The value of the integral int_{-2}^{2}(1+2 si | vedclass

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Question

(C) Let $I = \int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$. We can split the integral as: $I = \int_{-2}^{2} e^{|x|} d x + 2 \int_{-2}^{2} \sin x e^{|x|} d x

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$2(e^{2}-1)$

Vedclass · Answer

(C) Let $I = \int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$. We can split the integral as: $I = \int_{-2}^{2} e^{|x|} d x + 2 \int_{-2}^{2} \sin x e^{|x|} d x$. Consider the first part: $f(x) = e^{|x|}$. Since $f(-x) = e^{|-x|} = e^{|x|} = f(x)$, $f(x)$ is an even function. Thus, $\int_{-2}^{2} e^{|x|} d x = 2 \int_{0}^{2} e^{x} d x = 2[e^{x}]_{0}^{2} = 2(e^{2}-1)$. Consider the second part: $g(x) = \sin x e^{|x|}$. Since $g(-x) = \sin(-x) e^{|-x|} = -\sin x e^{|x|} = -g(x)$, $g(x)$ is an odd function. Thus, $\int_{-2}^{2} \sin x e^{|x|} d x = 0$. Therefore, $I = 2(e^{2}-1) + 2(0) = 2(e^{2}-1)$.

The value of the integral $\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$ is equal to

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