Definite Integration
Improper + Exponential
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^\infty x^2 e^{-x}\,dx\) [JEE Advanced 2005]</p>
Step-by-Step Solution
Key Concept: Gamma function: \int_0^\infty x^2e^(-x)dx = \Gamma(3) = 2\! = 2.
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<p>\(\int_0^\infty x^2 e^{-x}dx=\Gamma(3)=2\!=\boxed{2}\)</p>
<p><em>Direct IBP:</em> \(=[-x^2e^{-x}]_0^\infty+2\int_0^\infty xe^{-x}dx=0+2[-xe^{-x}-e^{-x}]_0^\infty=0+2(0+1)=2\). ✓</p>
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Correct Answer: A