Q. If $I_{n} = \int_{}^{}x^{n}.e^{cx}\ dx\ \ for\ \ n \geq 1,\ then\ c.I_{n} + n.I_{n - 1}$ is equal to
A) $x^{n}\ e^{cx}$
B) $x^{n}$
C) $e^{cx}$
D) $x^{n} + e^{cx}$
Explanation
(a)
Given, $I_{n} = \int_{}^{}{x^{n}.e^{cx}}dx$
$= \frac{e^{cx}}{c}.x^{n} - \int_{}^{}\frac{e^{cx}}{c}.nx^{n - 1}dx$
$\Rightarrow I_{n} = \frac{e^{cx}.x^{n}}{c} - \frac{n}{c}\ I_{n - 1}$
$\Rightarrow cI_{n} + nI_{n - 1} = e^{cx}.x^{n}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.