Giải bài 4.16 trang 13

Đề bài

Tính các tích phân sau:

a) $\underset{0}{\overset{1}{\int }}\left(3^x-2e^x\right)dx$;

b) $\underset{0}{\overset{1}{\int }}\frac{{\left(e^x-1\right)}^2}{2e^x}dx$.

Lời giải chi tiết

a) Ta có $\underset{0}{\overset{1}{\int }}\left(3^x-2e^x\right)dx$

$={\left(\frac{3^x}{\ln 3}-2e^x\right)|}_0^1$

$=\frac{3}{\ln 3}-2e-\frac{1}{\ln 3}+2$

$=2-2e+\frac{2}{\ln 3}$.

b) Ta có $\underset{0}{\overset{1}{\int }}\frac{{\left(e^x-1\right)}^2}{2e^x}dx$

$=\underset{0}{\overset{1}{\int }}\frac{e^{2x}-2e^x+1}{2e^x}dx$

$=\underset{0}{\overset{1}{\int }}\frac{e^{2x}-2e^x+1}{2e^x}dx$

$=\underset{0}{\overset{1}{\int }}\left(\frac{e^x}{2}-1+\frac{1}{2e^x}\right)dx$

$=\frac{1}{2}\underset{0}{\overset{1}{\int }}{e}^{x}dx-\underset{0}{\overset{1}{\int }}dx-\frac{1}{2}\underset{0}{\overset{1}{\int }}{\left(e^{-x}\right)}^{\prime }dx$

$={\frac{1}{2}{e}^x|}_0^1-{x|}_0^1-\frac{1}{2}{e^{-x}|}_0^1$

$=\frac{e-1}{2}-1-\frac{e^{-1}}{2}+\frac{1}{2}$

$=\frac{e-e^{-1}-2}{2}$.