# 2. Area Under a Curve by Integration

by M. Bourne

We met areas under curves earlier in the Integration section (see 3. Area Under A Curve), but here we develop the concept further. (You may also be interested in Archimedes and the area of a parabolic segment, where we learn that Archimedes understood the ideas behind calculus, 2000 years before Newton and Leibniz did!)

**It is important to sketch
the situation before you start.**

We wish to find the area under the curve `y = f(x)` from `x = a` to `x = b`.

We can have several situations:

**Case 1:** Curves which
are entirely **above the ***x***-axis.**

*x*

The curve y = f(x), completely above the x-axis. Shows a "typical" rectangle, Δ*x* wide and y high.

In this case, we find the area by simply finding the integral:

`"Area"=int_a^bf(x)dx`

Where did this formula come from?

## Area Under a Curve from First Principles

In the diagram above, a "typical rectangle" is shown with width `Δx` and height `y`. Its area is `yΔx`.

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

`sum_{x=a}^\b(y)Deltax`

Now if we let `Δx → 0`, we can find the exact area by integration:

`"Area"=int_a^bf(x)dx`

This follows from the Riemann Sums, from the Introduction to Integration chapter.

### Example of Case 1

### Need Graph Paper?

Find the area underneath the curve `y = x^2+ 2` from `x = 1` to `x = 2`.

Answer

The curve *y* = *x*^{2} + 2, showing the portion under the curve from *x* = 1 to *x* = 2.

`text[Area] = int_a^b f(x) dx`

`=int_1^2(x^2+2)dx`

`=[(x^3)/(3)+2x]_1^2`

`=[(8/3 + 4 )-(1/3 + 2)]`

`=13/3\ text[units]^2`

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## Case 2: Curves which are entirely **below the ***x***-axis **

*x*

We consider the case where the curve is below the `x`-axis for the range of `x`* *values being considered.

In this case, the integral gives a **negative number.** We need to take the **absolute value** of this to find our area:

`"Area"=|int_a^bf(x)dx|`

### Example of Case 2

Find the area bounded by `y = x^2 − 4`, the `x`-axis and the lines `x = -1` and `x = 2`.

Answer

The curve *y* = *x*^{2} − 4, showing the portion under the curve from *x* = −1 to *x* = 2.

The required area is totally below the `x`-axis in this example, so we need to use absolute value signs.

`text[Area] = |int_a^bf(x) dx|`

`=|int_-1^2(x^2-4) dx|`

`=|[x^3/3 - 4x]_-1^2|`

`=|[(8/3-8)-(-1/3+4)]|`

`=|-9|`

`=9\ text[units]^2`

**Case 3**: Part of the curve **is below** the *x*-axis, part of it is **above** the *x*-axis

In this case, we have to sum the individual parts, taking the absolute value for the section where the curve is below the `x`-axis (from `x = a` to `x = c`).

`"Area"=|int_a^cf(x)dx|+int_c^bf(x)dx`

If we don't do it like this, the "negative" area (the part below the `x`-axis) will be subtracted from the "positive" part, and our total area will not be correct.

### Example of Case 3

What is the area bounded by the curve `y = x^3`, `x = -2` and `x = 1`?

Answer

The curve *y* = *x*^{3}, showing the portion under the curve from *x* = −2 to *x* = 1.

We can see from the graph that the portion between `x = -2` and `x = 0` is below the x-axis, so we need to take the absolute value for that portion.

`text[Area]= |int_-2^0x^3 dx|+int_0^1x^3 dx`

`=|[x^4/4]_-2^0|+[x^4/4]_0^1`

`=|(0-16/4)|+(1/4-0)`

`=4+1/4`

`=4.25\ text[units]^2`

### Don't do it like this!

If you just blindly find the integral from the lower limit to the upper limit, you won't get the actual area in such cases.

`text[Integral]= int_(-2)^1x^3 dx`

`=[x^4/4]_(-2)^1`

`=(1/4-16/4)`

`=-15/4`

`=-3.25`

This is **not** the correct answer for the area under the curve. It **is** the value of the integral, but clearly an area cannot be negative.

It's always best to sketch the curve before finding areas under curves.

### Summary (so far)

In each of Case 1, Case 2 and Case 3, we are summing elements left to right, like this:

We are (effectively) finding the area by **horizontally
**adding the areas of the rectangles, width `dx` and
heights `y` (which we find by substituting values of `x` into `f(x)`).

So

`A=int_a^bf(x)dx`

(with absolute value signs where necessary, if the curve goes under the `x`-axis).

## Case 4: Certain curves are much easier to sum vertically

In some cases, it is easier to find the area if we take **vertical** sums. Sometimes the only possible way is to sum vertically.

In this case, we find the area is the sum of the rectangles, heights `x = f(y)` and width `dy`.

If we are given `y = f(x)`, then we need to re-express this as `x = f(y)` and we need to **sum from bottom to top.**

So, in case 4 we have:

`A=int_c^df(y)dy`

### Example of Case 4

Find the area of the region bounded by the curve `y=sqrt(x-1)`, the `y`-axis and the lines `y = 1` and `y = 5`.

Answer

Sketch first:

*x*

*dy*

*x*=

*y*

^{2}+ 1

The curve *x* = *y*^{2} + 1, showing the portion "under" the curve from *y* = 1 to *y* = 5.

In this case, we express *x* as a function of *y:*

`y=sqrt{x-1}`

`y^2=x-1`

`x=y^2+1`

So the area is given by:

`A=int_1^5(y^2+1) dy=[y^3/3+y]_1^5`

`=45 1/3\ text[sq units]`

**Note:** For this particular example, we could have also summed it horizontally (integrating `y` and using `dx`), but we would need to break it up into sections first.

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