Question:
If \(\displaystyle\int x^5e^{-x^2}\,dx=g(x)e^{-x^2}+c\), where \(c\) is a constant of integration, then \(g(-1)\) is equal to :
If \(\displaystyle\int x^5e^{-x^2}\,dx=g(x)e^{-x^2}+c\), where \(c\) is a constant of integration, then \(g(-1)\) is equal to :
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Substitute \(t=x^2\) and integrate \(t^2e^{-t}\) by parts, then write the result as \(g(x)e^{-x^2}\).
Updated On: Nov 5, 2026
- \(1\)
- \(-1\)
- \(-\frac{5}{2}\)
- \(-\frac{1}{2}\)
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The Correct Option is C
Solution and Explanation
Put \(t=x^2\), \(dt=2x\,dx\). Then \(\int x^5e^{-x^2}dx=\frac12\int t^2e^{-t}dt\).
\(\int t^2e^{-t}dt=-e^{-t}(t^2+2t+2)\) (integration by parts twice).
So the integral \(=-\frac12e^{-x^2}(x^4+2x^2+2)+c\), hence \(g(x)=-\frac12(x^4+2x^2+2)\).
\(g(-1)=-\frac12(1+2+2)=-\frac52\).
Answer: (3) \(-\frac{5}{2}\)
\(\int t^2e^{-t}dt=-e^{-t}(t^2+2t+2)\) (integration by parts twice).
So the integral \(=-\frac12e^{-x^2}(x^4+2x^2+2)+c\), hence \(g(x)=-\frac12(x^4+2x^2+2)\).
\(g(-1)=-\frac12(1+2+2)=-\frac52\).
Answer: (3) \(-\frac{5}{2}\)
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