The integral value int_{-2}^{0} [x^3 + 3x^2 + | vedclass

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(B) Let $I = \int_{-2}^{0} [x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)] \, dx$. Substitute $t = x + 1$,so $dt = dx$. When $x = -2$,$t = -1$. When $x = 0

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $I = \int_{-2}^{0} [x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)] \, dx$. Substitute $t = x + 1$,so $dt = dx$. When $x = -2$,$t = -1$. When $x = 0$,$t = 1$. Also,$x^3 + 3x^2 + 3x + 3 = (x+1)^3 + 2 = t^3 + 2$. Thus,$I = \int_{-1}^{1} [t^3 + 2 + t\cos(t)] \, dt$. We can split this into $I = \int_{-1}^{1} t^3 \, dt + \int_{-1}^{1} 2 \, dt + \int_{-1}^{1} t\cos(t) \, dt$. Since $t^3$ and $t\cos(t)$ are odd functions,their integrals over the symmetric interval $[-1, 1]$ are $0$. Therefore,$I = 0 + \int_{-1}^{1} 2 \, dt + 0 = [2t]_{-1}^{1} = 2(1 - (-1)) = 2(2) = 4$.

The integral value $\int_{-2}^{0} [x^3 + 3x^2 + 3x + 3 + (x + 1)\cos(x + 1)] \, dx$ is

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