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 the*ancho*multiplied 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_{UE}is 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,* norte*rectangles

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)

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 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)

t (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

C(t) (onzas) | 0 | 5.3 | 8.8 | 11.2 | 12.8 | 13.8 | 14.5 |

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)

t (minutes) | 0 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |

Vermont) | 7,0 | 9.2 | 9.5 | 7,0 | 4.5 | 2.4 | 2.4 | 4.3 | 7.3 |

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)

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 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 x_{0}, X_{1}, X_{2}, X_{3}…..X_{norte}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 arise 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 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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## FAQs

### How do you use the midpoint rule in calculus? ›

1: The midpoint rule approximates the area between the graph of f(x) and the x-axis by summing the areas of rectangles with midpoints that are points on f(x). **Use the midpoint rule to estimate ∫10x2dx using four subintervals**. Compare the result with the actual value of this integral.

**How do you know if midpoint rule is over or underestimate? ›**

**If the graph is concave up the trapezoid approximation is an overestimate, and the midpoint is an underestimate**. If the graph is concave down, then trapezoids give an underestimate and the midpoint an overestimate.

**Are midpoint Riemann sums the most accurate? ›**

Though still just an estimate, **the midpoint rule is typically more accurate than the right and left Riemann sums**. Here's an example of the rule being used in a math problem: Estimate the area under the curve f(x)=x3−6x+8 over the interval [-2,3] with 5 rectangles using the midpoint rule.

**Why is the midpoint method more accurate? ›**

The advantage of the midpoint method is that **one obtains the same elasticity between two price points whether there is a price increase or decrease**. This is because the formula uses the same base for both cases.

**What is the correct midpoint formula? ›**

Midpoint formula

If we have coordinates (x₁,y₁) and (x₂,y₂) , then the midpoint of these coordinates is determined by **(x₁ + x₂)/2, (y₁ + y₂)/2** .

**Is midpoint rule always more accurate? ›**

**The Midpoint rule is always more accurate than the Trapezoid rule**. ... For example, make a function which is linear except it has nar- row spikes at the midpoints of the subdivided intervals. Then the approx- imating rectangles for the midpoint rule will rise up to the level of the spikes, and be a huge overestimate.

**How do you know if a midpoint Riemann sum is an overestimate or underestimate? ›**

Overestimation and underestimation

In general, **if the function is always increasing or always decreasing on an interval**, we can tell whether the Riemann sum approximation will be an overestimation or underestimation based on whether it's a left or a right Riemann sum.

**What is the order of accuracy of the midpoint rule? ›**

. The midpoint method is **second order** accurate (i.e., the global error is second order).

**Is Simpsons rule more accurate than midpoint? ›**

In most cases I was correct that **the Simpsons gave the most accurate approximation**. However, I incorrectly predicted that the Trapezoidal would do better than the Midpoint. The Midpoint gave more accuracy than the Trapezoidal.

**How do you increase the accuracy of a Riemann sum? ›**

The Riemann sum is only an approximation to the actual area underneath the graph of f. To make the approximation better, we can **increase the number of subintervals n**, which makes the subinterval width Δx=(b−a)/n decrease.

### Is a midpoint sum an underestimate? ›

It takes a little more work to see, but because the function is concave up, **the midpoint estimate will be an underestimate**.

**Why is midpoint better than Euler? ›**

This instability can be controlled by careful timestep control. However, Euler's low order and the fact that the next simplest method--the Midpoint method--is **stable, more accurate, and only marginally more complicated to program**, mean that the Euler method is never used in real calculations.

**Why is it important to learn midpoint formula? ›**

Midpoint formula is used **to find the centre point of a straight line**. Sometimes you will need to find the number that is half of two particular numbers. For that, you find the average of the two numbers.

**What is the midpoint of 30 to 40? ›**

Hence, the midpoint of class 30-40 is **35**.

**What is the midpoint of 40 50? ›**

i.e, **45**. Was this answer helpful?

**Is midpoint rule better than Trapezoidal? ›**

The midpoint rule uses the midpoint of the rectangles for the estimate. A midpoint rule is a much better estimate of the area under the curve than either a left- or right- sum. As a rule of thumb, midpoint sums are twice as good than trapezoid estimates. Midpoint Riemann sum.

**Is the midpoint method stable? ›**

The midpoint method is quite attractive then because it produces a bounded solution, which is to say that it is stable, for hλ ∈ [−i, i]. Unfortunately, **it is not stable for any hλ with Re(λ) < 0**. For the test equation the midpoint method becomes y n + 1 = y n - 1 + 2 zy n , where it is convenient to introduce z = hλ.

**Does midpoint mean both sides are equal? ›**

Lesson Summary. **The midpoint of a line segment is a point that divides the line segment into two equal halves**. The midpoint theorem states that in a triangle, the segment that is formed by connecting the midpoints of two sides must be parallel to the third side and also half the length of the third side.

**Is it better to overestimate or underestimate a project? ›**

Whilst obviously accurate estimates are the best outcome, **over-estimation is less bad than underestimation**. Underestimation can impact dependencies and the overall quality of the project.

**How do you know if you under or over approximation? ›**

Compute f (t). **If f (t) > 0 for all t in I, then f is concave up on I, so L(x0) < f(x0), so your approximation is an under-estimate**. If f (t) < 0 for all t in I, then f is concave down on I, so L(x0) > f(x0), so your approximation is an over-estimate.

### How do you know if a function is an overestimate or underestimate? ›

Recall that one way to describe a concave up function is that it lies above its tangent line. So the concavity of a function can tell you whether the linear approximation will be an overestimate or an underestimate. 1. **If f(x) is concave up in some interval around x = c, then L(x) underestimates in this interval**.

**Why is Simpsons 3/8 more accurate? ›**

Simpson's 3/8 rule is similar to Simpson's 1/3 rule, the only difference being that, for the 3/8 rule, the interpolant is a cubic polynomial. Though **the 3/8 rule uses one more function value**, it is about twice as accurate as the 1/3 rule.

**Which is the most accurate method of numerical integration? ›**

If the functions are known analytically instead of being tabulated at equally spaced intervals, the best numerical method of integration is called **Gaussian quadrature**. By picking the abscissas at which to evaluate the function, Gaussian quadrature produces the most accurate approximations possible.

**What is the limitation of Simpsons rule? ›**

Limitations of Simpson's rule

**It is obviously inaccurate, i.e. there will always be a difference between it and the actual integral** (except in some cases, such as the area under straight lines). Integrals allow you to get exact answers in terms of fundamental constants, which Simpson's method does not allow.

**Why is Riemann sums important in studying calculus? ›**

A Riemann sum is an approximation of a region's area, obtained by adding up the areas of multiple simplified slices of the region. It is applied in calculus **to formalize the method of exhaustion, used to determine the area of a region**. This process yields the integral, which computes the value of the area exactly.

**Can you get a negative answer for Riemann sum? ›**

So, the Riemann sum is the sum of signed areas of rectangles: rectangles that lie above the -axis contribute positive values, and **rectangles that lie below the -axis contribute negative values to the Riemann sum**.

**Is Riemann sum the same as midpoint? ›**

**The Midpoint Riemann Sum is one for which we evaluate the function we're integrating at the midpoint of each interval, and use those values to determine the heights of the rectangles**. Our example function is going to be f(x)=x2+1, where we integrate over the interval [0,3].

**Can midpoint be negative? ›**

The Midpoint changes the paradigm of the story. The Midpoint requires a definitive and story-altering response from the characters. **The Midpoint should be negative if the First Plot Point was positive**; it should be positive if the First Plot Point was negative.

**Is Riemann sum a midpoint? ›**

**In a midpoint Riemann sum, the height of each rectangle is equal to the value of the function at the midpoint of its base**. We can also use trapezoids to approximate the area (this is called trapezoidal rule).

**What is midpoint in calculus? ›**

Midpoint is **the point which is in between an equal distance from both the points and it lies on the line segment which connects the two points**. You need to find out the midpoint by just adding the two values of both points together and then dividing it by 2.

### Why do we use the midpoint method? ›

The advantage of the Midpoint Method is that **one obtains the same elasticity between two price points whether there is a price increase or decrease**. This is because the formula uses the same base (average quantity and average price) for both cases.