Question

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

Answer 1

`f(x)=3x^3-4x^2+2x-11`

Explanation:

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.