What is #f(x) = int (3x-2)^2-5x+1 dx# if #f(2) = 1 #?

1 Answer
Bill K.
May 29, 2016

#f(x)=3x^3-17/2 x^2+5x+1#

Explanation:

The indefinite integral is

#int ((3x-2)^2-5x+1)\ dx=int (9x^2-17x+5)\ dx#

#=3x^3-17/2 x^2+5x+C#

Therefore, #f(x)=3x^3-17/2 x^2+5x+C# for some #C#. Since #f(2)=1#, we must choose the value of #C# so that #1=3*2^3-17/2 * 2^2+5*2+C#. This implies that #1=24-34+10+C# so that #C=1#.

Hence, the answer is #f(x)=3x^3-17/2 x^2+5x+1#.

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