1.11 Numerical Integration

The absolute error is the difference of the two values; the relative error is the absolute error divided by the exact value; the percent error is one hundred times the relative error.

Relative error: $\approx 0.08147$; absolute error: $0.113$; percent error: $\approx 8.147\%$.

You should have four rectangles in one drawing, and four trapezoids in another.

Midpoint rule:

Trapezoidal rule:

Sketch the second derivative–it's quadratic.

$M=6.25$, $L=2$

You don't have to find the actual, exact maximum the second derivative achieves–you only have to give a reasonable “ceiling" that it never breaks through.

One reasonable answer is $M=3$.

To compute the upper bound on the error, find an upper bound on the fourth derivative of cosine, then use Theorem 1.11.12 in the CLP-2 text.

To find the actual error, you need to find the actual value of $A$.

(a) $\dfrac{\pi^5}{180\cdot8}$ (b) $0$ (c) $0$

Find a function with $f''(x)=3$ for all $x$ in $[0,1]$.

Possible answers: $f(x) = \dfrac{3}{2}x^2+Cx+D$ for any constants $C$, $D$.

You're allowed to use common sense for this one.

my mother

For part (b), consider Question 7.

(a) true (b) false

Draw a sketch.

True. Because $f(x)$ is positive and concave up, the graph of $f(x)$ is always below the top edges of the trapezoids used in the trapezoidal rule.

The error bound for the approximation is given in Theorem 1.11.12 in the CLP-2 text. You want this bound to be zero.

Any polynomial of degree at most 3 will do. For example, $f(x)=5x^3-27$, or $f(x)=x^2$.

Follow the formulas in Equations 1.11.2, 1.11.6, and 1.11.9 in the CLP-2 text.

Midpoint:

$\displaystyle \int_0^{30} \frac{1}{x^3+1}\,\dee{x}\approx\left[\tfrac{1}{\left(2.5\right)^3+1} +\tfrac{1}{\left(7.5\right)^3+1} +\tfrac{1}{\left(12.5\right)^3+1} +\tfrac{1}{\left(17.5\right)^3+1} +\tfrac{1}{\left(22.5\right)^3+1} +\tfrac{1}{\left(27.5\right)^3+1} \right]5$

Trapezoidal:

$\displaystyle\int_0^{30} \frac{1}{x^3+1}\,\dee{x} \approx\left[ \frac{1/2}{0^3+1}+ \frac{1}{5^3+1}+ \frac{1}{10^3+1}+ \frac{1}{15^3+1}+ \frac{1}{20^3+1}+ \frac{1}{25^3+1}+ \frac{1/2}{30^3+1} \right]5$

Simpson's:

$\displaystyle\int_0^{30} \frac{1}{x^3+1}\,\dee{x}\approx \Big[\frac{1}{{0}^3+1}\!+\frac{4}{{5}^3+1}\!+\frac{2}{{10}^3+1}\!+\frac{4}{{15}^3+1}\!+\frac{2}{{20}^3+1}\!+\frac{4}{{25}^3+1}\!+ \frac{1}{{30}^3+1}\Big]\frac{5}{3}$

See Section 1.11.1 in the

CLP-2 text. You should be able to simplify your answer to an exact value (in terms of $\pi$).

$\dfrac{2\pi}{3}$

See Section 1.11.2 in the

CLP-2 text. To set up the volume integral, see Example 1.6.6 in the

CLP-2 text. Note the dimensions given for the cross sections are diameters, not radii.

$1720\pi\approx 5403.5\ {\rm cm}^3$

See Section 1.11.3 in the

CLP-2 text, and compare to Question 13. Note the table gives diameters, not radii.

$\displaystyle\frac{\pi}{12}(16.72)\approx4.377\ {\rm m}^3$

See §1.11.3 in the

CLP-2 text. To set up the volume integral, see Example 1.6.6 in the

CLP-2 text, or Question 14.

Note that the table gives the circumference, not radius, of the tree at a given height.

$\dfrac{12.94}{6\pi} \approx0.6865\ {\rm m}^3$

(a) 363,500 (b) 367,000

(a) $\dfrac{49}{2}$ (b) $\dfrac{77}{3}$

The main step is to find an appropriate value of $M$. It is not necessary to find the smallest possible $M$.

Let $f(x) = \sin(x^2)$. Then $f'(x) = 2x \cos(x^2)$ and

Since $|x^2|\le1$ when $|x|\leq 1$, and $\left|\sin\theta\right|\le1$ and $\left|\cos\theta\right|\leq 1$ for all $\theta$, we have

We can therefore choose $M=6$, and it follows that the error is at most

The main step is to find $M$. This question is unusual in that its wording requires you to find the smallest possible allowed $M$.

$\dfrac{3}{100}$

The main steps in part (b) are to find the smallest possible values of $M$ and $L$.

(a) $\dfrac{1/3}3 \Big( (-3)^5 + 4\Big( \frac13-3 \Big)^5 + 2\Big( \frac23-3 \Big)^5 + 4(-2)^5 + 2\Big( \frac43-3 \Big)^5 + 4\Big( \frac53-3 \Big)^5 + (-1)^5 \Big)$

(b) Simpson's Rule results in a smaller error bound.

As usual, the biggest part of this problem is finding $L$. Don't be thrown off by the error bound being given slightly differently from Theorem 1.11.12 in the CLP-2 text: these expressions are equivalent, since $\De x = \frac{b-a}{n}$.

$\dfrac{8}{15}$

The function $e^{-2x} = \dfrac{1}{e^{2x}}$ is positive and decreasing, so its maximum occurs when $x$ is as small as possible.

$\displaystyle\frac{1}{180\times 3^4} =\frac{1}{14580}$

(a) $\displaystyle T_4 =\frac{1}{4}\left[\left(\frac{1}{2}\times 1\right)+\frac{4}{5}+\frac{2}{3}+ \frac{4}{7}+\left(\frac{1}{2}\times\frac{1}{2}\right)\right]$,

(b) $\displaystyle S_4 =\frac{1}{12}\left[1+\left(4\times\frac{4}{5}\right)+\left(2\times \frac{2}{3}\right)+\left(4\times \frac{4}{7}\right)+\frac{1}{2}\right]$

(c) $\displaystyle\Big|I -S_4\Big| \le \frac{24}{180\times 4^4}=\frac{1}{1920}$

The “best ... approximations that you can" means using the maximum number of intervals, given the information available.

The final sentence in part (b) is just a re-statement of the error bounds we're familiar with from Theorem 1.11.12 in the CLP-2 text. The information $\big|s^{(k)}(x)\big|\le \dfrac{k}{1000}$ gives you values of $M$ and $L$ when you set $k=2$ and $k=4$, respectively.

(a) $T_4=8.03515$, $S_4\approx 8.03509$

(b) $\displaystyle\Big|\int_a^b f(x)\ \dee{x} -T_n\Big| \le \frac{2}{1000}\frac{8^3}{12(4)^2} \le0.00533$, $\displaystyle\Big|\int_a^b f(x)\ \dee{x} -S_n\Big| \le \frac{4}{1000}\frac{8^5}{180(4)^4}\le0.00284$

Set the error bound to be less than $0.001$, then solve for $n$.

Any $n\ge 68$ works.

See Section 1.11.3 in the

CLP-2 text. To set up the volume integral, see Example 1.6.2 in the

CLP-2 text.

Since the cross-sections of the pool are semi-circular disks, a section that is $d$ metres across will have area $\frac{1}{2}\pi\left(\frac{d}{2}\right)^2$ square feet. Based on the drawing, you may assume the very ends of the pool have distance 0 feet across.

$\dfrac{472}{3}\approx 494 \ {\rm ft}^3$

See Example 1.11.14 in the

CLP-2 text.

Don't get caught up in the interpretation of the integral. It's nice to see how integrals can be used, but for this problem, you're still just approximating the integral given, and bounding the error.

When you find the second derivative to bound your error, pay attention to the difference between the integrand and $g(r)$.

(a) $0.025635$ (b) $1.8\times 10^{-5}$

See Example 1.11.15 in the

CLP-2 text. You'll want to use a calculator for the approximation in (a), and for finding the appropriate number of intervals in (b). Remember that Simpson's rule requires an even number of intervals.

(a) $\approx 0.6931698$ (b) $n\ge 12$ with $n$ even

See Example 1.11.15 in the

CLP-2 text.

Rather than calculating the fourth derivative of the integrand, use the graph to find the largest absolute value it attains over our interval.

(a) $0.01345$ (b) $n\ge 28$ with $n$ even

See Example 1.11.14 in the

CLP-2 text.

You'll have to differentiate $f(x)$. To that end, you may also want to review the fundamental theorem of calculus and, in particular, Example 1.3.5 in the

CLP-2 text.

You don't have to find the best possible value for $M$. A reasonable upper bound on $|f''(x)|$ will do.

To have five decimal places of accuracy, your error must be less than 0.000005. This ensures that, if you round your approximation to five decimal places, they will all be correct.

$n\ge 259$

To find the maximum value of $|f''(x)|$, check its critical points and endpoints.

(a) When $0 \leq x \leq1$, then $x^2 \leq 1$ and $x+1 \geq 1$, so $|f''(x)| = \dfrac{x^2}{|x+1|}\leq \dfrac{1}{1}=1$.
(b) $\dfrac{1}{2}$ (c) $n \geq 65$ (d) $n \geq 46$

In using Simpson's rule to approximate $\displaystyle\int_1^{x\vphantom{\frac{1}{2}}} \frac{1}{t}~\dee{t}$ with $n$ intervals, $a=1$, $b=x$, and $\De x = \dfrac{x-1}{n}$.

$\displaystyle\frac{x-1}{12}\left[1+\frac{16}{x+3}+\frac{4}{x+1}+\frac{16}{3x+1}+\frac{1}{x}\right]$

First, we use Simpson's rule with $n=4$ to approximate $\int_1^2 \frac{1}{1+x^2}\,\dee{x}$. The choice of this method (what we're approximating, why $n=4$, etc.) is explained in the solutions–here, we only show that it works.

For ease of notation, define $A=0.321748$.

Now, we bound the error associated with this approximation. Define $N(x) = 24(5x^4-10x^2+1)$ and $D(x) = (x^2+1)^5$, so $N(x)/D(x)$ gives the fourth derivative of $\frac{1}{1+x^2}$. When $1 \le x \le 2$, $|N(x)| \le N(2)=984$ (because $N(x)$ is increasing over that interval) and $|D(x)| \geq D(1) = 2^5$ (because $D(x)$ is also increasing over that interval), so $\left| \ddiff{4}{}{x}\left\{\frac{1}{1+x^2}\right\}\right| = \left| \frac{N(x)}{D(x)}\right| \leq \frac{984}{2^5}=30.75$. Now we find the error bound for Simpson's rule with $L=30.75$, $b=2$, $a=1$, and $n=4$.

So,