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

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

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

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

Let $u = x - 3$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

Let $u = x - 3$ . Find $\frac{d u}{d x}$ .

Differentiate $x - 3$ .

$\frac{d}{d x} [x - 3]$

By the Sum Rule, the derivative of $x - 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [- 3]$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

Rewrite the problem using $u$ and $d u$ .

$\int u^{2} d u$

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

$\frac{1}{3} u^{3} + C$

Replace all occurrences of $u$ with $x - 3$ .

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

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

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