Home > Integration > Integration: the area under a curve

## Integration: the area under a curve

### Definite Integration

When we work out a definite integral we end up with a value as opposed to a function

A definite integral is in the form $I=\int_a^b f(x)dx$

To calculate $I$

We integrate $f(x)$ (find the inverse derivative)

We substitute in $b\text{ and }a$ into this new function

Then substitute the second value from the first

#### Notation

$\left[F(x)\right]_a^b=F(b)-F(a)$

### Area under a curve

To find the area under a curve we find the definite integral between the two bounds (ends)

#### Proof

Examples

Find the area (A) under the curve $y=4$ and the lines $x=2\text{ and }x=5$

$A=\int_2^5 4dx$

$\implies A=\left[4x\right]_2^5=(4\times 5)-(4\times2)=12$

This is to be expected as all we had was rectangle that was 4 high and 3 wide

Find the area (A) under the curve $f(x)=3+2x$ and the lines $x=0\text{ and }x=3$$A=\int_0^3 (3+2x)dx$

$\implies A=\left[3x+x^2\right]_0^3=(3\times 3+3^2)-(3\times 0+0^2)=18$

(note: we do not require the constant of integration because it will be cancelled out in the subtraction)

Find the area (A) under the curve $y=3x^2+4x+3$ and the lines $x=1\text{ and }x=4$

$A=\int_1^3 (3x^2+4x+3)dx$

$\implies A=\left[x^3+2x^2+3x\right]_1^4=(64+32+12)-(1+2+3)=102$

### Areas below the x axis

Let’s look at an example

$A=\int_2^5 (-4)dx=\left[-4x\right]_2^5=(-4\times 5)-(-4\times2)=-12$

Here, we have the same rectangle as in the first example, but this time it is below the x axis.  The result is that the area is given as negative.

We would report the area as positive, but understand that it is below the axis.

Example

Find the area (A) between the x axis the lines $x=0\text{ and } x=4$ and the curve $y=11x-x^2-30$

$A=\int_0^4 (11x-x^2-30)dx=\left[\frac{11x^2}{2}-\frac{x^3}{3}-30x\right]_0^4$

$\implies A=(\frac{11\times 16}{2}-\frac{64}{3}-30\times 4)-(0)=53\frac{1}{3}$

### When areas are above and below the axis

Let’s consider the line $y=2x-4$ and go from $x=0\text{ to } x=4$

$A= \int_0^4 (2x-4)dx=\left[x^2-4x\right]_0^4=(16-16)-(0-0)=0$

If we look at the graph we can see that there is an area, so what has happened.

Well the area below the axis has cancelled out the area above.

To calculate the desired area we must find the area from 0 to 2 and then from 2 to 4.

$A_1= \int_0^2 (2x-4)dx=\left[x^2-4x\right]_0^2=(-4)-(0-0)=-4$

So $A_1$ has an area of 4 (area is positive)

$A_2= \int_2^4 (2x-4)dx=\left[x^2-4x\right]_2^4=(16-16)-(4-8)=4$

So $A_2$ has an area of 4

$\therefore A=A_1+A_2 = 8$

Examples

Find the area between the x axis, $x=0,x=2$ and the curve $f(x)=x^2-1$

First we find if the function equals zero between $x=0\text{ and }x=2$

let $f(x)=0\implies x^2-1=0\implies x^2=1\implies x=\pm 1$

Since it crosses the axis at $x=1$ we need to integrate from 0 to 1 and then from 1 to 2.  We will call these areas $A_1\text{ and } A_2$.

$A_1=\int_0^1 (x^2-1)dx=\left[\frac{x^3}{3}-x\right]_0^1=(\frac{1}{3}-1)-(0-0)=-\frac{2}{3}$

Since $A_1>0\implies A_1=\frac{2}{3}$

$A_2=\int_1^2 (x^2-1)dx=\left[\frac{x^3}{3}-x\right]_1^2=(\frac{8}{3}-2)-(\frac{1}{3}-1)=\frac{4}{3}$

So the area required $= A_1+A_2=\frac{2}{3} + \frac{4}{3} =2$

Find the area between the x axis, $x=0,x=4$ and the curve $f(x)=x^2+2x-3$

First we find if the function equals zero between $x=0\text{ and }x=4$

let $f(x)=0\implies x^2+2x-3=0\implies (x-1)(x+3)=0\implies x=-1\text{ or }x=3$

Since it crosses the axis at $x=1$ we need to integrate from 0 to 1 and then from 1 to 4.  We will call these areas $A_1\text{ and } A_2$.

$A_1=\int_0^1 (x^2+2x-3)dx=\left[\frac{x^3}{3}+x^2-3x\right]_0^1=(\frac{1}{3}+1-3)-(0+0-0)=-\frac{5}{3}$

$A_2=\int_1^4 (x^2+2x-3)dx=\left[\frac{x^3}{3}+x^2-3x\right]_1^4=(\frac{64}{3}+16-12)-(\frac{1}{3}+1-3)=27$

So the area required $=A_1+A_2=\frac{5}{3}+27=28\frac{2}{3}$