +3\cdot 4\cdot \frac{\pi }{4}=5\pi \end{array}$$
$$\begin{array}{rl}& (2){\int }_0^{\pi }x{\sin }^{4}xdx=\frac{\pi }{2}{\int }_0^{\pi }{\sin }^{4}xdx=\frac{\pi }{2}\cdot 2{\int }_0^{\frac{\pi }{2}}{\sin }^{4}xdx\\ & =\pi \cdot \frac{3!!}{4!!}\cdot \frac{\pi }{2}=\frac{3}{16}{\pi }^2\\ & \\ & (3){\int }_0^{2\pi }x{\cos }^{4}xdx=\pi {\int }_0^{2\pi }{\cos }^{4}xdx=\pi \cdot 4{\int }_0^{\frac{\pi }{2}}{\cos }^{4}xdx\\ & =4\pi \cdot \frac{3!!}{4!!}\cdot \frac{\pi }{2}=4\pi \cdot \frac{3}{8}\cdot \frac{\pi }{2}=\frac{3}{4}{\pi }^2\end{array}$$
$$\begin{array}{rl}& (4)={\int }_0^{2\pi }x(1-\cos x{)}^{2}dx-{\int }_0^{2\pi }\sin x(1-\cos x{)}^{2}dx\\ & =\pi {\int }_0^{2\pi }(1-\cos x{)}^{2}dx-\frac{1}{3}(1-\cos x{)}^3{|}_0^{2\pi }\\ & =\pi {\int }_0^{2\pi }\left(1-2\cos x+{\cos }^{2}x\right)dx\\ & =\pi {\int }_0^{2\pi }\left(1+{\cos }^{2}x\right)dx\\ & =\pi \left(2\pi +4{\int }_0^{\frac{\pi }{2}}{\cos }^{2}xdx\right)\\ & =\pi \left(2\pi +4\cdot \frac{\pi }{4}\right)=3{\pi }^2\end{array}$$

$$\begin{array}{rl}& \text{证：}(1){\int }_{-a}^{a}f(x)g(x)dx\stackrel{u=-x}{=}{\int }_a^{-a}f(-u)g(-u)d(-u)\\ & ={\int }_{-a}^{a}f(-u)g(u)du=\frac{1}{2}{\int }_{-a}^a\left(f(x)+f(-x)\right)g(x)dx\\ & =\frac{A}{2}{\int }_{-a}^{a}g(x)dx=A{\int }_0^{a}g(x)dx\end{array}$$
$$\begin{array}{rl}& \arctan x+\arctan \frac{1}{x}=\frac{\pi }{2}\\ & \\ & \tan \alpha =x,\alpha =\arctan x\\ & \tan \beta =\frac{1}{x},\beta =\arctan \frac{1}{x}\\ & \\ & \arctan e^x+\arctan e^{-x}=\frac{\pi }{2}\end{array}$$
$$\begin{array}{rl}& (2){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}|\sin x|\arctan e^{x}dx=\frac{\pi }{2}{\int }_0^{\frac{\pi }{2}}\sin xdx=\frac{\pi }{2}\end{array}$$
$$\begin{array}{rl}& \boxed{【小结】}\text{对于对称区间上的积分，如果被积函数本身不具有奇偶性，同时也不便于直接}\\ & \text{计算，则可以使用公式}{\int }_{-a}^{a}f(x)dx={\int }_0^a\left[f(x)+f(-x)\right]dx\text{进行计算。对于该公式，如果}\\ & \text{是解答题，则使用的时候需要首先进行证明。}\end{array}$$