What is #f(x) = int (x-2)^2-5x+4 dx# if #f(2) = 1 #?
1 Answer
Feb 7, 2016
#f(x) = 1/3 (x-2)^3 -5/2 x^2 + 4x + 3 #
Explanation:
#intax^n dx =(ax^(n+1))/(n+1) + c # is the standard integral , where c is the constant of integration
further
#int(ax+b)^n dx = (ax+b)^(n+1)/(a(n+1)) + c# These can be applied 'term by term' to the above integral
# f(x) = 1/3 (x-2)^3 - 5/2 x^2 + 4x + c # using f(2) = 1 , allows c , to be found
# 1/3 (2-2)^3 - 5/2. 4 + 8 + c = 1# 0 - 10 + 8 + c = 1 → c = 3
# rArr f(x) = 1/3 (x-2)^3 - 5/2 x^2 + 4x + 3 #
