定积分例题
first
\[\begin{align}
\Phi(x)=\int_{0}^{x^{2}} \sin t d t, \enspace \Phi^{\prime}(x)=?
\\ \\
设: u=x^{2}
\\ \\
\text { 则: } G(u)=\int_{0}^{u} \sin t d t=\Phi(x)
\\ \\
链式法则: G^{\prime}(u)=G^{\prime}(u) \cdot(u)^{\prime}
\\ \\
微积分基本公式: G^{\prime}(u)=\left(\int_{0}^{u} \sin t d t\right)^{\prime}=\sin (u)=\sin x^{2}
\\ \\
(u)^{\prime}=\left(x^{2}\right)^{\prime}=2 x
\\ \\
\therefore
\Phi^{\prime}(x)=\left(\int_{0}^{x} \sin t d t\right)^{\prime}=2 x \sin x^{2}
\end{align}
\]
second
\[\begin{eqnarray}
\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \cos t^{2} d t}{x}=?
\\ \\
\because \lim _{x \rightarrow 0} \int_{0}^{x} \cos t^{2} d t=0
\\ \\
\therefore 题目是\frac{0}{0} 型极限，\text { 适用于洛必达法则 }
\\ \\
\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \cos t^{2} d t}{x}=\lim _{x \rightarrow 0} \frac{\left(\int_{0}^{x} \cos t^{2} d t\right)^{\prime}}{(x)^{\prime}}
\\ \\
\left(\int_{0}^{x} \cos ^{2}t d t\right)^{\prime}=\cos x^{2}
\\ \\
\Rightarrow \lim _{x \rightarrow 0} \frac{\cos x^{2}}{1}=\lim _{x \rightarrow 0} \cos x^{2}=1
\end{eqnarray}
\]
third
\[\begin{align}
\int_{0}^{1}x^{2}dx=?
\\ \\
\because(\frac{1}{u+1}x^{u+1})^{\prime}=x^{u}
\\ \\
\therefore(\frac{1}{3}x^{3})^{\prime}=x^{2}
\\ \\
\int x^{2}dx=\frac{x^{3}}{3}+C
\\ \\
根据牛顿莱布尼茨公式:
\\
\int_{0}^{1}x^{2}dx = [\frac{x^{3}}{3}]_{1}^{2}
\\
\frac{(1)^{3}}{3}-\frac{(0)^{3}}{3}=\frac{1}{3}
\end{align}
\]
fourth
\[\begin{align}
\int_{0}^{1}\frac1{1+x^{2}}dx=?
\\ \\
\because\int\frac{1}{1+x^{2}}dx=\arctan{\left(x\right)}+C
\\ \\
\therefore\int_{0}^{1}\frac{1}{1+x^{2}}dx=[\arctan(x)]_{0}^{1}
\\ \\
\arctan(1)-\arctan(0)=\frac{\pi}{4}-0=\frac{\pi}{4},
\quad
(0\leq x\leq1)
\end{align}
\]
fifth
\[\begin{align}
\int_{1}^{3}\left|x-2\right|dx=?
\\ \\
\int_{1}^{3}|x-2|dx\Rightarrow\int_{1}^{2}|x-2|dx+\int_{2}^{3}|x-2|dx
\\ \\
当:1\leq x\leq 2, \quad 2-x=|x-2|
\\ \\
当:2\leq x\leq3,\quad x-2=|x-2|
\\ \\
\therefore\int_{1}^{2}(2-x)dx+\int_{2}^{3}(x-2)dx
\\ \\
\int2dx=2x+C
\\ \\
\int xdx=\frac{1}{2}x^{2}+C
\\ \\
\int_{1}^{2}\left(2-x\right)dx=\left[2x-\frac{1}{2}x^{2}\right]_{1}^{2}
\\ \\
\int_{2}^{3}\left(x-2\right)dx=\left[ \frac{1}{2}x^{2}-2x \right]_{2}^{3}
\\ \\
[2x-\frac{1}{2}x^{2}]_{1}^{2}=
4-\frac{1}{2}(2)^{2}-(2-\frac{1}{2})=\frac{1}{2}
\\ \\
[\frac{1}{2}x^{2}-2x]_{2}^{3}=
\frac{9}{2}-6-(\frac{4}{2}-4)=\frac{1}{2}
\\ \\
\therefore\int_{1}^{3}|x-2|dx=
\frac{1}{2}+\frac{1}{2}=1
\end{align}
\]
sixth
\[\begin{align}
\int_{\frac{\pi}{3}}^{\pi} \sin \left(2 x+\frac{\pi}{3}\right) d x=?
\\ \\
\text { 首先求原函数: } F(x)
\\ \\
\text { 设: } t=2 x+\frac{\pi}{3}
\\ \\
\int \sin \left(2 x+\frac{\pi}{3}\right) d x=\int \sin t d\left(2 x+\frac{\pi}{3}\right)
\\ \\
d\left(2 x+\frac{\pi}{3}\right)=\left(2 x+\frac{\pi}{3}\right)^{\prime} \cdot d x
\\ \\
d t=2 d x \Rightarrow d x=\frac{1}{2} d t
\\ \\
\Rightarrow \int \sin t \cdot \frac{1}{2} d t=\frac{1}{2} \cdot-\cos (t)+C
\\ \\
F(x)=-\frac{1}{2} \cos \left(2 x+\frac{\pi}{3}\right)+C
\\ \\
据牛顿-莱布尼茨公式:
\\ \\
\int_{\frac{\pi}{3}}^{\pi} \sin \left(2 x+\frac{\pi}{3}\right) d x=\left[-\frac{1}{2} \cos \left(2 x+\frac{\pi}{3}\right)\right]_{\frac{\pi}{3}}^{\pi}
\\ \\
-\frac{1}{2} \cos \left(2 \cdot \pi+\frac{\pi}{3}\right)=
-\frac{1}{2} [
\cos (2 \pi) \cos \left(\frac{\pi}{3}\right)-\sin (2 \pi) \sin \left(\frac{\pi}{3}\right) ]
\\ \\
-\frac{1}{2}\left(1 \cdot \frac{1}{2}-0\right)=-\frac{1}{4}
\\ \\
-\frac{1}{2} \cos \left(\frac{2 \pi}{3}+\frac{\pi}{3}\right)=-\frac{1}{2} \cos (\pi)=-\frac{1}{2} \cdot-1=\frac{1}{2}
\\ \\
-\frac{1}{4}-\frac{1}{2}=-\frac{3}{4}
\end{align}
\]
