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

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Answer 1 · 2016-01-29T19:25:27

$f (x) = \frac{x^{3}}{3} - 3 x - 2$ $f(x)=x^3/3-3x-2$

$\int (x^{2} - 3 d x) = \frac{x^{3}}{3} - 3 x + C$ $int(x^2-3dx)=x^3/3-3x+C$ .

Now substituting the initial boundary condition in we get

$f (3) = - 2$ $f(3)=-2$

$therefore3^3/3-3(3)+C=-2$

$therefore C=-2$ .

$therefore$ the required function is $f(x)=x^3/3-3x-2$ .

What is f(x) = int x^2-3 dxf(x)=∫x2−3dxf(x) = int x^2-3 dx if f(3)=-2 f(3)=−2f(3)=-2 ?