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Midpoint Rule Calculator

Apply the midpoint rule to approximate definite integrals. Ideal when the function is undefined at endpoints.

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Basic Polynomial

Integrate x^2 from 0 to 1 using Simpson's rule. Exact answer: 1/3.

Key values: x^2 · [0, 1] · Simpson's rule

Trigonometric Integral

Integrate sin(x) from 0 to pi. Exact answer: 2.

Key values: sin(x) · [0, pi] · Exact: 2

Area Under a Bell Curve

Approximate the Gaussian integral e^(-x^2) from -3 to 3.

Key values: e^(-x^2) · [-3, 3] · Gaussian

Documentation

The Midpoint Rule

The midpoint rule evaluates the function at the center of each subinterval rather than at the endpoints. This simple change gives it a natural advantage for functions with endpoint singularities.

\int_a^b f(x)\,dx \approx h\sum_{i=0}^{n-1} f\!\left(a + \left(i + \tfrac{1}{2}\right)h\right)

Error Bound

|E_M| \leq \frac{(b-a)^3}{24n^2} \max_{x \in [a,b]} |f''(x)|

Same $O(h^2)$ convergence rate as the trapezoidal rule, but with half the error constant (denominator 24 vs. 12). For smooth functions, the midpoint rule is roughly twice as accurate as the trapezoidal rule for the same number of subdivisions.

Handling Endpoint Singularities

The midpoint rule's key advantage: it never evaluates $f(x)$ at the endpoints $a$ or $b$ . This makes it naturally robust for integrals like:

\int_0^1 \frac{1}{\sqrt{x}}\,dx = 2

Here $f(0)$ is undefined, so the trapezoidal and Simpson's rules fail without special handling. The midpoint rule works directly because it only evaluates at interior points like $x = 0.05, 0.15, 0.25, \ldots$

When to choose midpoint: Use it when the integrand has singularities at the integration bounds, or when you want a simple method that's more accurate than trapezoidal without the even- $n$ constraint of Simpson's.

Frequently Asked Questions

What is the midpoint rule for integration?

The midpoint rule approximates a definite integral by evaluating the function at the center of each subinterval rather than at the endpoints. The formula is $\sum f(\text{midpoint}_i) \times \Delta x$ summed over all subintervals.

Why is the midpoint rule better for functions with endpoint singularities?

The midpoint rule never evaluates $f(x)$ at the integration bounds $a$ or $b$ . This makes it naturally robust for integrals like $\int_0^1 \frac{1}{\sqrt{x}}\,dx$ , where $f(0)$ is undefined but the integral converges.

How accurate is the midpoint rule compared to the trapezoidal rule?

Both have $O(h^2)$ convergence, but the midpoint rule has half the error constant (denominator 24 vs. 12 in the error bound). For smooth functions, the midpoint rule is roughly twice as accurate as the trapezoidal rule for the same number of subdivisions.

Does the midpoint rule require an even number of subintervals?

No. Unlike Simpson's rule, the midpoint rule works with any number of subintervals, whether odd or even. This makes it the most flexible basic numerical integration method.

When should I choose the midpoint rule over Simpson's rule?

Choose the midpoint rule when the integrand has singularities at the bounds, when you need a simple method without parity constraints, or when you want a quick estimate that is more accurate than the trapezoidal rule.

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Midpoint Rule Calculator

Apply the midpoint rule to approximate definite integrals. Ideal when the function is undefined at endpoints.

Try an Example

Basic Polynomial

Trigonometric Integral

Area Under a Bell Curve

The Midpoint Rule

Error Bound

Handling Endpoint Singularities

Frequently Asked Questions

What is the midpoint rule for integration?

Why is the midpoint rule better for functions with endpoint singularities?

How accurate is the midpoint rule compared to the trapezoidal rule?

Does the midpoint rule require an even number of subintervals?

When should I choose the midpoint rule over Simpson's rule?

Related method Variants

Simpson's Rule

Trapezoidal Rule

Riemann Sum

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