Find the Function f(x)=x^2-6x

$f (x) = x^{2} - 6 x$

The function $F (x)$ can be found by evaluating the indefinite integral of the derivative $f (x)$ .

$F (x) = \int f (x) d x$

Split the single integral into multiple integrals.

$\int x^{2} d x + \int - 6 x d x$

By the Power Rule, the integral of $x^{2}$ with respect to $x$ is $\frac{1}{3} x^{3}$ .

$\frac{1}{3} x^{3} + C + \int - 6 x d x$

Since $- 6$ is constant with respect to $x$ , move $- 6$ out of the integral.

$\frac{1}{3} x^{3} + C - 6 \int x d x$

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$\frac{1}{3} x^{3} + C - 6 (\frac{1}{2} x^{2} + C)$

Simplify.

$\frac{1}{3} x^{3} - 6 (\frac{1}{2}) x^{2} + C$

Simplify.

Combine $- 6$ and $\frac{1}{2}$ .

$\frac{1}{3} x^{3} + \frac{- 6}{2} x^{2} + C$

Cancel the common factor of $- 6$ and $2$ .

Factor $2$ out of $- 6$ .

$\frac{1}{3} x^{3} + \frac{2 \cdot - 3}{2} x^{2} + C$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{1}{3} x^{3} + \frac{2 \cdot - 3}{2 (1)} x^{2} + C$

Cancel the common factor.

$\frac{1}{3} x^{3} + \frac{2 \cdot - 3}{2 \cdot 1} x^{2} + C$

Rewrite the expression.

$\frac{1}{3} x^{3} + \frac{- 3}{1} x^{2} + C$

Divide $- 3$ by $1$ .

$\frac{1}{3} x^{3} - 3 x^{2} + C$

The function $F$ if derived from the integral of the derivative of the function. This is valid by the fundamental theorem of calculus.

$F (x) = \frac{1}{3} x^{3} - 3 x^{2} + C$