How do you evaluate the integral $\int x \sqrt{x - 2}$ $int xsqrt(x-2)$ ?

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Answer 1 · 2017-01-29T08:46:03

The answer is $= \frac{2}{15} {(x - 2)}^{\frac{3}{2}} (3 x + 4) + C$ $=2/15(x-2)^(3/2)(3x+4)+ C$

We need

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

We solve this integral by substitution

Let $u = x - 2$ $u=x-2$

$d u = d x$ $du=dx$

and $x = u + 2$ $x=u+2$

Therefore,

$\int x \sqrt{x - 2} d x = \int (u + 2) \sqrt{u} d u$ $intxsqrt(x-2)dx=int(u+2)sqrtu du$

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

$= \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + 2 \cdot \frac{u^{\frac{3}{2}}}{\frac{3}{2}}$ $=u^(5/2)/(5/2)+2*u^(3/2)/(3/2)$

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

$= \frac{2}{15} u^{\frac{3}{2}} (3 u + 10)$ $=2/15u^(3/2)(3u+10)$

$= \frac{2}{15} {(x - 2)}^{\frac{3}{2}} (3 x - 6 + 10) + C$ $=2/15(x-2)^(3/2)(3x-6+10)+C$

$= \frac{2}{15} {(x - 2)}^{\frac{3}{2}} (3 x + 4) + C$ $=2/15(x-2)^(3/2)(3x+4)+ C$

How do you evaluate the integral ∫x√x−2int xsqrt(x-2)?