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

1 Answer
Rohan A.K
Jan 21, 2016

#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.

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