Use the substitution u=-x to evaluate int_{-2}^{2} frac{x^{2}}{e^{x}+1} d x

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Accepted Answer

Step:1 ：Let u = -x. Then, x = -u and d x = -d u. Also, when x = -2, u = 2, and when x = 2, u = -2.Step:2 ：Substitute u into the integral: int_{-2}^{2} frac{x^{2}}{e^{x}+1} d x = -int_{2}^{-2} frac{u^{2}}{e^{u}+1} d u.Step:3 ：Evaluate the integral: int_{-2}^{2} frac{x^{2}}{e^{x}+1} d x = -int_{2}^{-2} frac{u^{2}}{e^{u}+1} d u.Step:4 ：(boxed{-int_{2}^{-2} frac{u^{2}}{e^{u}+1} d u})

Answer

Step: 1 ：Let u = -x. Then, x = -u and d x = -d u. Also, when x = -2, u = 2, and when x = 2, u = -2.Step: 2 ：Substitute u into the integral: int_{-2}^{2} frac{x^{2}}{e^{x}+1} d x = -int_{2}^{-2} frac{u^{2}}{e^{u}+1} d u.Step: 3 ：Evaluate the integral: int_{-2}^{2} frac{x^{2}}{e^{x}+1} d x = -int_{2}^{-2} frac{u^{2}}{e^{u}+1} d u.Step: 4 ：(boxed{-int_{2}^{-2} frac{u^{2}}{e^{u}+1} d u})

Use the substitution $u = - x$ to evaluate $\int_{- 2}^{2} \frac{x^{2}}{e^{x} + 1} d x$

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Use the substitution $u = - x$ to evaluate $\int_{- 2}^{2} \frac{x^{2}}{e^{x} + 1} d x$