Definite Integration
Definite Integral + IBP
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^1 xe^{-x}\,dx\) [JEE Main 2018]</p>
<li>\(1-\dfrac{2}{e}\)</li>
<li>\(2-\dfrac{1}{e}\)</li>
<li>\(1-\dfrac{1}{e^2}\)</li>
<li>\(\dfrac{2}{e}\)</li>
Step-by-Step Solution
Key Concept: IBP: \intxe^(-x)dx = -xe^(-x) - e^(-x) + C = -e^(-x)(x+1)+C.
<div class='solution'>
<p>IBP: $u=x$, $dv=e^{-x}dx\Rightarrow du=dx$, $v=-e^{-x}$.</p>
<p>$$\int_0^1 xe^{-x}dx = [-xe^{-x}]_0^1+\int_0^1 e^{-x}dx = -e^{-1}+[-e^{-x}]_0^1 = -\frac{1}{e}+(-\frac{1}{e}+1)=1-\frac{2}{e}$$</p>
</div>
Correct Answer: A