How do you use substitution to integrate $x^{2} \cdot {(x - 2)}^{\frac{1}{2}}$ $x^2*(x-2)^(1/2)$ ?

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How do you use substitution to integrate $x^{2} \cdot {(x - 2)}^{\frac{1}{2}}$ $x^2*(x-2)^(1/2)$ ?

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Answer 1 · 2018-03-08T09:37:21

The answer is $= \frac{2}{7} {(x - 2)}^{\frac{7}{2}} + \frac{8}{5} {(x - 2)}^{\frac{5}{2}} + \frac{8}{3} {(x - 2)}^{\frac{3}{2}} + C$ $=2/7(x-2)^(7/2)+8/5(x-2)^(5/2)+8/3(x-2)^(3/2)+C$

Explanation:

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

Perform the substitution

$u = x - 2$ $u=x-2$ , $\Rightarrow$ $=>$ , $d x = d u$ $dx=du$

Therefore,

$\int x^{2} \sqrt{x - 2} d x = \int {(u + 2)}^{2} \sqrt{u} d u$ $intx^2sqrt(x-2)dx=int(u+2)^2sqrtudu$

$= \int (u^{2} + 4 u + 4) \sqrt{u} d u$ $=int(u^2+4u+4)sqrt(u)du$

$= \int (u^{\frac{5}{2}} + 4 u^{\frac{3}{2}} + 4 u^{\frac{1}{2}}) d u$ $=int(u^(5/2)+4u^(3/2)+4u^(1/2))du$

$= \int u^{\frac{5}{2}} d u + 4 \int u^{\frac{3}{2}} d u + 4 \int u^{\frac{1}{2}} d u$ $=intu^(5/2)du+4intu^(3/2)du+4intu^(1/2)du$

$= \frac{u^{\frac{7}{2}}}{\frac{7}{2}} + 4 \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + 4 \frac{u^{\frac{3}{2}}}{\frac{3}{2}}$ $=u^(7/2)/(7/2)+4u^(5/2)/(5/2)+4u^(3/2)/(3/2)$

$= \frac{2}{7} u^{\frac{7}{2}} + \frac{8}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}}$ $=2/7u^(7/2)+8/5u^(5/2)+8/3u^(3/2)$

$= \frac{2}{7} {(x - 2)}^{\frac{7}{2}} + \frac{8}{5} {(x - 2)}^{\frac{5}{2}} + \frac{8}{3} {(x - 2)}^{\frac{3}{2}} + C$ $=2/7(x-2)^(7/2)+8/5(x-2)^(5/2)+8/3(x-2)^(3/2)+C$

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How do you use substitution to integrate $x^{2} \cdot {(x - 2)}^{\frac{1}{2}}$ $x^2*(x-2)^(1/2)$ ?

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