The value of the integral _-2^2|x^3+x|e^x|x|+1,dx is equal to:

Question

The value of the integral $\displaystyle\int_{-2}^2\frac{|x^3+x|}{e^{x|x|}+1}\,dx$ is equal to:

Accepted Answer

$$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx} $$ ..... (i)

$$I = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx} $$ ..... (ii)

$$2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx} $$

$$2I =2 \int\limits_0^2 {({x^3} + x)dx} $$

$$I = \int\limits_0^2 {({x^3} + x)dx} $$

$$ = \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2$$

$$ = \left( {{{16} \over 4} + {4 \over 2}} \right) - 0$$

$= 4 + 2 = 6$

Solutions · 1