As we move on from applications of derivatives, we will now start look at area found under a curve. In doing so, we will start off with an example and try to determine the best way to approach.

**Find the area**

Find the area under the curve \(f(x)=2x^{2}+3\) on the interval \([0,4]\).

**Drawing**

Before just jumping in, we want to think about what we are really trying to do. In order to see this we will graph the given function on the interval.

Now, if we look at this area, we see that it isn’t a nice geometric shape like a triangle, rectangle, circle or some other shape we already know the formula for the area of. Therefore, we can’t just find this area. Since we can’t find this area directly, we will instead try to get a good approximation of the area.

**Approximating area**

One way we can approximate the area under the curve is by drawing a geometric shape that looks similar to the above graph, but that we already know the formula for the area. While there are many options, each one offering varying benefits, we will start by using a rectangle to approximate the area. The reason why we choose a rectangle is because it is the simplest two-dimensional shape to find the area of.

What rectangle should we draw? In this case, it is easy to see that one side of the rectangle should cover the interval \([0,4]\), but the height of the rectangle isn’t as obvious. Two easy choices to provide an approximation are found by using the function value at the left endpoint or the right endpoint of the interval. That is, we can use the height of \(f(0)=3\) or \(f(4)=35\). We have both of these rectangles drawn below.

If we look at the rectangle using the left endpoint for the height, we find the area of the rectangle is \(4* 3=12\), while the area of the rectangle using the right endpoint would be \(4*35=140\). If we look at the picture, we will not the \(12\) will be too small and \(140\) will be too large for the true value of the shape we want to find the area for. In this case, there is a large difference between these values, but this has still provided us with an upper and lower limit of the value of the area, so we are much better off than we were before.

**Improving the approximation**

Now that we have seen that we can use rectangles to approximate the area of the curve, we want to see if we can make this approximation better. We have already mentioned that other shapes may offer an alternative, but we will stick with rectangles. Therefore, the way we want to make this approximation better is to use more rectangles. That is, we can break the given interval up into smaller intervals. Then we can draw rectangles on each of the subintervals. If we add up the areas of all the rectangles, this will give us an approximation for the total area under the curve. Graphically, we can see what this would look like with 2 rectangles using the right endpoints to find the height.

**Calculating the approximation**

What we have found, is that we can use what is called a Riemann Sum to approximate the area under the curve. While we can use any number of rectangles with any point to find the height, we will go through one example using right endpoints as evaluation points and four rectangles. That is, we will find the right Riemann sum with 4 rectangles, \(R_{4}\).

As we do this, the first thing we need to do is figure out how wide each of the rectangles will be. Note that we want to cover the interval \([0,4]\) with \(4\) rectangles. Therefore, we need to cover a width of four with four rectangles. If we want each rectangle to be the same width, each one will need to be \(\frac{4}{4}=1\) wide. Now that we know how wide each one is, we want to draw a picture denoted the section that will be covered by each rectangle. We have done this below.

Now that we’ve found the width of the rectangles, we need to find the heights of the rectangles. In order to do this, we first need to find the evaluation points for each of the rectangles. If we look at the first rectangle, note that the left endpoint is the left endpoint of the entire interval, that is \(0\). Since the rectangle has a width of 1, this means that the first rectangle covers the subinterval \([0,1]\). Therefore, the right endpoint of this subinterval will be \(1\). Now, in order to find the height, we need to find the function value at \(1\). We, therefore, get that the height of the rectangle is \(f(1)=2(1)^{2}+3=5\).

If we continue this for the next rectangle, we will see that the second rectangle covers the subinterval \([1,2]\). The second evaluation point will then be \(2\), so the second rectangle height is \(f(2)=2(2)^{2}+3=11\). Again continuing this, we find that the third evalutation point is \(3\) and the height is \(f(3)=2(3)^{2}+3=21\). Finally, the fourth evaluation point will be \(4\), so the height will be \(f(4)=35\) (which we found above). We’ve drawn all four rectangles below.

Now that we know the width of each rectangle, \(1\), and the height of all four rectangles, we can find the sum of the areas. We now get that

\begin{align*}

R_{4}&=A_{1}+A_{2}+A_{3}+A_{4} \\

&=w*h_{1}+w*h_{2}+w*h_{3}+w*h_{4} \\

&=1*5+1*11+1*21+1*35 \\

&=72.

\end{align*}

That is, we say that the area under the curve is approximately \(72\).

**Conclusion**

As we look at this approximation, we can see from the picture that it will be an over approximation. If we wanted a lower bound, we could repeat the process using left endpoints, that is we could find the left Riemann Sum with four rectangles, \(L_{4}\). Now, if we wanted to make the approximation better, recall that we could repeat the process with more and more rectangle. As we move forward, we will see that if we use infinitely many rectangles, we will get the exact area under the curve. I do have a lab that walks you through this process in our products focusing on the Stopping Distance of a Motorcycle if you would like to see more about this.

I hope you enjoyed the post and that it helped to clarify Riemann sums. If you found it helpful, make sure to like the post and share it with anyone else that may find it useful on Social Media.