5.6 | Numerical Integration


The definite integral of \(f(x)\) on \([a,b]\) is denoted
$$
\int_a^b f(x) \; dx
$$
and informally represents the “area” of the region “bounded” by the graphs of \(y = f(x)\), there vertical lines \(x=a\) and \(x =b\), and the \(x\)-axis.
Right Rectangular Rule If \(f(x)\) is continuous on \([a,b]\),
$$
\int_a^b f(x) \; dx \approx \sum_{k = 1}^n f(a+k \Delta x) \Delta x
$$
and
$$
\int_a^b f(x) \; dx =\lim_{n \rightarrow \infty} \sum_{k = 1}^n f(a+k \Delta x) \Delta x
$$
where \(\Delta x = \frac{b-a}{n}. \)
Left Rectangular Rule If \(f(x)\) is continuous on \([a,b]\),
$$
\int_a^b f(x) \; dx \approx \sum_{k = 1}^n f(a+(k-1) \Delta x) \Delta x
$$
and
$$
\int_a^b f(x) \; dx =\lim_{n \rightarrow \infty} \sum_{k = 1}^n f(a+(k-1) \Delta x) \Delta x
$$
where \(\Delta x = \frac{b-a}{n}. \)
Midpoint Rule If \(f(x)\) is continuous on \([a,b]\),
$$
\int_a^b f(x) \; dx \approx \sum_{k = 1}^n f\left(a+\left(k-\frac{1}{2} \right) \Delta x \right) \Delta x
$$
and
$$
\int_a^b f(x) \; dx =\lim_{n \rightarrow \infty} \sum_{k = 1}^n f\left(a+\left(k-\frac{1}{2}\right) \Delta x \right) \Delta x
$$
where \(\Delta x = \frac{b-a}{n}. \)
Trapezoidal Rule If \(f(x)\) is continuous on \([a,b]\),
$$
\begin{aligned}
\int_a^b f(x) \; dx & \approx \sum_{k = 1}^n \frac{f(a+k \Delta x)+f(a + (k-1) \Delta x)}{2} \Delta x \\
& = \frac{\Delta x}{2} [ f(x_0) + 2 f(x_1) + \cdots + 2 f(x_{n-1}) + f(x_n)],
\end{aligned}
$$
where \(\Delta x = (b-a)/n\) and \(x_i = a+ i \Delta x,\) for \(0 \leq i \leq n\)
and
$$
\int_a^b f(x) \; dx =\lim_{n \rightarrow \infty} \sum_{k = 1}^n \frac{f(a+k \Delta x)+f(a + (k-1) \Delta x)}{2} \Delta x
$$
where \(\Delta x = \frac{b-a}{n}. \)
Error Bound for the Trapezoidal Rule:
If \(f^{\prime \prime}\) is continuous on \([a,b],\) then the error \(E_n\) in approximating \(\int_a^b f(x) \; dx\) by the Trapezoidal Rule satisfies
$$
|E_n| \leq \frac{M(b-a)^3}{12n^2}
$$
where \(M\) is a positive number such that \(|f^{\prime \prime}(x)| \leq M\) for all \(x\) in \([a,b]\).
Simpson’s Rule If \(f(x)\) is continuous on \([a,b]\) and \(n\) is even,
$$
\begin{aligned}
\int_a^b f(x) \; dx & \approx \sum _{k=1}^{n/2} \left( \frac{f(a+(2k-2) \Delta x) +4 f(a+(2 k-1)\Delta x )+f(a+2 k \Delta x )}{3} \right) \Delta x \\
& = \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)]
\end{aligned}
$$
and
$$
\begin{aligned}
\int_a^b f(x) \; dx & =\lim_{n \rightarrow \infty} \sum _{k=1}^{n/2} \left( \frac{f(a+(2k-2) \Delta x) +4 f(a+(2 k-1)\Delta x )+f(a+2 k \Delta x )}{3} \right) \Delta x \\
& =\lim_{n \rightarrow \infty} \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + \cdots + 4f(x_{n-1}) + f(x_n)]
\end{aligned}
$$
where \(\Delta x = \frac{b-a}{n}. \)
Error Bound for Simpson’s Rule
If \(f^{(4)}\) is continuous on \([a,b],\) then the error \(E_n\) in approximating \(\int_a^b f(x) \; dx\) by Simpson’s Rule satisfies
$$
|E_n| \leq \frac{M(b-a)^5}{180n^4}
$$
where \(M\) is a positive number such that \(|f^{(4)}(x)| \leq M\) for all \(x\) in \([a,b].\)

Use the right rectangular rule, left rectangular rule, and midpoint rule to approximate
$$
\int_{-1}^2 x \sin x +2 \; dx
$$
using 4 rectangles.


  • The right rectangular rule (also referred to as a right Riemann sum) with 4 rectangles:
    $$
    \begin{aligned}
    \int_{-1}^2 x \sin x +2 \; dx & \approx \sum_{k = 1}^4 f(-1+k \Delta x) \Delta x = \sum_{k = 1}^4 f\left(-1+k \frac{3}{4} \right) \frac{3}{4} \\
    &= \sum_{k = 1}^4 f\left(\frac{3k-4}{4} \right) \frac{3}{4} = \frac{3}{4}\sum_{k = 1}^4 f\left(\frac{3k-4}{4} \right) \\
    & = \frac{3}{4}\left(f\left(\frac{-1}{4} \right)+ f\left(\frac{1}{2} \right) + f\left(\frac{5}{4} \right) + f\left(2 \right) \right) \\
    & \approx 8.47979
    \end{aligned}
    $$
  • The left rectangular rule (also referred to as a left Riemann sum) with 4 rectangles:
    $$
    \begin{aligned}
    \int_{-1}^2 x \sin x +2 \; dx & \approx \sum_{k = 1}^4 f(-1+(k-1) \Delta x) \Delta x \\
    & = \sum_{k = 1}^4 f\left(-1+(k-1) \frac{3}{4} \right) \frac{3}{4} \\
    &= \sum_{k = 1}^4 f\left(\frac{3k-7}{4} \right) \frac{3}{4} = \frac{3}{4}\sum_{k = 1}^4 f\left(\frac{3k-7}{4} \right) \\
    & = \frac{3}{4}\left(f\left(-1 \right)+ f\left(\frac{-1}{4} \right) + f\left(\frac{1}{2} \right) + f\left(\frac{5}{4} \right) \right) \\
    & \approx 7.74695
    \end{aligned}
    $$
  • The midpoint rule with 4 rectangles:
    $$
    \begin{aligned}
    \int_{-1}^2 x \sin x +2 \; dx & \approx \sum_{k = 1}^4 f(-1+(k-1/2) \Delta x) \Delta x \\
    & = \sum_{k = 1}^4 f\left(-1+(k-1/2) \frac{3}{4} \right) \frac{3}{4} \\
    & = \sum_{k = 1}^4 f\left(\frac{6k-9}{8} \right) \frac{3}{4} = \frac{3}{4}\sum_{k = 1}^4 f\left(\frac{3k-7}{4} \right) \\
    & = \frac{3}{4}\left(f\left(-3/8 \right)+ f\left(\frac{3}{8} \right) + f\left(\frac{9}{8} \right) + f\left(\frac{15}{8} \right) \right) \\
    & \approx 8.00661
    \end{aligned}
    $$
Use the trapezoidal rule and Simpson’s rule to approximate
$$
\int_{-1}^2 x \sin x +2 \; dx
$$
using 4 trapezoids and 2 parabolas, respectively.


  • The trapezoidal rule with 4 trapezoids:
    $$
    \begin{aligned}
    \int_{-1}^2 x \sin x +2 \; dx & \approx \sum_{k = 1}^4 \frac{f\left(-1+k \frac{3}{4}\right)+f \left(-1 + (k-1) \frac{3}{4} \right)}{2} \frac{3}{4} \\
    & = \sum_{k = 1}^4 \frac{f\left(\frac{3k-4}{4}\right)+f \left(\frac{3k-7}{4} \right)}{2} \frac{3}{4} \\
    & = \frac{3}{8}\sum_{k = 1}^4
    \left[f\left(\frac{3k-4}{4}\right)+f \left(\frac{3k-7}{4} \right) \right] \\
    & = \frac{3}{8}\left[f \left(-1 \right)+2f \left(\frac{-1}{4} \right)+2f \left(\frac{1}{2} \right)+2f \left(\frac{5}{4} \right) +f\left(2\right)\right] \\
    & \approx 8.11337
    \end{aligned}
    $$
  • Simpson’s rule with 2 parabolas:
    $$
    \begin{aligned}
    \int_{-1}^2 x \sin x +2 \; dx & \approx \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + f(x_4)] \\
    & = \frac{\frac{3}{4}}{3} \left[f\left(-1+0 \cdot \frac{3}{4} \right) + 4 f\left(-1+1 \cdot \frac{3}{4} \right) + \cdots \right. \\
    & \quad \left. + f\left(-1+4 \cdot \frac{3}{4} \right) \right] \\
    & = \frac{1}{4} \left[f\left(-1\right) + 4 f\left( -\frac{1}{4} \right) + 2f\left(\frac{1}{2}\right) +4f\left(\frac{5}{4}\right) + f\left(2 \right) \right] \\
    & \approx 8.03295
    \end{aligned}
    $$
Use the Trapezoidal Rule to approximate
$$
\int_1^2 \frac{x^2}{4} + 1 \; dx
$$
with \(n = 4\). Then find the exact value of the definite integral and find the actual error.


$$
\begin{aligned}
\int_1^2 \frac{x^2}{4} + 1 \; dx
& = \frac{\Delta x}{2} [ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)] \\
& = \frac{1}{8}[f(1) + 2f(5/4)+2f(3/2)+2f(7/4)+f(2)] \\
& \approx \frac{1}{8} [ 1.25 + 2(1.39) + 2(1.56) + 2(1.76) +2] \\
& \approx 1.58
\end{aligned}
$$

$$
\int_1^2 \frac{x^2}{4} +1 \; dx = \left( \frac{x^3}{4} +x \right) \Big|_1^2 = 19/12 \approx 1.585333
$$

$$
|E_4| = |\text{actual} – \text{approx}| = |19/12 – 1.58| = 0.003
$$

Find a bound on the error in approximating
$$
\int_0^2 x \sqrt{x^2+1} \; dx
$$
with \(n = 4\).


$$
\begin{aligned}
f'(x) & = \frac{2 x^2+1}{\sqrt{x^2+1}} \\
f^{\prime \prime}(x) & = \frac{2 x^3+3
x}{\left(x^2+1\right)^{3/2}}\\
f^{\prime \prime \prime}(x) & = \frac{3}{\left(x^2+1\right)^{5/2}} \\
f^{(4)}(x) & = -\frac{15
x}{\left(x^2+1\right)^{7/2}}
\end{aligned}
$$

\(M\) is a bound on \(g(x) = |f^{(4)}(x)|\) on the interval \([0,2]\). From the graph of \(g(x)\), we see that we can choose \(M = 3.7\).

NI56E4

The best choice of \(M\) is \(g(1/\sqrt{6})\). Find the extreme values of \(g(x)\) on the interval \([0,2]\). The largest of these, in absolute value, will be the best choice for \(M\).
$$
g'(x) = \frac{15 \left(6
x^2-1\right)}{\left(x^2+1\right)^{9/2}}
$$
$$
g'(x) = 0 \Rightarrow x = \pm 1/\sqrt{6} \Rightarrow x = 1 /\sqrt{6}
$$
By the Extreme Value Theorem, the extreme values of \(g(x)\) occur either at the critical numbers or the endpoints.
$$
\begin{aligned}
g(0) & = 0 \\
g(1/\sqrt{6}) & = -\frac{3240}{343 \sqrt{7}} \approx -3.57028 \\
g(2) & = -\frac{6}{25 \sqrt{5}} \approx -0.107331
\end{aligned}
$$

Therefore, the maximum value for \(|g(x)|\) on \([0,2]\) is \(\frac{3240}{343 \sqrt{7}}\). This is the best choice for \(M\).

Therefore,
$$
|E_n| \leq \frac{M (b-a)^5}{180 n^4} \Rightarrow |E_4| \leq \frac{\frac{3240}{343 \sqrt{7}} 2^5}{180 \cdot 4^4} = \frac{9}{1372 \sqrt{7}} \approx 0.00247936
$$

E 5.6 Exercises