You are probably familiar with term midpoint rule. Have you faced problems for approximating the area under a curve using the midpoint rule, and never had an idea how to go about these types of questions? Well, let us break it down for you and make it easier to understand.

The midpoint rule, also known as the rectangle method or mid-ordinate rule, is used to approximate the area under a simple curve. There are other methods to approximate the area, such as the left rectangle or right rectangle sum, but the midpoint rule gives the better estimate compared to the two methods.

Midpoint Rule Formula

\int_{a}^{b}{f(x)dx \approx{M}_{n}=\sum_{i=1}^{n}{f\left(\dfrac{\dfrac{b-a}{n}i+\dfrac{b-a}{n}(i-1)\quad}{2}\right)\left(\dfrac{b-a}{n}\right)}}

The midpoint rule formula is very easy to work with although the formula seems to be a bit complex. Let’s break it down and talk about the components of this formula. \left( \dfrac { b-a }{ n } \right) serves as the width, of a certain representative rectangle.

\left(\dfrac{b-a}{n}\right)i is the width multiplied by the counter, i. This value equals the rightmost edge of each rectangle, which is a typical approach to the right-hand point approximation.

\left(\dfrac{b-a}{n}\right)(i-1) is the width multiplied by the counter(i-1). This value equals the leftmost edge of each rectangle, which is a typical approach to the left-hand point approximation.

At this point, you are probably starting to get the idea. Averaging what is on the left and right gives us something in the middle. That’s why this method is called the midpoint rule approximation method.

Sometimes the midpoint rule can be written as:

{M}_{n}=\dfrac{b-a}{n}\left(f\left(\dfrac{{x}_{0}{+x}_{1}}{2}\right)+f\left(\dfrac{{x}_{1}{+x}_{2}}{2}\right)+f\left(\dfrac{{x}_{2}{+x}_{3}}{2}\right)....+f\left(\dfrac{{x}_{n-1}{+x}_{n}}{2}\right)\right)

Which can be simplified to:

\int_{a}^{b}{f(x)dx \approx{M}_{n}=\sum_{i=1}^{n}{f\left(\dfrac{{x}_{i}+{x}_{i-1}}{2}\right)\Delta x}}

This can be compared to the first formula, that is:

\left(\dfrac{b-a}{n}\right)i={x}_{i} where x_i is just some value of x that lies along the x-axis and serves as the right endpoint of the rectangle, \left(\dfrac{b-a}{n}\right)(i-1)={x}_{i-1} this serves as the left endpoint of the rectangle, and the subscript counter i, implies that there are many of them counting from 1 to n. This corresponds to, n rectangles.

Most of the time in the AP® Calculus AB and BC exams, you will be using the midpoint rule in a numeric setting. That is, you will always be given the number of rectangles or sub-intervals you will be using for the approximation.

Now let us look at an example to see how we can use the midpoint rule for approximation.

Example 1

Use the midpoint rule to approximate the area under a curve given by the function f(x)=x^2+5 on the interval [0,4] and n=4.

Solution:

The entire distance along the x-axis is 4, that is:

b-a=4-0=4

Recall that the width of the rectangle is given by:

width=\dfrac{b-a}{n}=\dfrac{4}{4}=1

Now we will find the values halfway across, this will give us the midpoint of the sub-intervals.

The midpoints of the 4 subintervals are

\dfrac{1}{2},\dfrac{3}{2},\dfrac{5}{2},\dfrac{7}{2}

We know that the area of a rectangle is given by the length times the width.

Area=w\times l

So in this case, we will use the following area as an approximation for the area under the curve:

Area \approx1[f(\dfrac{1}{2})+f(\dfrac{3}{2})+f(\dfrac{5}{2})+f(\dfrac{7}{2})]

Now we will plug in the values of x into our function to determine how tall each of the rectangles are (the heights).

Area \approx1[(\dfrac{1}{4}+5)+(\dfrac{9}{4}+5)+(\dfrac{25}{4}+5)+(\dfrac{49}{4}+5)]

Evaluating we obtain:

Area \approx1(5.25+7.25+11.25+17.25)

Area \approx1(41)

Area \approx41

By working out the exact integral we obtain:

\int_{0}^{4}{(x^2+5)dx}=41.33

We can thus confirm how close the midpoint rule approximation is compared to the exact integral.

You see how easy it is to use the midpoint rule. Let us try another example then check at some past AP® Calculus questions that required us to apply midpoint rule.

Example 2

Approximate the exact area under a curve between 0 and 1, when n=10 and the function of the curve is

f(x)=\sqrt{x^{3}+2}

Solution

This is no different from our previous examples. The first step is to find the entire distance along the x-axis.

b-a=1-0=1

Then we find the width of the rectangles.

width=\dfrac{b-a}{n}=\dfrac{1}{10}

Next step is to find the midpoints of the sub-intervals.

The midpoints will be:

0.05,0.15,0.25,0.35,0.45,0.55,0.65,0.75,0.85,0.95

Then we will use these points together with the width to approximate the area.

Area \approx \dfrac{1}{10}(f(0.05)+f(0.15)+f(0.25)...+f(0.75)+f(0.85)+f(0.95))

You can probably guess what follows in the next step. We will find the heights of the rectangles by plugging in the value of x into the function.

Area=\dfrac{1}{10}(\sqrt{0.05^3+2}+\sqrt{0.15^3+2}+\sqrt{0.25^3+2}...+\sqrt{0.85^3+2}+\sqrt{0.95^3+2})

Evaluating we obtain:

Area \approx 2.2488

The exact integral gives:

\int_{0}^{1}{(x^3+2)dx}=2.25

Hopefully you have seen that nothing changes in the steps to approximate the areas for different types of functions given. However complex the function is, the steps will remain the same.

We will now take a look at Free Response Questions in AP® Calculus AB and BC exams from the past years. We will just do parts of the problem, which requires midpoint approximation.

2012 AP® Calculus BC Free Response Question 4(b)

x	1	1.1	1.2	1.3	1.4
f’(x)	8	10	12	13	14.5

f is twice differentiable, x > 0, f(1) = 15, f”(1) = 20

Use a midpoint Riemann sum with two sub-intervals of equal length and values to approximate \int_{1}^{1.4}{f'(x)dx}. Use the approximation for \int_{1}^{1.4}{f'(x)dx} to estimate the value of f(1.4). Show the computation that leads to your answer.

Solution

From the table, the width

=\dfrac{1.4-1}{2}=\dfrac{0.4}{2}=0.2

\int_{1}^{1.4}{f'(x)dx}\approx 0.2(f'(1.1)+f'(1.3))

\int_{1}^{1.4}{f'(x)dx}\approx 0.2(10+13)=4.6

Then we approximate the value of f(1.4)

f(1.4)=f(1)+\int_{1}^{1.4}{f'(x)dx}

f(1.4)=15+4.6

f(1.4)=19.6

2013 AP® Calculus BC Free Response Question 3(c)

t (minutes)	0	1	2	3	4	5	6
C(t) (ounces)	0	5.3	8.8	11.2	12.8	13.8	14.5

Hot water is dripping through a coffee maker, filling a large cup with coffee. The amount of coffee in the cup at time t, is given by a differentiable function C, where t is measured in minutes. Selected values of C(t), measured in ounces, are given in the table above.

3c) Use a midpoint sum with three sub-intervals of equal length indicated by the data in the table to approximate the value of \dfrac{1}{6}\int_{0}^{6}{C(t)dt}. Using correct units, explain the meaning of \dfrac{1}{6}\int_{0}^{6}{C(t)dt} in context of the problem.

Solution

The width will be:

=\dfrac{b-a}{n}=\dfrac{6-0}{3}=2

Next step will be to approximate the value of the integral

\dfrac{1}{6}\int_{0}^{6}{C(t)dt}\approx \dfrac{1}{6}[2(C(1)+C(3)+C(5))]

=\dfrac{1}{6}[2(5.3+11.2+13.8)]

=\dfrac{1}{6}(60.6)=10.1 ounces

\dfrac{1}{6}\int_{0}^{6}{C(t)dt} is the average amount of coffee in the cup, in ounces, over the time interval from 0 to 6 minutes.

2004 AP® Calculus BC Free Response Question 3(a)

t (minutes)	0	5	10	15	20	25	30	35	40
V(t)	7.0	9.2	9.5	7.0	4.5	2.4	2.4	4.3	7.3

A test plane flies in a straight line with positive velocity v(t), in miles per minute at time t minutes, where v is a differentiable function of t. Selected values of v(t) for are shown in the table above.

3a) Use a midpoint Riemann sum with four sub-intervals of equal length and values from the table to approximate \int_{0}^{40}{V(t)dt}. Show the computation that leads to your answer. Using correct units, explain the meaning of \int_{0}^{40}{V(t)dt} in terms of the plane’s flight.

Solution

Midpoint Riemann sum is given by:

10[v(5)+v(15)+v(25)+v(35)]

=10[9.2+7.0+2.4+4.3]

=229mi

Now we can interpret \int_{0}^{40}{V(t)dt} as the total distance in miles during the 40 minutes of plane’s flight.

2006 AP® Calculus BC Free Response Question 4(b)

t(seconds)	0	10	20	30	40	50	60	70	80
V(t) (feet per second)	5	14	22	29	35	40	44	47	49

Rocket A has positive velocity v(t) after being launched upward from an initial height of 0 feet at time t = 0 seconds. The velocity of the rocket is recorded for selected values of t over the interval 0 to 80 seconds as shown in the table above.

4b) Using correct units explain the meaning of \int_{10}^{70}{V(t)dt} in terms of the rocket’s flight. Use midpoint Riemann sum with 3 sub-intervals of equal length to approximate \int_{10}^{70}{V(t)dt}.

Solution

The integral \int_{10}^{70}{V(t)dt} represents the distance in feet traveled by rocket A from t = 10 seconds to t = 70 seconds.

The midpoint Riemann sum is given by:

=20[v(20)+v(40)+v(60)]=20[22+35+44]

=2020ft

Conclusion

After working the examples and questions from previous AP® Calculus exams, you notice how simple using the midpoint rule is! All we need to know is the formula and how to substitute in actual numbers. Then, you will be well prepared to solve any question requiring the application of the midpoint rule.

Points to note:

1. The midpoint formula is given by:

{M}_{n}=\dfrac{b-a}{n}\left(f\left(\dfrac{{x}_{0}{+x}_{1}}{2}\right)+f\left(\dfrac{{x}_{1}{+x}_{2}}{2}\right)+f\left(\dfrac{{x}_{2}{+x}_{3}}{2}\right)....+f\left(\dfrac{{x}_{n-1}{+x}_{n}}{2}\right)\right)

Where x₀, x₁, x₂, x₃…..x_n are points on the x-axis

2. The width is given by:

\left(\dfrac{b-a}{n}\right)

Problems that require the application of the midpoint rule can come in two ways:

• Using the midpoint rule to approximate the area under a curve.
• Using the midpoint rule to approximate the value of an integral.

With these points in mind you will never have any trouble solving questions that require the use of midpoint rule.

Let’s put everything into practice. Try this AP® Calculus practice question:

Looking for more AP® Calculus practice?

Check out our other articles on AP® Calculus.

You can also find thousands of practice questions on Albert.io. Albert.io lets you customize your learning experience to target practice where you need the most help. We’ll give you challenging practice questions to help you achieve mastery of AP® Calculus.

Start practicing here.

Are you a teacher or administrator interested in boosting AP® Calculus student outcomes?

Learn more about our school licenses here.