Calculus Practice Questions Answers

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Exercise 1

令 $x=\frac{1}{t},dx=-\frac{1}{t^2}dt$

$\begin{array}{l}\int \frac{\sqrt{1+\mathrm{t}+t^2}}{t^2}dt\\ \\ =\frac{\sqrt{1+t+t^2}}{t}-\frac{1}{2}\int \frac{2t+}{t\sqrt{1+t+t^2}}dt\\ \\ =\sqrt{1+x+x^2}-\int \frac{1}{\sqrt{1+t+t^2}}dt-\frac{1}{2}\int \frac{1}{t\sqrt{1+t+t^2}}dt\\ \\ =\sqrt{1+x+x^2}-\ln \left(t+\frac{1}{2}+\sqrt{1+t+t^2}\right)+\frac{1}{2}\int \frac{-\frac{1}{t^2}}{\sqrt{{\left(\frac{1}{t}\right)}^2+\frac{1}{t}+1}}dt\\ \\ =\sqrt{1+x+x^2}-\ln \left(\frac{2+x+2\sqrt{1+x+x^2}}{2x\_}\right)+\frac{1}{2}\ln \left(x+\frac{1}{2}+\sqrt{1+x+x^2}\right)\\ \\ =\sqrt{1+x+x^2}+\ln x+\frac{1}{2}\ln \frac{2x+1+2\sqrt{1+x+x^2}}{{\left(2+x+2\sqrt{1+x+x^2}\right)}^2}+C\end{array}$

Exercise 2 :

（1）原式 $=\underset{n\to \infty }{\lim }\sum _{i=1}^n\frac{2}{n}\cdot \ln \left(1+\frac{i}{n}\right)=2{\int }_0^1\ln (1+x)dx=2x\_\ln \left(1+x\right){|}_0^1-2{\int }_0^{1}1-\frac{1}{1+x}dx=4\ln 2-2$

（2）原式 $=\underset{n\to \infty }{\lim }\sum _{i=0}^n\frac{1}{n}\cdot \frac{1}{\sqrt{1+{\left(\frac{i}{n}\right)}^2}}={\int }_0^1\frac{1}{\sqrt{1+x^2}}dx={\int }_0^{\frac{}{4}}\sec tdt=\ln {\left|\sec t+\tan t\right|}_0^{\frac{}{4}}=\ln \left(1+\sqrt{2}\right)$

Exercise 3 :

（1）原式 $=a^3{\int }_0^{\frac{}{2}}{\cos }^{2t}(1\_-cost)dt=a^3{\int }_0^{\frac{}{2}}\frac{1+\cos 2}{2}dt-a^3{\int }_0^{\frac{}{2}}(1-si{n}^{2t})d\_\sin t=\frac{a^3}{4}-a^3+\frac{a^3}{3}=\frac{a^3}{4}-\frac{2a{\_}^3}{3}$

(2) Let $x=\tan t,dx={\sec }^{2}tdt$ $$\begin{array}{l}{\int }_0^{\frac{}{4}}\ln (1+tanx)dx={\int }_0^{\frac{}{4}}\ln (1+tan(\frac{}{4}-x))dx\\ \\ ={\int }_0^{\frac{}{4}}\ln (\frac{2}{1+\tan x})dx={\int }_0^{\frac{}{4}}\ln 2dx-{\int }_0^{\frac{}{4}}\ln (1+tanx)dx\to =\frac{1}{2}{\int }_0^{\frac{}{4}}\ln 2dx=\frac{}{8}\ln 2\end{array}$$