Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The value of the integral  <img alt="\mathrm \int_{-2}^{2} \frac{\left|x^{3}+x\right|}{\left({e}^{x|x|}+1\right)} \mathrm{d} x" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%2

The value of the integral  \mathrm \int_{-2}^{2} \frac{\left|x^{3}+x\right|}{\left({e}^{x|x|}+1\right)} \mathrm{d} x is equal to :

 

Option: 1

5 \mathrm{e}^{2}


Option: 2

3 \mathrm{e}^{-2}


Option: 3

4


Option: 4

6


Answers (1)

best_answer

\mathrm{I=\int_{-2}^{2} \frac{\left|x^{3}+x\right|}{e^{x|x|}+1} d x}                .............(1)

\mathrm{I(x)=I(a+b-x)} \\

\mathrm{\Rightarrow I =\int_{-2}^{2} \frac{\left|-x^{3}-x\right|}{e^{-x|-x|}+1} d x }\\

\mathrm{=\int_{-2}^{2} \frac{\left|x^{3}+x\right|}{e^{-x|x|}+1}=\int_{-2}^{2} \frac{e^{x|x|}\left|x^{3}+x\right|}{e^{x|x|}+1}dx}               ............(2)

Add eqn (1) and (2)

\mathrm{\Rightarrow 2 I =\int_{-2}^{2}\left|x^{3}+x\right| d x=2 \int_{0}^{2}\left|x^{3}+x\right| d x }\\

\mathrm{ \Rightarrow I=\int_{0}^{2}\left(x^{3}+x\right) d x=6}

Hence the correct aswer is option 4.

Posted by

Sanket Gandhi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question