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

1 Answer
Dec 6, 2017

#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#