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

Question

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

Scarlett Allison · Accepted Answer

#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#

Adam Adkins · Answer

undefined To find $ f(x) = \int (x-3)^2 - 3x + 4 \, dx $ given $ f(2) = 1 $, integrate the given expression and then use the given condition to solve for the constant of integration.

$$ f(x) = \int (x-3)^2 - 3x + 4 \, dx $$

$$ = \int (x^2 - 6x + 9 - 3x + 4) \, dx $$

$$ = \int (x^2 - 9x + 13) \, dx $$

$$ = \frac{x^3}{3} - \frac{9x^2}{2} + 13x + C $$

Given that $ f(2) = 1 $, substitute $ x = 2 $ and solve for $ C $.

$$ 1 = \frac{2^3}{3} - \frac{9 \cdot 2^2}{2} + 13 \cdot 2 + C $$

$$ 1 = \frac{8}{3} - \frac{36}{2} + 26 + C $$

$$ 1 = \frac{8}{3} - 18 + 26 + C $$

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

$$ 1 = \frac{8 + 24}{3} + C $$

$$ 1 = \frac{32}{3} + C $$

$$ C = 1 - \frac{32}{3} $$

$$ C = \frac{3}{3} - \frac{32}{3} $$

$$ C = \frac{-29}{3} $$

Therefore, the function $ f(x) = \int (x-3)^2 - 3x + 4 \, dx $ with $ f(2) = 1 $ is:

$$ f(x) = \frac{x^3}{3} - \frac{9x^2}{2} + 13x - \frac{29}{3} $$