Find the Function f(x)=2x^2-3x+4

$f (x) = 2 x^{2} - 3 x + 4$

Step 1

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

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

Step 2

Split the single integral into multiple integrals.

$\int 2 x^{2} d x + \int - 3 x d x + \int 4 d x$

Step 3

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

$2 \int x^{2} d x + \int - 3 x d x + \int 4 d x$

Step 4

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

$2 (\frac{1}{3} x^{3} + C) + \int - 3 x d x + \int 4 d x$

Step 5

Since $- 3$ is constant with respect to $x$ , move $- 3$ out of the integral.

$2 (\frac{1}{3} x^{3} + C) - 3 \int x d x + \int 4 d x$

Step 6

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

$2 (\frac{1}{3} x^{3} + C) - 3 (\frac{1}{2} x^{2} + C) + \int 4 d x$

Step 7

Apply the constant rule.

$2 (\frac{1}{3} x^{3} + C) - 3 (\frac{1}{2} x^{2} + C) + 4 x + C$

Step 8

Simplify.

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

$2 (\frac{x^{3}}{3} + C) - 3 (\frac{1}{2} x^{2} + C) + 4 x + C$

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

$2 (\frac{x^{3}}{3} + C) - 3 (\frac{x^{2}}{2} + C) + 4 x + C$

Simplify.

$\frac{2 x^{3}}{3} - \frac{3 x^{2}}{2} + 4 x + C$

Step 9

Reorder terms.

$\frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 4 x + C$

Step 10

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

$F (x) = \frac{2}{3} x^{3} - \frac{3}{2} x^{2} + 4 x + C$