Find the Function f(x)=2x^2+x-3

$f (x) = 2 x^{2} + x - 3$

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.

$\int 2 x^{2} d x + \int x d x + \int - 3 d x$

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$2 \int x^{2} d x + \int x d x + \int - 3 d x$

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

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

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

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

Combine $\frac{1}{3}$ and $x^{3}$ .

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

Apply the constant rule.

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

Simplify.

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

Reorder terms.

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

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

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