Giải bài 65 trang 30

Đề bài

Tính:

a) $\underset{0}{\overset{2}{\int }}{e}^{-5x}dx$;

b) $\underset{0}{\overset{1}{\int }}{3}^{x+2}dx$;

c) $\underset{-1}{\overset{1}{\int }}{3}^{2x}dx$.

Lời giải chi tiết

a) $\underset{0}{\overset{2}{\int }}{e}^{-5x}dx=\underset{0}{\overset{2}{\int }}{\left(e^{-5}\right)}^{x}dx={\frac{{\left(e^{-5}\right)}^x}{\ln e^{-5}}|}_0^2$

$={-\frac{e^{-5x}}{5}|}_0^2=-\frac{e^{-5.2}}{5}+\frac{e^{-5.0}}{5}=-\frac{1}{5e^{10}}+\frac{1}{5}$.

b) $\underset{0}{\overset{1}{\int }}{3}^{x+2}dx=\underset{0}{\overset{1}{\int }}{9.3}^{x}dx={\frac{{9.3}^x}{\ln 3}|}_0^1$

$=\frac{{9.3}^1}{\ln 3}-\frac{{9.3}^0}{\ln 3}=\frac{18}{\ln 3}$.

c) $\underset{-1}{\overset{1}{\int }}{3}^{2x}dx=\underset{-1}{\overset{1}{\int }}{\left(3^2\right)}^{x}dx=\underset{-1}{\overset{1}{\int }}{9}^{x}dx$

$={\frac{9^x}{\ln 9}|}_{-1}^1=\frac{9^1}{\ln 9}-\frac{9^{-1}}{\ln 9}=\frac{80}{9\ln 9}$.