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

Question

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

Paisley Johnson · Accepted Answer

#f(x)=3x^3-4x^2+2x-11# Now, the given function is apparently an integral of another function. That is #f(x)=int((3x-1)^2-2x+1)dx# Before integrating, let's expand and simplify the terms shall we? #(3x-1)^2-2x+1=9x^2+1-6x-2x+1=9x^2-8x+2# So, #int(9x^2-8x+2)dx=9x^3/3-8x^2/2+2x+c# #\impliesf(x)=3x^3-4x^2+2x+c# Given that at #x=2" "f(x)=1# So, #3*2^3-4*2^2+2*2+c=1\implies3*8-4*4+4+c=1# #:.24-16+4+c=1\impliesc=-11# Substituting for #c# we get the answer I have written above.

Adam Adkins · Answer

undefined To find $f(x)$ given that $f(2) = 1$, you need to integrate the given function and then use the information provided to solve for the constant of integration.

Given $f(x) = \int (3x-1)^2 - 2x + 1 \, dx$, integrate the function:

$$
\int (3x-1)^2 - 2x + 1 \, dx = \int (9x^2 - 6x + 1) - 2x + 1 \, dx = \int (9x^2 - 8x + 2) \, dx
$$

Integrate each term separately:

$$
\int (9x^2 - 8x + 2) \, dx = 3x^3 - 4x^2 + 2x + C
$$

Now, given that $f(2) = 1$, substitute $x = 2$ into the expression:

$$
1 = 3(2)^3 - 4(2)^2 + 2(2) + C
$$

$$
1 = 24 - 16 + 4 + C
$$

$$
1 = 12 + C
$$

$$
C = -11
$$

Therefore, $f(x) = 3x^3 - 4x^2 + 2x - 11$.