Question

$\underset{0}{\overset{\infty }{\int }}(2^{-x}-3^{-x})dx$ is equal to

• A. $\frac{3}{\ln 3}-\frac{2}{\ln 2}$
• B. $\frac{1}{\ln 3}-\frac{1}{\ln 2}$
• C. $\frac{1}{\ln 2}-\frac{1}{\ln 3}$
• D. $\frac{2}{\ln 2}-\frac{3}{\ln 3}$

Answer 1

The correct option is C $\frac{1}{\ln 2}-\frac{1}{\ln 3}$

$I=\underset{0}{\overset{\infty }{\int }}(2^{-x}-3^{-x})dx={\left[\frac{2^{-x}}{-\ln 2}\right]}_0^{\infty }-{\left[\frac{3^{-x}}{-\ln 3}\right]}_0^{\infty }\{\because \int a^{x}dx=\frac{a^x}{\ln a}+C\}=\frac{1}{\ln 2}-\frac{1}{\ln 3}$