# Area between a curve and the x-axis

To calculate the area between a curve and the $$x$$-axis we must evaluate using definite integrals.

## Example

### Solution

$\int\limits_0^5 {({x^2}} - 4x + 5)dx$

$= \left[ {\frac{{{x^3}}}{3} - 2{x^2} + 5x} \right]_0^5$

$= \left( {\frac{{125}}{3} - 50 + 25} \right) - \left( 0 \right)$

$= \frac{{125}}{3} - 25$

$= \frac{{50}}{3}unit{s^2}$

You may have to work out the limits of integration before calculating the area under a curve.

## Example

Find the area under the curve $$y = 4x - {x^2}$$

### Solution

First, we need to find out where the curve cuts the $$x$$-axis. Remember, a curve cuts the $$x$$-axis when $$y = 0$$.

$4x - {x^2} = 0$

$$x(4 - x) = 0$$ - factorise taking out a common factor of $$x$$.

$$x = 0$$ or $$4 - x = 0$$

Therefore $$x = 0$$ or $$x = 4$$

The curve cuts the $$x$$-axis at $$(0,0)$$ and $$(4,0)$$

$$0$$ and $$4$$ are the limits of integration.

Hence the area is:

$\int\limits_0^4 {(4x - {x^2}} )dx$

$= \left[ {2{x^2} - \frac{{{x^3}}}{3}} \right]_0^4$

$= \left( {32 - \frac{{64}}{3}} \right) - \left( 0 \right)$

$= \frac{{32}}{3}unit{s^2}$