What is f(x) = int (3x+5)^2-x dx if f(1)=2 ?
1 Answer
Jan 28, 2016
Explanation:
The easiest way to integrate this is by expanding
int9x^2+29x+25dx
We can now integrate each term individually using the rule
intx^ndx=(x^(n+1))/(n+1)+C
This gives us the indefinite integral of
=(9x^(2+1))/(2+1)+(29x^(1+1))/(1+1)+(25x^(0+1))/(0+1)+C
=3x^3+29/2x^2+25x+C
We can now find
f(1)=3(1^3)+29/2(1^2)+25(1)+C=2
85/2+C=2
C=-81/2
Thus,
f(x)=3x^3+29/2x^2+25x-81/2