int_{-2}^0 (x^3+3x^2+3x+3+(x+1) cos(x+1)) , d | vedclass

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

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$4$

Vedclass · Answer

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

$\int_{-2}^0 (x^3+3x^2+3x+3+(x+1) \cos(x+1)) \, dx$ is equal to:

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