# Area Under A Curve

For this lesion you must have some basic knowledge of integration. In this maths entry we shall be evaluating the area under the curve between certain limits.

There are some applications where you will be required to find the area under the curve. This could include an exam situation as well. In most cases when working with graphs such as the velocity-time graphs, you’re often required to find the area under the curve on the graph. Although you might have never realised that you do in fact find the area under the graph to solve certain problems.

The area under a velocity-time graph represents the distance or the displacement of an object. Here is an example of a velocity-time graph of a car travelling.

In 10 seconds we can see on the graph that the car accelerates from 0 to 10s. Suppose we wanted to find for how long the car has travelled.

We can do this by making an estimate of the area under the graph. We estimate by dividing the area under the required time into rectangular strips similarly to how the Simpson rule works. Look at the two following graphs.

For this graph the estimation of the area will be less than the actual area we want, remember this is only an estimate. This is due to the left over gaps at the top which are not included in the calculation.

If we look at it from the graph above the estimation will be a bit more than the area we want. This is due to the unwanted areas on top of the curve.

The area that we’re looking for lies between the above estimations, the more narrow or small the rectangles are the closer the estimate is to the true area that we’re after.

## Example

Here is a very simple and straight forward example; Work out the area under the curve y=x between x=0 and x=3.

The following graph shows a graph which illustrates the problem.

You can see that the graph forms a triangle which makes it easy to find the area below it. Just use the standard area of the triangle formula as follows;

How about if we wanted to find the area between x=0 and x=5 as shown on the graph below’

Observe carefully; The area becomes,

This is pretty much the same as the formula above. We can conclude here that for the curve y=x, the area below the curve is given by;

If you integrate y=x, you will get; x^{x}/2 as shown below.

The **c** at the end is unnecessary because when x=0, we know area=0. That is;

That leaves the area as;

We can see that we can find the area under the curve just by integration. Let’s try it with another common graph. y=x^{2}

because when x=0 A=0 which also means, 0=0+c c=0. So the formula for the area of y=x^{2} is;

for example from x=0 and x=3;

## Example

Here is an example; Find the area under the curve y=3x^{2} between x=0 and x=5. The following graph shows the area we’re trying to find;

We know that we can find the area under the curve by integration;

If you want to find the value of c, use the fact that area=0 when x=0; so 0=0^{3} +c, c=0. That leaves the area equal to;

Now we need to substitute 5 into the area formula we’ve found to find the area between x=0 and x=5 on the graph. That is;

It’s that simple!

## Example

So far we have been finding the area between x=0 and another large x value. Suppose we wanted to find the area between two x values where neither one of them is zero. This example shows this;

Find the area under the curve of y=3x^{2} between x=2 and x=7.

We know by integration we can find this area. So;

The c is unnecessary because it will result in zero. The above formula will give us the area between x=0 and another x value. In the question we have limits for the area we’re trying to find. If we substitute x=7 it will give the area shown on the graph below, that is between x=0 and x=7.

If we substitute in x=2, the result will be for this area shown on the graph.

We’re trying to find the area between x=2 and x=7 on the graph which is A_{1}-A_{2}. This is shown on the graph, which means the area is;

We have found this area using the longer way. We could have simply have written;

This is how you write the area under the graph. We put the limits (x values) on the integration sign. Now we integrate and put the answer in brackets as follows;

We finally substitute in the x value limits and subtract as follows;

## Example 3

Here is an example which sums up what we have done above quickly.

Find the area under the curve y=x^{2}+5 between x=2 and x=7.

We know by integration we find the area formula, so we integrate to find that;

We write with the x values substituted in the brackets.

It’s that simple. Finding areas under the curve can be done simply by integration as you’ve seen. Although you can find the areas under the curve this way, there are some problems which require a bit more thinking and being able to sketch, visualise and having a good knowledge of graphs. I will create an entry for such problems soon.

Example 3 is incorrect, ((7^3/3)+(5*4))= 448/3 which is not the value given. 448/3-38/3 = 410/3 units^2

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Thanks Aaron,

That must have been a graphical error while creating the images. It will be fixed.

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