The value of the integral int_{-2}^0 (x^3 + 3 | vedclass

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(D) Let $I = \int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$. We can rewrite the expression as: $I = \int_{-2}^0 ((x + 1)^3 + 4 + (x +

Vedclass · Accepted Answer

$8$

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(D) Let $I = \int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$. We can rewrite the expression as: $I = \int_{-2}^0 ((x + 1)^3 + 4 + (x + 1) \cos(x + 1)) \, dx$. Substitute $t = x + 1$,then $dt = dx$. When $x = -2$,$t = -1$. When $x = 0$,$t = 1$. So,$I = \int_{-1}^1 (t^3 + 4 + t \cos t) \, dt$. We know that $t^3$ and $t \cos t$ are odd functions,and the integral of an odd function over a symmetric interval $[-a, a]$ is $0$. Therefore,$I = \int_{-1}^1 t^3 \, dt + \int_{-1}^1 4 \, dt + \int_{-1}^1 t \cos t \, dt$. $I = 0 + \int_{-1}^1 4 \, dt + 0$. $I = 4 [t]_{-1}^1 = 4(1 - (-1)) = 4(2) = 8$. The correct value is $8$.

The value of the integral $\int_{-2}^0 (x^3 + 3x^2 + 3x + 5 + (x + 1) \cos(x + 1)) \, dx$ is equal to:

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