intlimits_{-1}^{1} frac{x^4}{1 + e^{x^7}} dx | vedclass

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(C) Let $I = \int_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get: $I = \int_{-1}^{1

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(C) Let $I = \int_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,we get: $I = \int_{-1}^{1} \frac{(-x)^4}{1 + e^{(-x)^7}} dx = \int_{-1}^{1} \frac{x^4}{1 + e^{-x^7}} dx$. Multiplying the numerator and denominator by $e^{x^7}$: $I = \int_{-1}^{1} \frac{x^4 e^{x^7}}{e^{x^7} + 1} dx$. Adding the two expressions for $I$: $2I = \int_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx + \int_{-1}^{1} \frac{x^4 e^{x^7}}{1 + e^{x^7}} dx = \int_{-1}^{1} \frac{x^4(1 + e^{x^7})}{1 + e^{x^7}} dx$. $2I = \int_{-1}^{1} x^4 dx$. Since $x^4$ is an even function,$2I = 2 \int_{0}^{1} x^4 dx$. $I = \int_{0}^{1} x^4 dx = \left[ \frac{x^5}{5} \right]_{0}^{1} = \frac{1}{5} - 0 = \frac{1}{5}$.

$\int\limits_{-1}^{1} \frac{x^4}{1 + e^{x^7}} dx = $

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