f(x) = int (x+3)^2 -3x dxf(x)=∫(x+3)2−3xdx if f(-1) = 0f(−1)=0.
f(-1)f(−1) is given to find the constant of integration and we shall use it after the integration.
int (x+3)^2 - 3x dx∫(x+3)2−3xdx
To simplify things we can expand (x+3)^2(x+3)2
(x+3)^2 = x^2+6x+9(x+3)2=x2+6x+9
int (x+3)^2 - 3x dx ∫(x+3)2−3xdx
=int x^2 + 6x + 9 - 3x dx=∫x2+6x+9−3xdx
=int x^2 +3x + 9 dx=∫x2+3x+9dx
= x^(2+1)/(2+1) + 3x^(1+1)/(1+1) + 9x + C=x2+12+1+3x1+11+1+9x+C
=x^3/3 - 3x^2/2 + 9x + C=x33−3x22+9x+C
This is f(x) = x^3/3 + 3x^2/2 + 9x + Cf(x)=x33+3x22+9x+C
We are given f(-1) = 6f(−1)=6 Let us use this
f(-1) = (-1)^3/3 +3(-1)^2/2 + 9(-1) + Cf(−1)=(−1)33+3(−1)22+9(−1)+C
6 = -1/3+3/2 - 9 + C6=−13+32−9+C
6 = -2/6+9/6-54/6 + C6=−26+96−546+C
6=-47/6+C6=−476+C
6+47/6=C6+476=C
36/6 + 47/6 = C366+476=C
83/6 = C836=C
f(x) = x^3/3 +3/2x^2 +9x + 83/6f(x)=x33+32x2+9x+836
