Question

What is the antiderivative of the function, f'(x)=`sqrtx`(6+5x), and f(1)=10?

Answer 1

`f(x)=4x^(3/2)+2x^(5/2)+4`

Explanation:

STEP 1 : Find the antiderivative of the function by using: if f(x)=`ax^n`, then F(x)=`a/(n+1)` `x^(n+1)`
`f(x)=6*2/3x^(3/2)+5*2/5x^(5/2)+c` = `4x^(3/2)+2x^(5/2)+c`

STEP2 : Substitute f(1)=10 into the F(x) found to find c
f(1)=10=4+2+c
c=4

STEP 3 : Write out the entire F(x) with the constant found
`f(x)=4x^(3/2)+2x^(5/2)+4`