Find the Function f(x)=x^2-5x+6

$f (x) = x^{2} - 5 x + 6$

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.

$\int x^{2} d x + \int - 5 x d x + \int 6 d x$

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

$\frac{1}{3} x^{3} + C + \int - 5 x d x + \int 6 d x$

Since $- 5$ is constant with respect to $x$ , move $- 5$ out of the integral.

$\frac{1}{3} x^{3} + C - 5 \int x d x + \int 6 d x$

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

$\frac{1}{3} x^{3} + C - 5 (\frac{1}{2} x^{2} + C) + \int 6 d x$

Apply the constant rule.

$\frac{1}{3} x^{3} + C - 5 (\frac{1}{2} x^{2} + C) + 6 x + C$

Simplify.

Combine $\frac{1}{2}$ and $x^{2}$ .

$\frac{1}{3} x^{3} + C - 5 (\frac{x^{2}}{2} + C) + 6 x + C$

Simplify.

$\frac{x^{3}}{3} - \frac{5 x^{2}}{2} + 6 x + C$

Reorder terms.

$\frac{1}{3} x^{3} - \frac{5}{2} x^{2} + 6 x + C$

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

$F (x) = \frac{1}{3} x^{3} - \frac{5}{2} x^{2} + 6 x + C$