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#