How do you integrate $\int x \sqrt{4 - x}$ $int xsqrt(4-x)$ using substitution?

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How do you integrate $\int x \sqrt{4 - x}$ $int xsqrt(4-x)$ using substitution?

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Answer 1 · 2016-11-03T08:37:30

$\int . x \sqrt{4 - x} . d x = - \frac{2}{15} (3 x + 8) {(4 - x)}^{\frac{3}{2}} + C$ $int color(white)(.)xsqrt(4-x)color(white)(.) dx = -2/15(3x+8)(4-x)^(3/2) + C$

Explanation:

Let $u = 4 - x$ $u = 4-x$

Then $d u = - d x$ $du = -dx$

$\int . x \sqrt{4 - x} . d x = \int . (u - 4) \sqrt{u} . d u$ $int color(white)(.)xsqrt(4-x)color(white)(.) dx = int color(white)(.)(u-4)sqrt(u)color(white)(.) du$

$\int . x \sqrt{4 - x} . d x = \int . u^{\frac{3}{2}} - 4 u^{\frac{1}{2}} . d u$ $color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = int color(white)(.)u^(3/2) - 4 u^(1/2)color(white)(.) du$

$\int . x \sqrt{4 - x} . d x = \frac{2}{5} u^{\frac{5}{2}} - \frac{8}{3} u^{\frac{3}{2}} + C$ $color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = 2/5 u^(5/2) - 8/3 u^(3/2) + C$

$\int . x \sqrt{4 - x} . d x = \frac{2}{5} {(4 - x)}^{\frac{5}{2}} - \frac{8}{3} {(4 - x)}^{\frac{3}{2}} + C$ $color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = 2/5 (4-x)^(5/2) - 8/3 (4-x)^(3/2) + C$

$\int . x \sqrt{4 - x} . d x = (\frac{2}{5} (4 - x) - \frac{8}{3}) {(4 - x)}^{\frac{3}{2}} + C$ $color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = (2/5 (4-x) - 8/3)(4-x)^(3/2) + C$

$\int . x \sqrt{4 - x} . d x = \frac{2}{15} (3 (4 - x) - 20) {(4 - x)}^{\frac{3}{2}} + C$ $color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = 2/15(3 (4-x) - 20)(4-x)^(3/2) + C$

$\int . x \sqrt{4 - x} . d x = - \frac{2}{15} (3 x + 8) {(4 - x)}^{\frac{3}{2}} + C$ $color(white)(int color(white)(.)xsqrt(4-x)color(white)(.) dx) = -2/15(3x+8)(4-x)^(3/2) + C$

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How do you integrate $\int x \sqrt{4 - x}$ $int xsqrt(4-x)$ using substitution?

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