The value of the integral int_{-2}^{2} frac{| | vedclass

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(B) Let $f(x) = \frac{|x^{3}+x|}{e^{x|x|}+1}$.We use the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$.Here $a = 2$,so $\int_{-2}^

Vedclass · Accepted Answer

$6$

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(B) Let $f(x) = \frac{|x^{3}+x|}{e^{x|x|}+1}$.We use the property $\int_{-a}^{a} f(x) dx = \int_{0}^{a} (f(x) + f(-x)) dx$.Here $a = 2$,so $\int_{-2}^{2} f(x) dx = \int_{0}^{2} (f(x) + f(-x)) dx$.Since $|x^3+x| = |x(x^2+1)| = |x|(x^2+1)$,for $x > 0$,$|x^3+x| = x(x^2+1) = x^3+x$.Also,$f(-x) = \frac{|(-x)^3+(-x)|}{e^{-x|-x|}+1} = \frac{|-(x^3+x)|}{e^{-x^2}+1} = \frac{x^3+x}{e^{-x^2}+1}$.Thus,$f(x) + f(-x) = \frac{x^3+x}{e^{x^2}+1} + \frac{x^3+x}{e^{-x^2}+1} = (x^3+x) \left( \frac{1}{e^{x^2}+1} + \frac{e^{x^2}}{1+e^{x^2}} \right) = (x^3+x) \left( \frac{1+e^{x^2}}{1+e^{x^2}} \right) = x^3+x$.Therefore,$\int_{0}^{2} (x^3+x) dx = \left[ \frac{x^4}{4} + \frac{x^2}{2} \right]_{0}^{2} = \left( \frac{16}{4} + \frac{4}{2} \right) - 0 = 4 + 2 = 6$.

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

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