What is f(x) = int (3x+5)^2-x dx if f(1)=2 ?

1 Answer
Jan 28, 2016

f(x)=3x^3+29/2x^2+25x-81/2

Explanation:

The easiest way to integrate this is by expanding (3x+5)^2 and subtracting x to get

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 C, the constant of integration for this particular function, since we know that f(1)=2.

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