What is the solution for f^'(x)=x^2-x,f(3)= 19/2 ?

The solution for f^'(x)=x^2-x,f(3)= 19/2 is f(x)=(x^3)/3-(x^2)/2+5

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f^'(x)=x^2-x,f(3)= 19/2

f^{^{'}} (x) = x^{2} - x, f (3) = \frac{19}{2}

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + 5

Solution steps

f^{'} (x) = x^{2} - x

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

f (x) = \int x^{2} - x d x

\int x^{2} - x d x = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + c_{1}

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + c_{1}

Apply initial conditions:

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + 5

f (x) = \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + 5

\frac{\partial}{\partial x} (\frac{x}{x + 4})

\int 6 cos (3 x) d x

x \to 0 lim (\frac{sin ( 10 x )}{sin ( 3 x )})

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\frac{\partial}{\partial x} (x^{3} + x^{2} y^{2} + y^{3} + x + y)

What is the solution for f^'(x)=x^2-x,f(3)= 19/2 ?
The solution for f^'(x)=x^2-x,f(3)= 19/2 is f(x)=(x^3)/3-(x^2)/2+5