Question

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

Answer 1

let `I=` `intsqrt(x^3)dx`
let `x=t^2(=>x^3=t^6)`
`=>dx/dt=2t` or `dx=2tdt`
putting this in the integral we get
`intsqrt(x^3)dx=intsqrt(t^6)*2tdt`
`=intt^3*2tdt=int2t^4dt`
`=2*t^5/5+C`
substitute for `t`
then required integral ,`I=2*(sqrtx)^5/5+C=2x^(5/2)+C`
where `C` is constant of integration