How to Use the Midpoint Rule to Approximate Definite Integrals

Geometric Interpretation

The Midpoint Rule can be understood visually as replacing the region under a curve with a collection of rectangles. Each rectangle covers one subinterval, and the value of the function at the center of that subinterval determines its height.

The resulting sum of rectangle areas approximates the area represented by the definite integral.

An important feature of this approach is that increasing \( n \) makes the rectangles narrower. As the width of the rectangles decreases, they can generally follow the curve more closely, improving the approximation.

Why Is It Called the Midpoint Rule?

The name comes directly from where the function is evaluated. For each subinterval, the function is evaluated at the midpoint rather than at the left or right endpoint.

If a subinterval is

$$[x_{i – 1}, x_i],$$

then its midpoint is

$$\frac{x_{i – 1}+x_i}{2}.$$

The corresponding rectangle therefore has height

$$f(\frac{x_{i – 1} + x_i}{2}).$$

This simple choice of sample point is what distinguishes the Midpoint Rule from other numerical integration methods.

The Midpoint Rule is particularly useful because it provides a relatively simple approximation. Understanding why this happens requires a closer look at the function’s geometry and curvature, which we examine next.

Why Does the Midpoint Rule Work?

The Midpoint Rule works because a definite integral can be interpreted as the limit of a sum of small areas. When the interval of integration is divided into sufficiently small subintervals, the area under a smooth curve over each subinterval can be approximated by a rectangle. The Midpoint Rule chooses the function value at the center of each subinterval to determine the height of that rectangle. This choice lets the rectangle capture the function’s behavior.

For a refresher on smooth functions, please refer to the articles Newton’s Method Explained for Beginners, The Ultimate Resource for Understanding the Secant Method, and Interpolation Using Lagrange Polynomials in Numerical Analysis: A Complete Tutorial.

Approximating the Area Under a Curve

Consider the definite integral

$$\int_a^b f(x) dx.$$

Geometrically, this integral represents the signed area between the graph of \( f(x) \) and the \( x \)-axis over the interval \( [a,b] \). For many functions, however, the boundary of this region is curved, making the exact area difficult to determine directly.

The Midpoint Rule replaces the curved region with a collection of rectangles. If the interval is divided into \( n \) equal subintervals, each rectangle has width

$$\Delta x = \frac{b – a}{n}.$$

The height of each rectangle is determined by evaluating the function at the midpoint of its corresponding subinterval. Therefore, the area of the \( i \)-th rectangle is approximately

$$f(x_i^*)\Delta x.$$

Adding these areas produces

$$M_n = \Delta x\sum_{i = 1}^{n}f(x_i^*).$$

As \( n \) increases, \( \Delta x \) becomes smaller, and the rectangles generally provide a closer approximation to the shape of the region under the curve.

This idea is closely connected to the definition of the definite integral as a limit of Riemann sums. In fact, as the number of subintervals approaches infinity, the Midpoint Rule approximation approaches the exact value of the integral under appropriate conditions:

$$\lim_{n\to\infty}M_n = \int_a^b f(x)dx.$$

Thus, the Midpoint Rule works because sufficiently fine partitions allow the sum of the rectangle areas to converge to the area represented by the integral.

The Role of Curvature

The accuracy of the Midpoint Rule is also strongly influenced by the curvature of the function. Curvature describes how quickly the slope of a function changes and is related to its second derivative, \( f”(x) \). For a review of curvature, please refer to the article Concavity of Functions for Beginners: Everything You Need to Know. For a review of derivatives, please refer to the article How to Differentiate a Function Step by Step: A Beginner’s Guide.

If a function is linear on a subinterval, the midpoint provides a particularly effective approximation. In fact, for a linear function, the average value of the function over the subinterval is exactly its value at the midpoint. Consequently, the Midpoint Rule produces the exact integral for any linear function, regardless of the number of subintervals.

For curved functions, however, the midpoint value will generally not equal the average value of the function across the subinterval. The resulting difference contributes to the approximation error.

This is why functions with significant curvature can require more subintervals to obtain a desired level of accuracy. The more rapidly the curvature changes, the more the interval may need to be divided.

The second derivative quantifies this effect. The Midpoint Rule error bound depends on the maximum value of \( |f”(x)| \) over the interval. We will examine this error bound in detail later in the article.

Connection to Taylor Series

Now we use Taylor series to see how the midpoint rule works. For a review of Taylor series, please refer to the articles How to Calculate Limits in Calculus: Everything You Need to Know and Finite Difference Method for Numerical Differentiation: A Step-by-Step Guide.

The effectiveness of the Midpoint Rule can also be understood using a Taylor series. Suppose \( f(x) \) is sufficiently smooth and consider a subinterval centered at \( m \). For a point \( x \) near \( m \), Taylor’s theorem gives

$$f(x) = f(m) + f'(m)(x – m) + \frac{f”(m)}{2!}(x – m)^2 + \cdots.$$

The Midpoint Rule uses \( f(m) \) to approximate the function throughout the subinterval. When the function is integrated symmetrically around \( m \), the terms involving odd powers of \( x – m \) cancel. As a result, the first significant error term is associated with the second derivative.

This helps explain why the Midpoint Rule error decreases as the number of subintervals increases. In particular, the error is proportional to

$$\frac{1}{n^2}.$$

Thus, if the number of subintervals is doubled, the error will generally decrease by approximately a factor of four, assuming the function satisfies the conditions required for the error estimate.

When Does the Midpoint Rule Succeed?

The Midpoint Rule tends to be most successful when the function being integrated is smooth and does not change too dramatically over each subinterval. Although no numerical integration method is guaranteed to produce a highly accurate approximation with an arbitrary choice of \( n \), the Midpoint Rule behaves predictably when its underlying assumptions are satisfied.

Smooth Functions

The Midpoint Rule generally performs well when \( f(x) \) is smooth throughout the interval of integration. A smooth function changes gradually, allowing the rectangles the method constructs to follow the curve closely.

Increasing the Number of Subintervals

One of the simplest ways to improve a Midpoint Rule approximation is to increase the number of subintervals. As the number of subintervals increases, the subinterval lengths decrease. The resulting rectangles become narrower and can therefore follow the curve more closely. This means increasing the number of subintervals can substantially improve accuracy.

Functions With Moderate Curvature

The amount of curvature in a function has a direct effect on how well the Midpoint Rule performs.

The Midpoint Rule is exact for a linear function, regardless of the number of subintervals. This occurs because the value of a linear function at the midpoint of an interval is equal to its average value over that interval.

For nonlinear functions, the situation is different. As the curvature becomes more pronounced, the midpoint value may differ more substantially from the average value of the function over a subinterval. In such cases, additional subintervals may be needed.

The second derivative provides a useful measure of this behavior. If

$$|f”(x)|$$

remains small throughout \( [a,b] \), the Midpoint Rule will generally have a smaller error bound than it would for a function with a much larger second derivative.

When Can the Midpoint Rule Fail?

The Midpoint Rule is a reliable numerical integration method for many problems, but it is not always accurate. A function can have discontinuities or significant curvature that make the midpoint approximation unreliable. Understanding these limitations is important because increasing the number of subintervals is not always enough to solve every problem.

Too Few Subintervals

One of the most common reasons the Midpoint Rule produces a poor approximation is simply using too few subintervals.

Recall that

$$\Delta x = \frac{b-a}{n}.$$

When \( n \) is small, each rectangle covers a relatively large portion of the interval. If the function changes significantly across one of these large subintervals, a single midpoint value may not adequately represent the function throughout that region.

Increasing \( n \) generally improves the approximation for a sufficiently smooth function because it makes the rectangles narrower. However, the appropriate value of \( n \) depends on the desired accuracy and the behavior of the function.

Large Curvature

The Midpoint Rule is particularly sensitive to the function’s curvature. As discussed earlier, the second derivative \( f”(x) \) plays an important role in determining the method’s error.

If

$$|f”(x)|$$

is large over some portion of the interval, the function may bend significantly between the endpoints of a subinterval. A rectangle based on the midpoint may then provide a relatively poor approximation of the area.

This does not mean that the Midpoint Rule cannot be used for functions with substantial curvature. Instead, it may require more subintervals to achieve the desired accuracy.

Discontinuities

Discontinuities can create more serious problems. Suppose \( f (x) \) is discontinuous somewhere inside the interval \( [a,b] \). The function may suddenly jump from one value to another or become undefined.

The Midpoint Rule may still produce a numerical value if none of the chosen midpoints happens to coincide with the discontinuity. However, that numerical value may not provide a reliable approximation to the integral.

In these situations, you may need to split the integral at the discontinuity and analyze each portion independently.

For a review of continuous functions, please refer to the articles A Complete Introduction to the Bisection Method with Worked Examples and Solutions and How to Solve Equations Using Fixed Point Iteration: A Step-by-Step Guide.

How to Calculate the Error in the Midpoint Rule

Knowing the Midpoint Rule approximation is useful, but it is even more valuable when we can measure its accuracy. The error tells us how far the Midpoint Rule approximation is from the exact value of the definite integral. In some problems, the exact integral is available, allowing us to calculate the actual error directly. In other situations, the exact value may be difficult or impossible to obtain. In those cases, an error bound can estimate how large the error can be.

Absolute Error

Let

$$I = \int_a^b f(x)dx$$

represent the exact value of the integral, and let \( M_n \) represent the Midpoint Rule approximation. The absolute error is

$$\epsilon = |I – M_n|.$$

The absolute error is useful because it tells us the magnitude of the difference between the approximation and the exact answer.

The Midpoint Rule Error Bound

In many practical applications, the exact value of the integral is unknown. Fortunately, the Midpoint Rule has a theoretical error bound that can estimate the maximum possible error.

If \( f”(x) \) is continuous on \( [a,b] \), then the Midpoint Rule error satisfies

$$\epsilon \leq \frac{(b – a)^3}{24n^2} \max_{a\leq x\leq b}|f”(x)|.$$

This formula provides an upper bound for the absolute error.

The actual error may be much smaller, but it cannot exceed the error bound.

The appearance of the second derivative is significant. It reflects that the error is influenced by the function’s curvature.

Finding the Maximum of \( |f”(x)| \)

The most challenging part of finding the error bound is often determining

$$\max_{a\leq x\leq b}|f”(x)|.$$

To find this maximum, first calculate the second derivative \( f”(x) \). Then determine where its absolute value is largest on the interval.

Depending on the function, this may require finding critical points and evaluating the function at those points and at the endpoints, as discussed in How to Find Intervals of Increase and Decrease: A Step-by-Step Guide.

Step-by-Step Procedure for Using the Midpoint Rule

The Midpoint Rule approximates a definite integral when finding its exact antiderivative is difficult or unnecessary. Use the following procedure for any Midpoint Rule problem.

Step 1: Identify the Interval of Integration

Begin by identifying the lower and upper limits of integration.

For the integral

$$\int_a^b f(x)dx,$$

the interval is

$$[a,b].$$

The value \( a \) is the left endpoint, while \( b \) is the right endpoint.

Step 2: Determine the Number of Subintervals

Next, identify the number of subintervals \( n \).

Step 3: Calculate the Width of Each Subinterval

The width of each subinterval is

$$\Delta x = \frac{b – a}{n}.$$

Step 4: Find the Subintervals

Once \( \Delta x \) has been calculated, divide the original interval into \( n \) equal pieces.

Step 5: Find the Midpoint of Each Subinterval

The midpoint is the average of the endpoints for each subinterval.

Step 6: Evaluate the Function at Each Midpoint

Next, substitute each midpoint into \( f(x) \).

Step 7: Multiply Each Function Value by the Subinterval Width

This gives the area of each rectangle.

Step 8: Add the Rectangle Areas

Finally, add all the rectangle areas together.

In the next section, we will put this procedure into practice with a couple of worked examples, showing how the Midpoint Rule can be used to approximate definite integrals.

Worked Out Examples

The first example shows how to use the midpoint rule.

Example 1: Use the midpoint rule to approximate \( \int_0^4 x^2 dx \). Use 4 subintervals and find the absolute error.

Solution: The length of the subintervals is

$$\Delta x = \frac{4 – 0}{4}.$$

Subtracting gives

$$\Delta x = \frac{4}{4}.$$

Dividing we obtain

$$\Delta x = 1.$$

The subintervals are then

$${[0,1],[1,2],[2,3],[3,4]}.$$

The first midpoint is

$$x_1^* = \frac{0 + 1}{2}.$$

Adding gives

$$x_1^* = \frac{1}{2}.$$

Dividing we obtain

$$x_1^* = 0.5.$$

Consider the function

$$f(x) = x^2.$$

Evaluating the function at \( x_1^* = 0.5 \) we obtain

$$f(0.5) = 0.5^2.$$

Squaring we find

$$f(0.5) = 0.25.$$

The second midpoint is

$$x_2^* = \frac{1 + 2}{2}.$$

Adding gives

$$x_2^* = \frac{3}{2}.$$

Dividing we obtain

$$x_2^* = 1.5.$$

Evaluating the function at \( x_2^* = 1.5 \) we obtain

$$f(1.5) = 1.5^2.$$

Squaring we find

$$f(1.5) = 2.25.$$

The third midpoint is

$$x_3^* = \frac{2 + 3}{2}.$$

Adding gives

$$x_3^* = \frac{5}{2}.$$

Dividing we obtain

$$x_3^* = 2.5.$$

Evaluating the function at \( x_3^* = 2.5 \) we obtain

$$f(2.5) = 2.5^2.$$

Squaring we find

$$f(2.5) = 6.25.$$

The fourth midpoint is

$$x_4^* = \frac{3 + 4}{2}.$$

Adding gives

$$x_4^* = \frac{7}{2}.$$

Dividing we obtain

$$x_4^* = 3.5.$$

Evaluating the function at \( x_4^* = 3.5 \) we obtain

$$f(3.5) = 3.5^2.$$

Squaring we find

$$f(3.5) = 12.25.$$

By the midpoint rule, our approximation is then

$$M_4 = 1(0.25) + 1(2.25) + 1(6.25) + 1(12.25).$$

Multiplying gives

$$M_4 = 0.25 + 2.25 + 6.25 + 12.25.$$

Adding we get

$$M_4 = 21.$$

Integrating gives

$$\frac{x^3}{3}|_0^4.$$

Applying the Fundamental Theorem of Calculus, we obtain

$$\frac{4^3}{3} – \frac{0^3}{3}.$$

Cubing we get

$$\frac{64}{3} – \frac{0}{3}.$$

Dividing this is

$$21.333333 – 0.$$

Subtracting gives

$$21.333333.$$

The absolute error is then

$$\epsilon = |21.333333 – 21|.$$

Subtracting gives

$$\epsilon = |0.333333|.$$

Taking the absolute value, we arrive at a final answer of

$$\epsilon = 0.333333.$$

The next example illustrates one of the primary applications of numerical analysis, namely solving definite integrals without an elementary antiderivative.

Example 2: Use the midpoint rule to approximate \( \int_0^2 e^{x^2} dx \). Use 4 subintervals and find the error bound.

Solution: The length of the subintervals is

$$\Delta x = \frac{2 – 0}{4}.$$

Subtracting gives

$$\Delta x = \frac{2}{4}.$$

Dividing we obtain

$$\Delta x = 0.5.$$

The subintervals are then

$${[0,0.5],[0.5,1],[1,1.5],[1.5,2]}.$$

The first midpoint is

$$x_1^* = \frac{0 + 0.5}{2}.$$

Adding gives

$$x_1^* = \frac{0.5}{2}.$$

Dividing we obtain

$$x_1^* = 0.25.$$

Consider the function

$$f(x) = e^{x^2}.$$

Evaluating the function at \( x_1^* = 0.25 \) we obtain

$$f(0.25) = e^{0.25^2}.$$

Squaring we find

$$f(0.25) = e^{0.0625}.$$

Exponentiating gives

$$f(0.25) = 1.064494.$$

The second midpoint is

$$x_2^* = \frac{0.5 + 1}{2}.$$

Adding gives

$$x_2^* = \frac{1.5}{2}.$$

Dividing we obtain

$$x_2^* = 0.75.$$

Evaluating the function at \( x_2^* = 0.75 \) we obtain

$$f(0.75) = e^{0.75^2}.$$

Squaring we find

$$f(0.75) = e^{0.5625}.$$

Exponentiating gives

$$f(0.75) = 1.755054.$$

The third midpoint is

$$x_3^* = \frac{1 + 1.5}{2}.$$

Adding gives

$$x_3^* = \frac{2.5}{2}.$$

Dividing we obtain

$$x_3^* = 1.25.$$

Evaluating the function at \( x_3^* = 1.25 \) we obtain

$$f(1.25) = e^{1.25^2}.$$

Squaring we find

$$f(1.25) = e^{1.5625}.$$

Exponentiating gives

$$f(1.25) = 4.770733.$$

The fourth midpoint is

$$x_4^* = \frac{1.5 + 2}{2}.$$

Adding gives

$$x_4^* = \frac{3.5}{2}.$$

Dividing we obtain

$$x_4^* = 1.75.$$

Evaluating the function at \( x_4^* = 1.75 \) we obtain

$$f(1.75) = e^{1.75^2}.$$

Squaring we find

$$f(1.75) = e^{3.0625}.$$

Exponentiating gives

$$f(1.75) = 21.380942.$$

By the midpoint rule, our approximation is then

$$M_4 = 0.5(1.064494) + 0.5(1.755054) + 0.5(4.770733) + 0.5(21.380942).$$

Multiplying gives

$$M_4 = 0.532247 + 0.877527 + 2.385366 + 10.690471.$$

Adding we get

$$M_4 = 14.485612.$$

Taking the derivative, we get

$$f'(x) = (e^{x^2})’.$$

Using the chain rule, this is

$$f'(x) = (x^2)’e^{x^2}.$$

Taking the derivative, we get

$$f'(x) = 2xe^{x^2}.$$

Taking the derivative again, we get

$$f”(x) = (2xe^{x^2})’.$$

Using the product rule, this is

$$f”(x) = (2x)’e^{x^2} + 2x(e^{x^2})’.$$

Using the chain rule gives

$$f”(x) = (2x)’e^{x^2} + 2x(x^2)’e^{x^2}.$$

Taking the derivatives, we get

$$f”(x) = 2e^{x^2} + 2x(2x)e^{x^2}.$$

Distributing gives

$$f”(x) = 2e^{x^2} + 4x^2e^{x^2}.$$

Factoring out \( e^{x^2} \) this is

$$f”(x) = (2 + 4x^2)e^{x^2}.$$

Taking the derivative once again, we get

$$f”‘(x) = ((2 + 4x^2)e^{x^2})’.$$

Using the product rule, this is

$$f”‘(x) = (2 + 4x^2)’e^{x^2} + (2 + 4x^2)(e^{x^2})’.$$

Using the sum rule and chain rule gives

$$f”‘(x) = ((2)’ + (4x^2)’)e^{x^2} + (2 + 4x^2)(x^2)’e^{x^2}.$$

Taking the derivatives, we get

$$f”‘(x) = (0 + 8x)e^{x^2} + (2 + 4x^2)(2x)e^{x^2}.$$

Adding gives

$$f”‘(x) = 8xe^{x^2} + (2 + 4x^2)(2x)e^{x^2}.$$

Factoring 2 from the first factor in the second term, we obtain

$$f”‘(x) = 8xe^{x^2} + 2(1 + 2x^2)(2x)e^{x^2}.$$

This is equivalent to

$$f”‘(x) = 8xe^{x^2} + 4x(1 + 2x^2)e^{x^2}.$$

Factoring out \( 4xe^{x^2} \) this is

$$f”‘(x) = 4xe^{x^2}(2 + 1 + 2x^2).$$

Simplifying gives

$$f”‘(x) = 4xe^{x^2}(3 + 2x^2).$$

Setting the third derivative equal to 0, we obtain

$$0 = 4xe^{x^2}(3 + 2x^2).$$

Since \( 4e^{x^2}(3 + 2x^2) \) is nonzero for all real x we can divide by it to get

$$0 = x.$$

Evaluating the second derivative at \( x = 0 \) we obtain

$$f”(0) = (2 + 4(0^2))e^{0^2}.$$

Squaring we find

$$f”(0) = (2 + 4(0))e^0.$$

Exponentiating gives

$$f”(0) = (2 + 4(0))1.$$

Multiplying we get

$$f”(0) = (2 + 0)1.$$

Adding we obtain

$$f”(0) = (2)1.$$

Multiplying this is

$$f”(0) = 2.$$

Evaluating the second derivative at \( x = 2 \) we obtain

$$f”(2) = (2 + 4(2^2))e^{2^2}.$$

Squaring we find

$$f”(2) = (2 + 4(4))e^4.$$

Exponentiating gives

$$f”(2) = (2 + 4(4))54.598150.$$

Multiplying we get

$$f”(2) = (2 + 16)54.598150.$$

Adding we obtain

$$f”(2) = (18)54.598150.$$

Multiplying this by

$$f”(2) = 982.766700.$$

The error bound is then

$$\epsilon \leq \frac{(2 – 0)^3}{24(4)^2} (982.766700).$$

Subtracting gives

$$\epsilon \leq \frac{(2)^3}{24(4)^2} (982.766700).$$

Cubing and squaring this is

$$\epsilon \leq \frac{8}{24(16)} (982.766700).$$

Multiplying we get

$$\epsilon \leq \frac{8}{384} (982.766700).$$

Dividing we find

$$\epsilon \leq 0.020833(982.766700).$$

Multiplying, we arrive at a final answer of

$$\epsilon \leq 20.474306.$$

Conclusion

The Midpoint Rule is a numerical integration technique for approximating definite integrals. Instead of finding an exact antiderivative, the method divides the interval into equal subintervals, evaluates the function at the midpoint of each subinterval, and uses those values to estimate the area under the curve.

As we have seen, using the midpoint rule involves several steps: identify the interval and number of subintervals, calculate \( \Delta x \), determine the midpoint of each subinterval, evaluate the function at those midpoints, and add the resulting rectangle areas. The accuracy of the approximation depends on the function’s behavior and the number of subintervals.

However, the Midpoint Rule is not appropriate for every situation. Discontinuities and regions of substantial curvature can make the method less reliable or invalidate its error bound. Recognizing these limitations is just as important as knowing how to perform the calculation.

Further Reading

Examples of Improper Integrals with Solutions: A Beginner’s Guide  – Integrals with discontinuities can also be approximated using the midpoint rule. This guide shows how to evaluate these integrals.

Frequently Asked Questions