Giải bài 4.14 trang 13

Đề bài

Tính các tích phân sau:

a) $\underset{0}{\overset{2}{\int }}\left|2x-1\right|dx$;

b) $\underset{-2}{\overset{3}{\int }}\left|x-1\right|dx$.

Lời giải chi tiết

a) Ta có $\underset{0}{\overset{2}{\int }}\left|2x-1\right|dx=\underset{0}{\overset{\frac{1}{2}}{\int }}\left|2x-1\right|dx+\underset{\frac{1}{2}}{\overset{2}{\int }}\left|2x-1\right|dx=\underset{0}{\overset{\frac{1}{2}}{\int }}\left(1-2x\right)dx+\underset{\frac{1}{2}}{\overset{2}{\int }}\left(2x-1\right)dx={\left(x-x^2\right)|}_0^{\frac{1}{2}}+{\left(x^2-x\right)|}_{\frac{1}{2}}^2$

$=\frac{1}{2}-\frac{1}{2^2}+\left[2^2-2-\left(\frac{1}{2^2}-\frac{1}{2}\right)\right]=\frac{5}{2}$.

b) Ta có $\underset{-2}{\overset{3}{\int }}\left|x-1\right|dx=\underset{-2}{\overset{1}{\int }}\left|x-1\right|dx+\underset{1}{\overset{3}{\int }}\left|x-1\right|dx=\underset{-2}{\overset{1}{\int }}\left(1-x\right)dx+\underset{1}{\overset{3}{\int }}\left(x-1\right)dx$

$={\left(x-\frac{x^2}{2}\right)|}_{-2}^1+{\left(\frac{x^2}{2}-x\right)|}_1^3=\frac{13}{2}$.