# 3. Area Between 2 Curves using Integration

by M. Bourne

Area bounded by the curves `y_1` and `y_2`, & the lines `x=a` and `x=b`, including a typical rectangle.

We are trying to find the area between 2 curves, `y_1 = f_1(x)` and `y_2 = f_2(x)`, and the lines `x = a` and `x = b`.

We see that if we subtract the area under lower curve

`y_1 = f_1(x)`

from the area under the upper curve

`y_2 = f_2(x)`,

then we will find the required area. This can be achieved in one step:

`A=int_a^b(y_2-y_1)dx`

## Alternative Way to Find The Formula (from first principles)

Another way of deriving this formula is as follows (the thinking here is important for understanding how we develop the later formulas in this section).

Each "typical" rectangle indicated has width `Δx` and height `y_2 − y_1`, so its area is `(y_2 − y_1)Δx`.

If we add all these typical rectangles, starting from `a`* *and finishing at `b`*, *the area is approximately:

`sum_(x=a)^b(y_2-y_1)Delta x`

Now if we let Δ*x* → 0, we can find the exact area by integration:

`A=int_a^b(y_2-y_1)dx`

## Summing vertically to find area between 2 curves

Likewise, we can sum **vertically**
by re-expressing both functions so that they are functions of *y*
and we find:

`A=int_c^d(x_2-x_1)dy`

Notice the `c` and `d` as the limits on the integral (to remind us we are summing vertically) and the `dy`. It reminds us to express our function in terms of `y`.

### Example

### Need Graph Paper?

Find the area between the curves `y = x^2 + 5x` and `y = 3 − x^2` between `x = -2` and `x = 0`.

Answer

Sketching first:

Graphs of `y=x^2 + 5x` and `y=3-x^2`, showing the portion between `-2 < x < 0`.

From the graph, we see that `y=3-x^2` is above `y=x^2 + 5x` in the region of interest, so we'll use:

`y_2=3-x^2`, and

`y_1=x^2 + 5x`

So we need to find:

`text[Area]=int_a^b(y_2-y_1) dx`

`=int_-2^0 [(3-x^2)-(x^2+5x)] dx`

`=int_-2^0 [(-2x^2-5x+3)] dx`

`=[-2/3x^3-5/2x^2+3x]_-2^0`

`=0-[16/3-10-6]`

`=10 2/3\ text(sq units)`

Some of the shaded area is above the `x`-axis and some of it is below. Don't worry about taking absolute value - the formula takes care of that automatically.

## Exercises

**1. **Find the area bounded by `y = x^3`,
`x = 0` and `y = 3`.

Answer

Sketch first:

Graph of `y=x^3`, showing the portion bounded by `x = 0` and `y=3`.

We will use:

`text[Area]=int_c^df(y) dy`

and use **horizontal elements**. (In this example we could have added horizontally as well, but will do it vertically to illustrate the method.)

In this case, `c = 0` and `d = 3`.

We need to express `x` in terms of `y`:

`y = x^3` so `x = y^(1//3)`

So

`text[Area]=int_c^df(y) dy`

`=int_0^3 (y^(1//3)) dy`

`=[3/4 y^(4//3)]_0^3`

`=3/4 [(3)^(4//3)-(0)^(4//3)]`

`=3.245\ text[sq units]`

**2. **Find the area bounded by
the curves

`y = x^2 + 5x` and `y = 3 − x^2`.

(This is an extension of the Example above.)

Answer

Sketch first:

Area bounded by the curves `y=x^2 + 5x` and `y=3-x^2`, including a typical rectangle.

We need to use: `A=int_a^b(y_2-y_1) dx`

We note that `y = 3 − x^2` is above `y = x^2+ 5x` so we take

`y_2= 3 − x^2` and `y_1= x^2+ 5x`

Points of intersection occur where:

`x^2 + 5x = 3 − x^2`

`2x^2 + 5x − 3 = 0`

`(x + 3)(2x − 1) = 0`

So `x = -3` or `x = 0.5`

We take **vertical elements** (indicated by the vertical rectangle in the graph above).

So the area is given by:

`text[Area]=int_a^b(y_2-y_1) dx`

`=int_-3^0.5 ([3-x^2]-[x^2+5x]) dx`

`=int_-3^0.5(3-5x-2x^2) dx`

`=[3x-(5x^2)/(2)-(2x^3)/(3)]_-3^0.5`

`=14.29\ text[sq units]`

**3. **Find the area bounded by the
curves

`y = x^2`, `y = 2 − x` and `y = 1`.

Answer

Sketch first:

Area bounded by `y = x^2`, `y = 2 − x` and `y = 1`, including a typical rectangle.

We take **horizontal elements** in this case.

So we need to solve `y = x^2` for `x`:

`x = ±sqrt(y)`

We need the left hand portion, so `x = − sqrt(y)`.

Notice that `x = 2 − y` is to the **right** of `x = -sqrt(y)` so we choose*
* `x_2= 2 − y` and `x_1= -sqrt(y.)`

The intersection of the graphs occurs at `(-2,4)` and `(1,1)`.

So we have: `c = 1` and `d = 4`.

`text[Area]=int_c^d(x_2-x_1) dy`

`=int_1^4([2-y]-[-sqrt[y]]) dy`

`=int_1^4(2-y+sqrt[y])dy`

`=[2y-(y^2)/(2)+2/3 y^(3//2)]_1^4`

`=(16/3)-(13/6)`

`=19/6\ text[sq units]`