Definite Integration
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^\infty x\,e^{-x^2}\,dx\) [JEE Main 2020]</p>
Step-by-Step Solution
Key Concept: Let t = x^2, dt = 2x dx. \int_0^\infty (1/2)e^(-t) dt = 1/2.
<div class='solution'>
<p>Let $t=x^2\Rightarrow dt=2x\,dx$:</p>
<p>$$\int_0^\infty xe^{-x^2}dx = \frac{1}{2}\int_0^\infty e^{-t}dt = \frac{1}{2}[-e^{-t}]_0^\infty = \frac{1}{2}(0+1)=\boxed{\frac{1}{2}}$$</p>
Correct Answer: A