Numerical Integration

Calculus 2

Overview

When working with Riemann sums, we use a left endpoint, right endpoint, or midpoint to set up our rectangles. These rectangles are occasionally equally spaced ( $\Delta x = (b-a)/n$ ), but this may not always be the case.

The approximations we found using these rectangles were decent, but why settle for decent?

This leads us to ask the questions:

What ways are we able to improve the area approximation a Riemann sum gives us?
Would shapes other than rectangles prove to be more useful?

The Trapezoidal Rule

With the Trapezoidal Rule, we use trapezoids instead of rectangles to better fit the region. That is, we move from using rectangles to approximate area (which corresponds to using constant functions to fit $f(x)$ ) to using trapezoids to approximate area (which corresponds to using linear functions to fit $f(x)$ ).

Recall that the area of a trapezoid with heights $h_1$ and $h_2$ and width $w$ is given by:

A = \frac{1}{2} w (h_1 + h_2) = w \left( \frac{h_1 + h_2}{2} \right).

The area of an arbitrary trapezoid is given by:

\begin{aligned} A_k &= \frac{1}{2} (f(x_k) + f(x_{k+1})) \cdot \Delta x \\ &= \frac{1}{2} (f(x_k) + f(x_{k+1})) \frac{b-a}{n} \\ &= \frac{b-a}{2n}(f(x_k) + f(x_{k+1})). \end{aligned}

Therefore, an approximation of the area of the entire region can be found by adding up/accumulating the area of all trapezoids of the form above. That is:

\begin{aligned} A &= \int_a^b f(x)dx\\ &\approx \frac{b-a}{2n}(f(x_0)+f(x_1))\\ & \ \ \ +\frac{b-a}{2n}(f(x_1)+f(x_2))\\ & \ \ \ \ \ \ +\ . \ . \ . \ \\ & \ \ \ \ \ \ \ \ \ +\frac{b-a}{2n}(f(x_{n-2})+f(x_{n-1}))\\ & \ \ \ \ \ \ \ \ \ \ \ \ +\frac{b-a}{2n}(f(x_{n-1})+f(x_{n}))\\ &= \frac{b-a}{2n}( \ (f(x_0)+f(x_1)) + (f(x_1)+f(x_2)) + \ . \ . \ . \ + (f(x_{n-2})+f(x_{n-1})) + (f(x_{n-1})+f(x_{n})) \ )\\ &= \frac{b-a}{2n}( \ f(x_0) + 2 \cdot f(x_1) + 2 \cdot f(x_2) + \ . \ . \ . \ + 2 \cdot f(x_{n-2}) + 2 \cdot f(x_{n-1}) + f(x_{n}) \ ) \end{aligned}

Note the 1-2--2-1 pattern for the Trapezoidal Rule.

Example 1

Use the Trapezoidal Rule with $n=4$ to estimate $\int_1^2 x^2 \, dx.$ Compare the estimate with the exact value.

View Answer

$\frac{75}{32} \approx 2.344$ (compared to $\frac{7}{3} \approx 2.333$ ).

Simpson's Rule

With Simpson's Rule, we use parabolas instead of trapezoids to better fit the region. That is, we move from using trapezoids to approximate area (which corresponds to using linear functions to fit $f(x)$ ) to using parabolas to approximate area (which corresponds to using quadratic functions to fit $f(x)$ ).

We won't go through the full derivation for this rule. However, the area of the parabolic section given by the points $(x_k,f(x_k))$ , $(x_{k+1},f(x_{k+1}))$ , and $(x_{k+1},f(x_{k+1}))$ is given by:

\begin{aligned} A_k &= \frac{1}{3}(f(x_k)+4f(x_{k+1})+f(x_{k+2}))\cdot \Delta x\\ &= \frac{1}{3}(f(x_k)+4f(x_{k+1})+f(x_{k+2})) \frac{b-a}{n}\\ &= \frac{b-a}{3n}(f(x_k)+4f(x_{k+1})+f(x_{k+2})) \end{aligned}

Therefore, an approximation of the area of the entire region can be found by adding up/accumulating the area of all parabolic sectors of the form above. That is:

\begin{aligned} A &= \int_a^b f(x)dx\\ &\approx \frac{b-a}{3n}(f(x_0)+4f(x_1)+f(x_2))\\ & \ \ \ +\frac{b-a}{3n}(f(x_2)+4f(x_3)+f(x_4))\\ & \ \ \ \ \ \ +\ . \ . \ . \ \\ & \ \ \ \ \ \ \ \ \ +\frac{b-a}{3n}(f(x_{n-2})+4f(x_{n-1})+f(x_{n}))\\ &= \frac{b-a}{3n}(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+ \ . \ . \ . \ +2f(x_{n-2})+4f(x_{n-1})+f(x_n)) \end{aligned}

Note the 1-4-2--2-4-1 pattern for Simpson's Rule.

Also note that the number of subintervals, $n$ , must be even to apply Simpson's Rule because each parabolic arc uses two subintervals.

Example 2

Use Simpson's Rule with $n=4$ to estimate $\int_0^2 5x^4 \, dx.$ Compare the estimate with the exact value.

View Answer

$\frac{385}{12} \approx 32.083$ (compared to $32$ ).

Error Bounds

When using approximation methods, it is important to determine how accurate the results are. While we may not know the exact error, we can compute an upper bound for the error.

Error Formulas

Example 3

Find an upper bound for the error in estimating $\int_0^2 5x^4 \, dx$ using Simpson's Rule with $n=4$ . What value of $n$ should we pick so that the error is within $0.001$ of the true value?

View Answer

$|E_S| \leq \frac{1}{12}$ ; $n \geq 4 \sqrt[4]{\frac{2 \cdot 5^3}{3}} \approx 12.086$ .

Trigonometric Substitution

Improper Integrals