The value of the integral int_{-1}^{1}left{fr | vedclass

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(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2015}}{e^{|x|}(x^2 + \cos x)} + \frac{1}{e^{|x|}} \right\} dx$. Split the integral into two parts: $I = \i

Vedclass · Accepted Answer

$2\left(1-e^{-1}\right)$

Vedclass · Answer

(D) Let $I = \int_{-1}^{1} \left\{ \frac{x^{2015}}{e^{|x|}(x^2 + \cos x)} + \frac{1}{e^{|x|}} \right\} dx$. Split the integral into two parts: $I = \int_{-1}^{1} \frac{x^{2015}}{e^{|x|}(x^2 + \cos x)} dx + \int_{-1}^{1} \frac{1}{e^{|x|}} dx$. Let $f(x) = \frac{x^{2015}}{e^{|x|}(x^2 + \cos x)}$ and $g(x) = \frac{1}{e^{|x|}}$. Check for symmetry: $f(-x) = \frac{(-x)^{2015}}{e^{|-x|}((-x)^2 + \cos(-x))} = \frac{-x^{2015}}{e^{|x|}(x^2 + \cos x)} = -f(x)$. Thus, $f(x)$ is an odd function. $g(-x) = \frac{1}{e^{|-x|}} = \frac{1}{e^{|x|}} = g(x)$. Thus, $g(x)$ is an even function. Since $f(x)$ is odd, $\int_{-1}^{1} f(x) dx = 0$. Since $g(x)$ is even, $\int_{-1}^{1} g(x) dx = 2 \int_{0}^{1} g(x) dx$. Therefore, $I = 0 + 2 \int_{0}^{1} e^{-x} dx = 2 [-e^{-x}]_{0}^{1}$. $I = 2 (-e^{-1} - (-e^{0})) = 2(1 - e^{-1})$.

The value of the integral $\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{\mid x \mid}\left(x^{2}+\cos x\right)}+\frac{1}{e^{\mid{x} \mid}}\right\} d x$ is equal to

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