Approximating Definite Integrals

Main idea

When you can't find an antiderivative, go back to what an integral means, area, and fill it with shapes you can measure: rectangles, trapezoids or little parabolas. The closer the shape hugs the curve, the faster the error shrinks as you add slices.

Midpoint and trapezoid

M_n = {\htmlData{sym=0}{\Delta x}}\sum_{i=1}^n f({\htmlData{sym=1}{\bar x_i}}), \qquad T_n = \frac{{\htmlData{sym=0}{\Delta x}}}{2}\left[f({\htmlData{sym=2}{x_0}}) + 2f(x_1) + \cdots + 2f(x_{n-1}) + f({\htmlData{sym=3}{x_n}})\right]

What the symbols mean

$\Delta x$
Width of each slice
(b − a)/n: split [a, b] into n equal pieces.
$\bar x_i$
Midpoint
The center of the i-th slice, where the rectangle's height is measured.
$x_0$
Left endpoint
x₀ = a. In general xᵢ = a + iΔx.
$x_n$
Right endpoint
xₙ = b.

Simpson's rule (n even)

S_n = {\htmlData{sym=0}{\frac{\Delta x}{3}}}\left[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)\right]

What the symbols mean

$\frac{\Delta x}{3}$
Simpson's weight
Comes from fitting a parabola through each group of three points.

Error bounds

|E_M| \le \frac{{\htmlData{sym=0}{K}}(b-a)^3}{24{\htmlData{sym=2}{n}}^2}, \quad |E_T| \le \frac{{\htmlData{sym=0}{K}}(b-a)^3}{12{\htmlData{sym=2}{n}}^2}, \quad |E_S| \le \frac{{\htmlData{sym=1}{L}}(b-a)^5}{180{\htmlData{sym=2}{n}}^4}

What the symbols mean

$K$
Bound on |f″|
The largest size of the second derivative on [a, b]: how curved the graph is.
$L$
Bound on |f⁽⁴⁾|
The largest size of the fourth derivative on [a, b].
$n$
Number of slices
More slices, smaller error.

Explanation

Numerical integration is simply estimating the area under a curve by filling it with simple shapes whose areas you can compute, then adding them up. It's good for every integral you can't do by hand, like $\int_0^1 e^{-x^2}\,dx$ , and for real data where you only have measurements and no formula at all. All we're really saying with a rule like $T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + \cdots + f(x_n)\right]$ is: $\Delta x$ (the width of each slice), the $f(x_i)$ (the curve's heights at the slice edges), and the weights $1, 2, 2, \dots, 1$ (how many shapes share each height). The error bounds look scary, but they just say how fast the estimate improves as you use more slices. It's all area of rectangles, trapezoids and parabolas; bear with it.

Remember where integrals came from: Riemann sums, adding up thin rectangles. Back then, the antiderivative was a shortcut that made the rectangles unnecessary. But when there's no antiderivative, like $\int_0^1 e^{-x^2}\,dx$ (one of the integrals from the strategy lesson that can't be done in elementary terms), you go back to adding up shapes. The question now is which shapes give the best answer for the least work.

Split $[a, b]$ into $n$ slices of width $\Delta x = \frac{b - a}{n}$ , with endpoints $x_i = a + i\,\Delta x$ .

Midpoint rule: a rectangle on each slice, as tall as $f$ at the slice's midpoint.
Trapezoid rule: join neighboring points with straight lines. Each interior point is a corner of two trapezoids, so it gets counted twice, and that's where the 2's come from.
Simpson's rule: fit a parabola through each group of three points. Each parabola covers two slices, so it needs an even $n$ , and the weights run $1, 4, 2, 4, \dots, 2, 4, 1$ .

How good are they? The error bounds tell the story. Doubling $n$ divides the midpoint and trapezoid errors by about 4, and Simpson's error by about 16. That makes sense: a straight line can't follow a curve's bend, so its error depends on $f''$ , while a parabola can bend too, so its error only depends on higher derivatives.

Here's something neat. Midpoint is usually about twice as accurate as trapezoid, with the opposite sign: on a curve that bends up, trapezoids overshoot and midpoint rectangles undershoot. So a weighted average should cancel most of the error, and Simpson's rule is exactly that average, $S_{2n} = \frac{2M_n + T_n}{3}$ .

A common mistake. It's tempting to think Simpson's rule is only exact for parabolas, since that's what it fits. It's actually exact for cubics too, because the error bound involves $f^{(4)}$ , which is zero for a cubic. The first worked example shows it hitting $\int_0^3 x^2\,dx = 9$ on the nose.

There's an animation further down that shows all three shapes on the same curve.

See it

The same area estimated three ways: rectangles, then trapezoids, then parabolas. Notice how the gaps between the shapes and the curve shrink at each stage. Flat tops miss the curve's slope, straight slanted tops miss its bend, and parabolas bend along with it.

Heights, test scores and measurement errors often follow a bell curve, and the chance of landing between the average and 2 standard deviations above it is the area under e^(−x²/2)/√(2π) from 0 to 2. That function has no elementary antiderivative, so the numbers in statistics tables come from estimates like this. As you slide n up, notice the trapezoids close in: 1 slice gives 0.453, off by 0.024, while 8 slices give 0.4767, within 0.0006 of the true 0.4772. Past the first couple of steps, each doubling of n cuts the error by about 4.

Worked examples

Example 1

Approximate $\displaystyle\int_0^3 x^2\,dx$ with $n = 6$ using the midpoint, trapezoid and Simpson rules. (The exact value is 9.)

1
$\Delta x = 0.5$ . The points are $0, 0.5, 1, \dots, 3$ with $f$ values $0, 0.25, 1, 2.25, 4, 6.25, 9$ .

Example 2

How accurate is $M_{10}$ for $\displaystyle\int_0^1 e^{-x^2}\,dx$ ?

1
The error bound needs $K$ , the biggest $|f''|$ on the interval. Here $f''(x) = (4x^2 - 2)e^{-x^2}$ , and on $[0, 1]$ its size is at most $2$ (at $x = 0$ ), so $K = 2$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Use the trapezoid rule with $n = 4$ to approximate $\displaystyle\int_0^2 x^3\,dx$ .

Problem 2+5 XP

Use Simpson's rule with $n = 6$ to approximate $\displaystyle\int_0^3 \frac{dx}{1 + x^4}$ . Give 4 decimals.

Problem 3+5 XP

Use the midpoint rule with $n = 4$ to approximate $\displaystyle\int_1^5 \frac{dx}{x^2}$ . Give 4 decimals.

Problem 4+5 XP

How large must $n$ be for Simpson's rule to approximate $\displaystyle\int_1^2 \frac{dx}{x}$ to within $0.0001$ ? (Use $|f^{(4)}(x)| = 24/x^5 \le 24$ on $[1, 2]$ .)

Problem 5+5 XP

For a smooth function and the same $n$ , which is usually the most accurate?

Problem 6+5 XP

A model rocket motor's thrust is measured every $0.5$ s: $0, 120, 180, 160, 100, 40, 0$ newtons at $t = 0, 0.5, \dots, 3$ . Use Simpson's rule to estimate its total impulse, $\int_0^3F\,dt$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

How many subintervals guarantee that the trapezoid rule approximates $\displaystyle\int_0^1e^{x^2}dx$ to within $10^{-4}$ ? Use the error bound $\frac{K(b - a)^3}{12n^2}$ , with $K = \max|f''|$ .

Where it's used

Your calculator's ∫ button, every physics engine and every GPS receiver integrates numerically. Simpson's rule's cousins, Gaussian quadrature for integrals and Runge–Kutta for differential equations, run inside nearly all scientific software.

Rating a Model Rocket MotorCalc II · Physics

Comparison Test for Improper Integrals Arc Length

Approximating Definite Integrals

Main idea

Midpoint and trapezoid

M_n = {\htmlData{sym=0}{\Delta x}}\sum_{i=1}^n f({\htmlData{sym=1}{\bar x_i}}), \qquad T_n = \frac{{\htmlData{sym=0}{\Delta x}}}{2}\left[f({\htmlData{sym=2}{x_0}}) + 2f(x_1) + \cdots + 2f(x_{n-1}) + f({\htmlData{sym=3}{x_n}})\right]

What the symbols mean

$\Delta x$
Width of each slice
(b − a)/n: split [a, b] into n equal pieces.
$\bar x_i$
Midpoint
The center of the i-th slice, where the rectangle's height is measured.
$x_0$
Left endpoint
x₀ = a. In general xᵢ = a + iΔx.
$x_n$
Right endpoint
xₙ = b.

Simpson's rule (n even)

S_n = {\htmlData{sym=0}{\frac{\Delta x}{3}}}\left[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)\right]

What the symbols mean

$\frac{\Delta x}{3}$
Simpson's weight
Comes from fitting a parabola through each group of three points.

Error bounds

|E_M| \le \frac{{\htmlData{sym=0}{K}}(b-a)^3}{24{\htmlData{sym=2}{n}}^2}, \quad |E_T| \le \frac{{\htmlData{sym=0}{K}}(b-a)^3}{12{\htmlData{sym=2}{n}}^2}, \quad |E_S| \le \frac{{\htmlData{sym=1}{L}}(b-a)^5}{180{\htmlData{sym=2}{n}}^4}

What the symbols mean

$K$
Bound on |f″|
The largest size of the second derivative on [a, b]: how curved the graph is.
$L$
Bound on |f⁽⁴⁾|
The largest size of the fourth derivative on [a, b].
$n$
Number of slices
More slices, smaller error.

Explanation

Split $[a, b]$ into $n$ slices of width $\Delta x = \frac{b - a}{n}$ , with endpoints $x_i = a + i\,\Delta x$ .

Midpoint rule: a rectangle on each slice, as tall as $f$ at the slice's midpoint.
Trapezoid rule: join neighboring points with straight lines. Each interior point is a corner of two trapezoids, so it gets counted twice, and that's where the 2's come from.
Simpson's rule: fit a parabola through each group of three points. Each parabola covers two slices, so it needs an even $n$ , and the weights run $1, 4, 2, 4, \dots, 2, 4, 1$ .

There's an animation further down that shows all three shapes on the same curve.

See it

Worked examples

Example 1

Approximate $\displaystyle\int_0^3 x^2\,dx$ with $n = 6$ using the midpoint, trapezoid and Simpson rules. (The exact value is 9.)

1
$\Delta x = 0.5$ . The points are $0, 0.5, 1, \dots, 3$ with $f$ values $0, 0.25, 1, 2.25, 4, 6.25, 9$ .

Example 2

How accurate is $M_{10}$ for $\displaystyle\int_0^1 e^{-x^2}\,dx$ ?

1
The error bound needs $K$ , the biggest $|f''|$ on the interval. Here $f''(x) = (4x^2 - 2)e^{-x^2}$ , and on $[0, 1]$ its size is at most $2$ (at $x = 0$ ), so $K = 2$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Use the trapezoid rule with $n = 4$ to approximate $\displaystyle\int_0^2 x^3\,dx$ .

Problem 2+5 XP

Use Simpson's rule with $n = 6$ to approximate $\displaystyle\int_0^3 \frac{dx}{1 + x^4}$ . Give 4 decimals.

Problem 3+5 XP

Use the midpoint rule with $n = 4$ to approximate $\displaystyle\int_1^5 \frac{dx}{x^2}$ . Give 4 decimals.

Problem 4+5 XP

How large must $n$ be for Simpson's rule to approximate $\displaystyle\int_1^2 \frac{dx}{x}$ to within $0.0001$ ? (Use $|f^{(4)}(x)| = 24/x^5 \le 24$ on $[1, 2]$ .)

Problem 5+5 XP

For a smooth function and the same $n$ , which is usually the most accurate?

Problem 6+5 XP

A model rocket motor's thrust is measured every $0.5$ s: $0, 120, 180, 160, 100, 40, 0$ newtons at $t = 0, 0.5, \dots, 3$ . Use Simpson's rule to estimate its total impulse, $\int_0^3F\,dt$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

How many subintervals guarantee that the trapezoid rule approximates $\displaystyle\int_0^1e^{x^2}dx$ to within $10^{-4}$ ? Use the error bound $\frac{K(b - a)^3}{12n^2}$ , with $K = \max|f''|$ .

Where it's used

Rating a Model Rocket MotorCalc II · Physics