Search any question & find its solution
Question:
Answered & Verified by Expert
If $I_n=\int x^n \cdot e^{c x} d x$ for $n \geq 1, \quad$ then $c \cdot I_n+n \cdot I_{n-1}$ is equal to
Options:
Solution:
1190 Upvotes
Verified Answer
The correct answer is:
$x^n e^{c x}$
Given that,
$$
\begin{aligned}
& I_n=\int x^n \cdot e^{c x} d x \\
& I_n=\frac{e^{c x}}{c} \cdot x^n-\int \frac{e^{c x}}{c} \cdot n x^{n-1} d x \\
& \Rightarrow \quad I_n=\frac{e^{c x} \cdot x^n}{c}-\frac{n}{c} I_{n-1} \\
&
\end{aligned}
$$
$\Rightarrow \quad c I_n+n I_{n-1}=e^{c x} \cdot x^n$
$$
\begin{aligned}
& I_n=\int x^n \cdot e^{c x} d x \\
& I_n=\frac{e^{c x}}{c} \cdot x^n-\int \frac{e^{c x}}{c} \cdot n x^{n-1} d x \\
& \Rightarrow \quad I_n=\frac{e^{c x} \cdot x^n}{c}-\frac{n}{c} I_{n-1} \\
&
\end{aligned}
$$
$\Rightarrow \quad c I_n+n I_{n-1}=e^{c x} \cdot x^n$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.