int_{-3}^0 x sqrt{x+4} , dx = | vedclass

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(A) Let $I = \int_{-3}^0 x \sqrt{x+4} \, dx$. Substitute $x+4 = t$,so $x = t-4$ and $dx = dt$. When $x = -3$,$t = 1$. When $x = 0$,$t = 4$. Substituti

Vedclass · Accepted Answer

$\frac{-94}{15}$

Vedclass · Answer

(A) Let $I = \int_{-3}^0 x \sqrt{x+4} \, dx$. Substitute $x+4 = t$,so $x = t-4$ and $dx = dt$. When $x = -3$,$t = 1$. When $x = 0$,$t = 4$. Substituting these into the integral: $I = \int_1^4 (t-4) \sqrt{t} \, dt = \int_1^4 (t^{3/2} - 4t^{1/2}) \, dt$. Integrating term by term: $I = \left[ \frac{t^{5/2}}{5/2} - 4 \cdot \frac{t^{3/2}}{3/2} \right]_1^4 = \left[ \frac{2}{5} t^{5/2} - \frac{8}{3} t^{3/2} \right]_1^4$. Evaluating at the limits: $I = \left( \frac{2}{5}(4)^{5/2} - \frac{8}{3}(4)^{3/2} \right) - \left( \frac{2}{5}(1)^{5/2} - \frac{8}{3}(1)^{3/2} \right)$. $I = \left( \frac{2}{5}(32) - \frac{8}{3}(8) \right) - \left( \frac{2}{5} - \frac{8}{3} \right)$. $I = \left( \frac{64}{5} - \frac{64}{3} \right) - \left( \frac{6-40}{15} \right) = \left( \frac{192-320}{15} \right) - \left( \frac{-34}{15} \right) = \frac{-128}{15} + \frac{34}{15} = -\frac{94}{15}$.

$\int_{-3}^0 x \sqrt{x+4} \, dx =$

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