Giải bài 1 trang 14

Đề bài

Tính các tích phân sau:

a) $\underset{0}{\overset{2}{\int }}\left(3x-2\right)\left(3x+2\right)dx$;

b) $\underset{1}{\overset{2}{\int }}{t}^2\left(5t^2-2\right)dt$;

c) $\underset{-1}{\overset{1}{\int }}\left(x-2\right)\left(x^2+2x+4\right)dx$.

Lời giải chi tiết

a) $\underset{0}{\overset{2}{\int }}\left(3x-2\right)\left(3x+2\right)dx=\underset{0}{\overset{2}{\int }}\left(9x^2-4\right)dx={\left(3x^3-4x\right)|}_0^2=\left({3.2}^3-4.2\right)-\left({3.0}^3-4.0\right)=16$.

b) $\underset{1}{\overset{2}{\int }}{t}^2\left(5t^2-2\right)dt=\underset{1}{\overset{2}{\int }}\left(5t^4-2t^2\right)dt={\left(t^5-\frac{2t^3}{3}\right)|}_1^2=\left(2^5-\frac{{2.2}^3}{3}\right)-\left(1^5-\frac{{2.1}^3}{3}\right)=\frac{79}{3}$.

c) $\underset{-1}{\overset{1}{\int }}\left(x-2\right)\left(x^2+2x+4\right)dx=\underset{-1}{\overset{1}{\int }}\left(x^3-8\right)dx={\left(\frac{x^4}{4}-8x\right)|}_{-1}^1=\left(\frac{1^4}{4}-8.1\right)-\left(\frac{{\left(-1\right)}^4}{4}-8.\left(-1\right)\right)=-16$.