How do you integrate $\int (x + 1) \sqrt{2 - x} d x$ $int (x+1)sqrt(2-x)dx$ ?

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How do you integrate $\int (x + 1) \sqrt{2 - x} d x$ $int (x+1)sqrt(2-x)dx$ ?

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Answer 1 · 2016-12-10T21:36:43

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

Explanation:

$I = \int (x + 1) \sqrt{2 - x} d x$ $I=int(x+1)sqrt(2-x)dx$

Let $u = 2 - x$ $u=2-x$ . This implies that $d u = - d x$ $du=-dx$ . Also note that $- u = x - 2$ $-u=x-2$ , so $- u + 3 = x + 1$ $-u+3=x+1$ . Then:

$I = \int (- u + 3) \sqrt{u} (- d u) = \int (u - 3) \sqrt{u} d u$ $I=int(-u+3)sqrtu(-du)=int(u-3)sqrtudu$

Expanding the square root as $u^{\frac{1}{2}}$ $u^(1/2)$ :

$I = \int (u (u^{\frac{1}{2}}) - 3 u^{\frac{1}{2}}) d u = \int (u^{\frac{3}{2}} - 3 u^{\frac{1}{2}}) d u$ $I=int(u(u^(1/2))-3u^(1/2))du=int(u^(3/2)-3u^(1/2))du$

Now using $\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C$ $intu^ndu=u^(n+1)/(n+1)+C$ :

$I = \frac{u^{\frac{5}{2}}}{\frac{5}{2}} - 3 (\frac{u^{\frac{3}{2}}}{\frac{3}{2}}) = \frac{2}{5} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}}$ $I=u^(5/2)/(5/2)-3(u^(3/2)/(3/2))=2/5u^(5/2)-2u^(3/2)$

Factoring and making it look nice:

$I = u^{\frac{3}{2}} (\frac{2}{5} u - 2) = \frac{u^{\frac{3}{2}} (2 u - 10)}{5} = \frac{2 u^{\frac{3}{2}} (u - 5)}{5}$ $I=u^(3/2)(2/5u-2)=(u^(3/2)(2u-10))/5=(2u^(3/2)(u-5))/5$

From $u = 2 - x$ $u=2-x$ :

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

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How do you integrate $\int (x + 1) \sqrt{2 - x} d x$ $int (x+1)sqrt(2-x)dx$ ?

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