Find the Function f(x)=x(x-1)

$f (x) = x (x - 1)$

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

Multiply $x (x - 1)$ .

$\int x \cdot x + x \cdot - 1 d x$

Step 3

Simplify.

Raise $x$ to the power of $1$ .

$\int x^{1} x + x \cdot - 1 d x$

Raise $x$ to the power of $1$ .

$\int x^{1} x^{1} + x \cdot - 1 d x$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int x^{1 + 1} + x \cdot - 1 d x$

Add $1$ and $1$ .

$\int x^{2} + x \cdot - 1 d x$

Move $- 1$ to the left of $x$ .

$\int x^{2} - 1 \cdot x d x$

Rewrite $- 1 x$ as $- x$ .

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

Step 4

Split the single integral into multiple integrals.

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

Step 5

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

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

Step 6

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

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

Step 7

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

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

Step 8

Simplify.

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

Step 9

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

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