Question

${\int }_0^{\infty }(a^{-x}-b^{-x})dx=$

• A. $\frac{1}{loga}-\frac{1}{logb}$
• B. $loga-logb$
• C. $loga+logb$
• D. $\frac{1}{loga}+\frac{1}{logb}$

Answer 1

The correct option is A $\frac{1}{loga}-\frac{1}{logb}$

${\int }_0^{\infty }(a^{-x}-b^{-x})dx={[\frac{a^{-x}}{log\left(a^{-1}\right)}-\frac{b^{-x}}{log\left(b^{-1}\right)}]}_0^{\infty }=-[\frac{1}{-loga}-\frac{1}{-logb}]$
$=\frac{1}{loga}-\frac{1}{logb}$