Giải bài 6 trang 15

Đề bài

Tính:

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

b) $B=\underset{-1}{\overset{0}{\int }}\left(x^3-6x\right)dx+\underset{0}{\overset{1}{\int }}\left(t^3-6t\right)dt$.

Lời giải chi tiết

a)

$\begin{array}{c}A=\underset{-1}{\overset{2}{\int }}\left(x-4x^2\right)dx+4\underset{-1}{\overset{2}{\int }}\left(x^2-1\right)dx=\underset{-1}{\overset{2}{\int }}\left(x-4x^2\right)dx+\underset{-1}{\overset{2}{\int }}\left(4x^2-4\right)dx=\underset{-1}{\overset{2}{\int }}\left(x-4x^2+4x^2-4\right)dx\\ =\underset{-1}{\overset{2}{\int }}\left(x-4\right)dx={\left(\frac{x^2}{2}-4x\right)|}_{-1}^2=\left(\frac{2^2}{2}-4.2\right)-\left(\frac{{\left(-1\right)}^2}{2}-4.\left(-1\right)\right)=-\frac{21}{2}\end{array}$

b)

$\begin{array}{c}B=\underset{-1}{\overset{0}{\int }}\left(x^3-6x\right)dx+\underset{0}{\overset{1}{\int }}\left(t^3-6t\right)dt=\underset{-1}{\overset{0}{\int }}\left(x^3-6x\right)dx+\underset{0}{\overset{1}{\int }}\left(x^3-6x\right)dx\\ =\underset{-1}{\overset{1}{\int }}\left(x^3-6x\right)dx={\left(\frac{x^4}{4}-3x^2\right)|}_{-1}^1=\left(\frac{1^4}{4}-{3.1}^2\right)-\left(\frac{{\left(-1\right)}^4}{4}-3.{\left(-1\right)}^2\right)=0\end{array}$