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  • Can someone help me with this, - Integral Calculus - JEE Main-11

If \int x^{5}e^{-x^{2}}dx=g(x)e^{-x^{2}}+C, where C is a constant 

of integration, then g(-1) is equal to :

  • Option 1)

    -1

  • Option 2)

    1

  • Option 3)

    -\frac{5}{2}

  • Option 4)

    -\frac{1}{2}

 

Answers (1)

best_answer

\int x^{5}e^{-x^{2}}dx=g(x)e^{-x^{2}}+C

I=\int x^{5}e^{-x^{2}}dx

Let -x^{2}=t

       -2xdx=dt

I=\frac{-1}{2}\int t^{2}e^{t}dt

Integrating by parts

 I=\frac{-1}{2}[t^{2}\int e^{t}dt-\int (2t)\int (e^{t}dt) dt] 

I=\frac{-1}{2}[t^{2} e^{t}dt-\int (2t e^{t}dt) dt]

I=\frac{-1}{2}[t^{2} e^{t}-2(te^{t}-e^{t})]+C

I=\frac{-e^{-x^{2}}}{2}[x^{4} +2x^{2}+2]+C

=> g(x)=\frac{-1}{2}[x^{4} +2x^{2}+2]

=> g(-1)=\frac{-1}{2}[(-1)^{4} +2(-1)^{2}+2]

                   =\frac{-5}{2}

So, option (3) is correct.

 

 


Option 1)

-1

Option 2)

1

Option 3)

-\frac{5}{2}

Option 4)

-\frac{1}{2}

Posted by

Plabita

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