How do you evaluate $\int \sqrt{x^{3}} \cdot d x$ $\int \sqrt { x ^ { 3} } \cdot d x$ ?

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How do you evaluate $\int \sqrt{x^{3}} \cdot d x$ $\int \sqrt { x ^ { 3} } \cdot d x$ ?

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Answer 1 · 2018-05-13T15:53:36

let $I =$ $I=$ $\int \sqrt{x^{3}} d x$ $intsqrt(x^3)dx$
let $x = t^{2} (\Rightarrow x^{3} = t^{6})$ $x=t^2(=>x^3=t^6)$
$\Rightarrow \frac{d x}{d t} = 2 t$ $=>dx/dt=2t$ or $d x = 2 t d t$ $dx=2tdt$
putting this in the integral we get
$\int \sqrt{x^{3}} d x = \int \sqrt{t^{6}} \cdot 2 t d t$ $intsqrt(x^3)dx=intsqrt(t^6)*2tdt$
$= \int t^{3} \cdot 2 t d t = \int 2 t^{4} d t$ $=intt^3*2tdt=int2t^4dt$
$= 2 \cdot \frac{t^{5}}{5} + C$ $=2*t^5/5+C$
substitute for $t$ $t$
then required integral , $I = 2 \cdot \frac{{(\sqrt{x})}^{5}}{5} + C = 2 x^{\frac{5}{2}} + C$ $I=2*(sqrtx)^5/5+C=2x^(5/2)+C$
where $C$ $C$ is constant of integration

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How do you evaluate $\int \sqrt{x^{3}} \cdot d x$ $\int \sqrt { x ^ { 3} } \cdot d x$ ?

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