How to Use the Midpoint Rule in AP® Calculus | albert.io (2023)

You are probably familiar with the term midpoint rule. Have you ever had trouble approximating the area under a curve using the midpoint rule and never had any idea how to solve these types of questions? Well, let's break it down to make it easier to understand.

The midpoint rule, also known as the rectangle method or ordinate rule, is used to approximate the area under a simple curve. There are other methods to approximate area, such as left rectangle or right rectangle sum, but the midpoint rule gives the best estimate compared to both 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}\direita)\esquerda(\dfrac{b-a}{n}\direita)}}

The midpoint rule formula is very easy to work with, although the formula seems 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)iand theanchomultiplied by the counter,UE.This value is equal to the rightmost edge of each rectangle, which is a typical approach for the right point approximation.\left(\dfrac{b-a}{n}\right)(i-1)is the width multiplied by the counter (i-1). This value is equal to the leftmost edge of each rectangle, which is a typical approach for the left point approximation.

At this point, you're probably starting to get the picture. The average of what is on the left and what is on the right gives us something in between. That is 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}\derecha)....+f\izquierda(\dfrac{{x}_{n-1}{+x}_{n}}{2}\derecha)\derecha)

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}\direita)\Delta x}}

This can be compared to the first formula, namely:

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

Most of the time on the AP® Calculus AB and BC exams, you will use the midpoint rule in a numeric setting. That is, you will always receive the number of rectangles or subintervals that you will use for the approximation.

Now, let's 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 functionf(x)=x^2+5not interval [0,4] at=4.

Solution:

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

b-a=4-0=4

Remember that the width of the rectangle is given by:

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

Now let's find the values ​​in the middle, this will give us the midpoint of the subintervals.

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 let's plug the x-values ​​into our function to determine the height of each of the rectangles (the heights).

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

Evaluating we get:

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

Area \approx.1(41)

Area \approx41

Calculating the exact integral, we obtain:

\int_{0}^{4}{(x^2+5)dx}=41,33

Thus, we can confirm how closely the midpoint rule approximation compares to the exact integral.

You see how easy it is to use the midpoint rule. Let's try another example, and then review some previous AP® Calculus questions that required us to apply the 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)=\square root{x^{3}+2}

Solution

This is not unlike our previous examples. The first step is to find the total 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}

The next step is to find the midpoints of the subintervals.

The midpoints will be:

0,05,0,15,0,25,0,35,0,45,0,55,0,65,0,75,0,85,0,95

So we'll use those points along with the width to approximate the area.

Área \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's next in the next step. We will find the heights of the rectangles by plugging 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 get:

Area \approx. 2.2488

The exact integral gives:

\int_{0}^{1}{(x^3+2)dx}=2,25

We hope you have seen that nothing changes in the steps to approximate the areas for different types of given functions. No matter how complex the function is, the steps will remain the same.

Now let's take a look at the free-response questions on the AP® Calculus AB and BC exams in recent years. We will only do part of the problem, which requires an approximation of the midpoint.

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

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

Use a midpoint Riemann sum with two subintervals of equal length and values ​​to approximate\int_{1}^{1.4}{f'(x)dx}🇧🇷 Use the approximation to\int_{1}^{1.4}{f'(x)dx}to estimate the value of f(1.4). Show the calculation that leads to your answer.

Solution

In the table, the width

=\dfrac{1,4-1}{2}=\dfrac{0,4}{2}=0,2

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

\int_{1}^{1,4}{f'(x)dx}\aprox. 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)

Hot water drips from a coffee pot, filling a large cup of 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 equal-length subintervals indicated by the data in the table to approximate the value of\dfrac{1}{6}\int_{0}^{6}{C(t)dt}🇧🇷 Using the correct units, explain the meaning of\dfrac{1}{6}\int_{0}^{6}{C(t)dt}in the context of the problem.

Solution

The width will be:

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

The 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)=10,1ounces

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

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

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

3a) Use a midpoint Riemann sum with four equal-length subintervals and table values ​​to approximate\int_{0}^{40}{V(t)dt}🇧🇷 Show the calculation that leads to your answer. Using the correct units, explain the meaning of\int_{0}^{40}{V(t)dt}in terms of flying the plane.

Solution

The Riemann sum of the midpoint is given by:

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

=10[9,2+7,0+2,4+4,3]

= 229 miles

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

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

Rocket A has a positive velocity v(t) after it is launched upward from an initial height of 0 ft at time t = 0 seconds. The rocket's velocity is recorded for selected values ​​of t in the range 0 to 80 seconds, as shown in the table above.

4b) Using the correct units, explain the meaning of\int_{10}^{70}{V(t)dt}in terms of rocket flight. Use the Riemann sum of the midpoint with 3 subintervals of equal length to approximate\int_{10}^{70}{V(t)dt}.

Solution

an 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 Riemann sum of the midpoint is given by:

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

= 2020 pies

conclusion

After working through the examples and questions from the AP® Calculus exams above, you realize how simple it is to use the midpoint rule! All we need to know is the formula and how to substitute in real numbers. Thus, you will be well prepared to answer any questions that require the application of the midpoint rule.

Points to consider:

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}\derecha)....+f\izquierda(\dfrac{{x}_{n-1}{+x}_{n}}{2}\derecha)\derecha)

where x₀, X₁, X₂, X₃…..X_norteare 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 arise in two ways:

• Using the midpoint rule to approximate thearea under a curve.
• Using the midpoint rule to approximate thevalue of an integral.

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

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

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