Giải bài tập 4.32 trang 8

Đề bài

Tính các tích phân sau:

a) $\underset{1}{\overset{4}{\int }}\left(x^3-2\sqrt{x}\right)dx$;

b) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\cos x-\sin x\right)dx$;

c) $\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{{\sin }^{2}x}$;

d) $\underset{1}{\overset{16}{\int }}\frac{x-1}{\sqrt{x}}dx$.

Lời giải chi tiết

a) $\underset{1}{\overset{4}{\int }}\left(x^3-2\sqrt{x}\right)dx$

$=\left(\frac{x^4}{4}-\frac{4x\sqrt{x}}{3}\right)|\begin{array}{c}4\\ 1\end{array}$

$=\frac{4^4}{4}-\frac{4.4\sqrt{4}}{3}-\frac{1}{4}+\frac{4.1\sqrt{1}}{3}$

$=\frac{653}{12}$.

b) $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\cos x-\sin x\right)dx$

$=\left(\sin x+\cos x\right)|\begin{array}{c}\frac{\pi }{2}\\ 0\end{array}$

$=\sin \frac{\pi }{2}+\cos \frac{\pi }{2}-\sin 0-\cos 0$

$=1-1=0$.

c) $\underset{\frac{\pi }{6}}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{{\sin }^{2}x}$

$=-\cot x|\begin{array}{c}\frac{\pi }{4}\\ \frac{\pi }{6}\end{array}$

$=-\cot \frac{\pi }{4}+\cot \frac{\pi }{6}$

$=-1+\sqrt{3}$.

d) $\underset{1}{\overset{16}{\int }}\frac{x-1}{\sqrt{x}}dx$

$=\underset{1}{\overset{16}{\int }}\left(x^{\frac{1}{2}}-x^{\frac{-1}{2}}\right)dx$

$=\left(\frac{2x\sqrt{x}}{3}-2\sqrt{x}\right)|\begin{array}{c}16\\ 1\end{array}$

$=\frac{2.16\sqrt{16}}{3}-2\sqrt{16}-\frac{2}{3}+2=36$.