What is $f (x) = \int {(x - 2)}^{3} d x$ $f(x) = int (x-2)^3 dx$ if $f (2) = - 3$ $f(2) = -3$ ?

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Answer 1 · 2016-02-07T18:50:22

$f (x) = \frac{1}{4} x^{4} - 2 x^{3} + 6 x^{2} - 8 x + 1$ $f(x)=1/4x^4-2x^3+6x^2-8x+1$

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

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

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

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

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

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

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

What is f(x) = int (x-2)^3 dxf(x)=∫(x−2)3dxf(x) = int (x-2)^3 dx if f(2) = -3 f(2)=−3f(2) = -3 ?