The value of int_{-2}^2(x cos x+sin x+1) d x | vedclass

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

Vedclass · Accepted Answer

$4$

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(D) Let $I = \int_{-2}^2(x \cos x+\sin x+1) d x$. We can split the integral as $I = \int_{-2}^2 x \cos x d x + \int_{-2}^2 \sin x d x + \int_{-2}^2 1 d x$. Recall the property of definite integrals: $\int_{-a}^a f(x) d x = 0$ if $f(x)$ is an odd function,and $2 \int_0^a f(x) d x$ if $f(x)$ is an even function. Let $f_1(x) = x \cos x$. Since $f_1(-x) = (-x) \cos(-x) = -x \cos x = -f_1(x)$,$f_1(x)$ is an odd function. Thus,$\int_{-2}^2 x \cos x d x = 0$. Let $f_2(x) = \sin x$. Since $f_2(-x) = \sin(-x) = -\sin x = -f_2(x)$,$f_2(x)$ is an odd function. Thus,$\int_{-2}^2 \sin x d x = 0$. Therefore,$I = 0 + 0 + \int_{-2}^2 1 d x = [x]_{-2}^2 = 2 - (-2) = 4$.

The value of $\int_{-2}^2(x \cos x+\sin x+1) d x$ is

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