Value of the definite integral int_{-3}^{1} ( | vedclass

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(A) Let $I = \int_{-3}^{1} (2(t+1)^5 - 5(t+1)^3 + t + 3) dt$. Substitute $z = t+1$,then $dz = dt$. When $t = -3$,$z = -2$. When $t = 1$,$z = 2$. $I =

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$8$

Vedclass · Answer

(A) Let $I = \int_{-3}^{1} (2(t+1)^5 - 5(t+1)^3 + t + 3) dt$. Substitute $z = t+1$,then $dz = dt$. When $t = -3$,$z = -2$. When $t = 1$,$z = 2$. $I = \int_{-2}^{2} (2z^5 - 5z^3 + (z-1) + 3) dz$ $I = \int_{-2}^{2} (2z^5 - 5z^3 + z + 2) dz$ Since $f(z) = 2z^5 - 5z^3 + z$ is an odd function,$\int_{-2}^{2} (2z^5 - 5z^3 + z) dz = 0$. Therefore,$I = \int_{-2}^{2} 2 dz = [2z]_{-2}^{2} = 2(2 - (-2)) = 2(4) = 8$.

Value of the definite integral $\int_{-3}^{1} (2(t+1)^5 - 5(t+1)^3 + t + 3) dt$ is equal to

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