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

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Answer 1 · 2016-02-07T18:50:22

#f(x)=1/4x^4-2x^3+6x^2-8x+1#

By the Binomial Theorem, #(x-2)^3=x^3-6x^2+12x-8#

#thereforeint(x-2)^3dx=int(x^3-6x^2+12x-8)dx#

#=1/4x^4-2x^3+6x^2-8x+C#.

Now substituting in the initial boundary condition of #f(2)=-3#, we get

#1/4(2)^4-2(2)^3+6(2)^2-8(2)+C=-3#

#therefore C=-3-4+16-24+16=1#.

#therefore f(x)=1/4x^4-2x^3+6x^2-8x+1#.