Browse by Stream
QnA
Home
Search for Exam, Articles, Questions
QnA
get app
Go To Your Study Dashboard
  1. Home
  2. Can someone help me with this, - Integral Calculus - JEE Main-11
# Engineering

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)
P Plabita

\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}

Similar Questions

Preparation Products

Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions