What is #f(x) = int (x-3)^2-3x+4 dx# if #f(2) = 1 #?

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Answer 1 · 2016-01-24T22:17:25

#f(x) = x^3/3 -(9x^2)/2 +13x -29/3#

First, expand the integrant as follow

#int((x-3)^2 -3x +4)dx = int(x^2-6x+9-3x+4)dx#

#=int(x^2-9x+13)dx#

Then we can integrate this using the power rule like this

#f(x) = x^3/3-(9x^2)/2+13x+C##

We are given #f(2) = 1# , substitute this into the #f(x)# to solve for C

#1= (2^3)/3-(9(2)^2)/2+13(2) +C#

#1= 8/3 -36/2 +26+C#

#1= 8/3 -18 +26+C#

#C= -29/3#

So #f(x) = x^3/3 -(9x^2)/2 +13x -29/3#