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

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Answer 1 · 2016-01-17T04:13:35

#f(x)=x^3-4x+2#

First, find the antiderivative of the function using the product rule in reverse:

#intax^ndx=(ax^(n+1))/(n+1)+C#

Thus,

#f(x)=(3x^(2+1))/(2+1)-(4x^(0+1))/(0+1)+C=(3x^3)/3-(4x)/1+C=x^3-4x+C#

Since #f(x)=x^3-4x+C#, and we know that #f(2)=2#, we can determine #C# (the constant of integration).

#f(2)=2^3-4(2)+C=2#

#8-8+C=2#

#C=2#

Thus,

#f(x)=x^3-4x+2#