If f(x) = begin{cases} e^{cos x}sin x, & |x| | vedclass

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(C) We are given the function $f(x) = \begin{cases} e^{\cos x}\sin x, & |x| \le 2 \ 2, & 	ext{otherwise} \end{cases}$.We need to evaluate the integr

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$2$

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(C) We are given the function $f(x) = \begin{cases} e^{\cos x}\sin x, & |x| \le 2 \ 2, & 	ext{otherwise} \end{cases}$.We need to evaluate the integral $I = \int_{-2}^{3} f(x) dx$.We can split the integral at $x = 2$ as follows:$I = \int_{-2}^{2} f(x) dx + \int_{2}^{3} f(x) dx$.For the interval $[-2, 2]$,$f(x) = e^{\cos x}\sin x$. Let $g(x) = e^{\cos x}\sin x$. Then $g(-x) = e^{\cos(-x)}\sin(-x) = e^{\cos x}(-\sin x) = -g(x)$. Thus,$g(x)$ is an odd function.By the property of definite integrals,if $g(x)$ is an odd function,then $\int_{-a}^{a} g(x) dx = 0$. Therefore,$\int_{-2}^{2} e^{\cos x}\sin x dx = 0$.For the interval $(2, 3]$,$f(x) = 2$. Thus,$\int_{2}^{3} 2 dx = [2x]_{2}^{3} = 2(3 - 2) = 2$.Adding these results,$I = 0 + 2 = 2$.

If $f(x) = \begin{cases} e^{\cos x}\sin x, & |x| \le 2 \\ 2, & \text{otherwise} \end{cases}$,then $\int_{-2}^{3} f(x) dx$ is equal to

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