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Question

Mathematics Question on integral

The value of the integral 22x3+x(exx+1)dx∫^2_{-2}\frac{ |x^3+x|}{(e^x|x|+1)}dx is equal to:

A

5e25e^2

B

3e23e^{–2}

C

4

D

6

Answer

6

Explanation

Solution

I=22x3+x(exx+1)dx.....(i)I = ∫^2_{-2}\frac{ |x^3+x|}{(e^x|x|+1)}dx.....(i)

I=22x3+x(exx+1)dx.....(ii)I = ∫^2_{-2}\frac{ |x^3+x|}{(e^x|x|+1)}dx.....(ii)

2I=22x3+xdx2I = ∫^2_{-2} |x^3+x|dx

2I=202(x3+x)dx2I = 2 ∫^2_0(x^3+x)dx

I=02(x3+x)dxI = ∫^2_0 (x^3+x)dx

= (164+42)0\bigg(\frac{16}{4}+\frac{4}{2}\bigg)-0

= 4+2=64+2 = 6

Hence, the correct option is (D): 66