int(3-x) sqrt{4-x} , dx = (Where C is a cons | vedclass

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(C) Let $I = \int(3-x) \sqrt{4-x} \, dx$. Substitute $u = 4-x$,then $du = -dx$,which means $dx = -du$. Also,$x = 4-u$,so $3-x = 3-(4-u) = u-1$. Substi

Vedclass · Accepted Answer

$\frac{2}{3}(4-x)^{3/2} - \frac{2}{5}(4-x)^{5/2} + C$

Vedclass · Answer

(C) Let $I = \int(3-x) \sqrt{4-x} \, dx$. Substitute $u = 4-x$,then $du = -dx$,which means $dx = -du$. Also,$x = 4-u$,so $3-x = 3-(4-u) = u-1$. Substituting these into the integral: $I = \int (u-1) \sqrt{u} (-du) = \int (1-u) \sqrt{u} \, du$ $I = \int (u^{1/2} - u^{3/2}) \, du$ $I = \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2} + C$ $I = \frac{2}{3} u^{3/2} - \frac{2}{5} u^{5/2} + C$ Substituting $u = 4-x$ back: $I = \frac{2}{3} (4-x)^{3/2} - \frac{2}{5} (4-x)^{5/2} + C$.

$\int(3-x) \sqrt{4-x} \, dx = $ (Where $C$ is a constant of integration.)

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