Giải bài tập 4 trang 20

Đề bài

Tính các tích phân sau:

a) $\underset{-2}{\overset{1}{\int }}\left|2x+2\right|dx$;

b) $\underset{0}{\overset{4}{\int }}\left|x^2-4\right|dx$;

c) $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left|\sin x\right|dx$.

Lời giải chi tiết

a) $\underset{-2}{\overset{1}{\int }}\left|2x+2\right|dx$

$=\underset{-2}{\overset{-1}{\int }}\left|2x+2\right|dx+\underset{-1}{\overset{1}{\int }}\left|2x+2\right|dx$

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

$=-{\left(x^2+2x\right)|}_{-2}^{-1}+{\left(x^2+2x\right)|}_{-1}^1$

$=-\left[\left(-1\right)-0\right]+\left[3-\left(-1\right)\right]=5$.

b) $\underset{0}{\overset{4}{\int }}\left|x^2-4\right|dx=\underset{0}{\overset{2}{\int }}\left|x^2-4\right|dx+\underset{2}{\overset{4}{\int }}\left|x^2-4\right|dx$

$=\underset{0}{\overset{2}{\int }}\left(4-x^2\right)dx+\underset{2}{\overset{4}{\int }}\left(x^2-4\right)dx$

$={\left(4x-\frac{x{}^3}{3}\right)|}_0^2+{\left(\frac{x{}^3}{3}-4x\right)|}_2^4$

$=\left(\frac{16}{3}-0\right)+\left[\frac{16}{3}-\left(-\frac{16}{3}\right)\right]=16$.

c) $\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left|\sin x\right|dx=\underset{-\frac{\pi }{2}}{\overset{0}{\int }}\left|\sin x\right|dx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left|\sin x\right|dx$

$=\underset{-\frac{\pi }{2}}{\overset{0}{\int }}\left(-\sin x\right)dx+\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin xdx$

$=-{\left(-\cos x\right)|}_{-\frac{\pi }{2}}^0+{\left(-\cos x\right)|}_0^{\frac{\pi }{2}}$

$=-\left[-1-0\right]+\left[0-\left(-1\right)\right]=2$.