Question

How do you find the antiderivative of `8sqrtx`?

Answer 1

`16/3x^(3/2)+C`

Explanation:

the antiderivitative of

`8sqrtx`

is given by

`int8sqrtxdx`

change to powers and use the power rule

`intx^ndx=x^(n+1)/(n+1)+C, " "n!=-1`

`int8sqrtxdx=8intx^(1/2)dx`

`=8xx(x^(1/2+1))/(1/2+1)+C`

`=8xxx^(3/2)/(3/2)+C`

`=8xx2/3x^(3/2)+C`

`=16/3x^(3/2)+C`