What is #f(x) = int (x-2)^2-5x+4 dx# if #f(2) = 1 #?

Question

What is #f(x) = int (x-2)^2-5x+4   dx# if #f(2) = 1   #?

Christopher Allers · Accepted Answer

#f(x) = 1/3 (x-2)^3 -5/2 x^2 + 4x + 3 #

#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 #

Cameron Alexander · Answer

undefined To find $ f(x) $, integrate $ (x - 2)^2 - 5x + 4 $ with respect to $ x $.

Given that $ f(2) = 1 $, you can find the constant of integration.

$ f(x) = \int (x - 2)^2 - 5x + 4 \, dx $

$ f(x) = \int (x^2 - 4x + 4) - 5x + 4 \, dx $

$ f(x) = \int x^2 - 9x + 8 \, dx $

$ f(x) = \frac{1}{3}x^3 - \frac{9}{2}x^2 + 8x + C $

Given $ f(2) = 1 $,

$ 1 = \frac{1}{3}(2)^3 - \frac{9}{2}(2)^2 + 8(2) + C $

$ 1 = \frac{8}{3} - 9(2) + 16 + C $

$ 1 = \frac{8}{3} - 18 + 16 + C $

$ 1 = \frac{8}{3} - 2 + C $

$ 1 = \frac{8}{3} - \frac{6}{3} + C $

$ 1 = \frac{2}{3} + C $

$ C = 1 - \frac{2}{3} $

$ C = \frac{3}{3} - \frac{2}{3} $

$ C = \frac{1}{3} $

Therefore, $ f(x) = \frac{1}{3}x^3 - \frac{9}{2}x^2 + 8x + \frac{1}{3} $.