Chapter 9, Engineering Applications of Integration Video Solutions, Engineering Mathematics Through Applications

Problem 1

[mechanics] A force, $F$, acting on a particle varies with time, $t$, according to the table below.$$
\begin{array}{|l|lllllll|}
\hline t(\mathrm{~s}) & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 \\
\hline F(\mathrm{~N}) & 3.2 & 5.6 & 7.0 & 7.7 & 8.4 & 9.9 & 11.6 \\
\hline
\end{array}
$$The impulse of this force is given by
$$
\int_{0}^{3} F \mathrm{~d} t
$$
Find an approximate value for the impulse.

Katie Mcalpine

Numerade Educator

14:10

Problem 2

Use the trapezium rule with four equal intervals to find the approximate values of
$\mathbf{a} \int_{0}^{1} e^{-x^{2}} \mathrm{~d} x$
b $\int_{0}^{\pi / 2} \sqrt{\cos (x)} \mathrm{d} x$

Oswaldo Jiménez

Numerade Educator

01:41

Problem 3

i Apply the trapezium rule to find the approximate value of
$$
\int_{0}^{1} x^{3} \mathrm{~d} x
$$
with a four equal intervals
b eight equal intervals.
ii Find the exact value of $\int_{0}^{1} x^{3} \mathrm{~d} x$.
iii Determine the percentage error between the exact value and the estimated values found in $\mathrm{i}$.
iv Comment upon your results to iii.

K B

Numerade Educator

04:45

Problem 4

[mechanics] The velocity, $v$, of a model for a new ship is tested in an experiment for 5 seconds and has the values shown in the table below.$$
\begin{array}{l|llllll}
\hline t \text { (s) } & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline v(\mathrm{~m} / \mathrm{s}) & 2.10 & 9.56 & 11.36 & 12.08 & 12.98 & 13.76
\end{array}
$$The distance travelled is given by
$$
\int_{0}^{5} v \mathrm{~d} t
$$
Estimate the distance.

Christian Harris

Numerade Educator

04:21

Problem 5

[fluid mechanics] A river is $15 \mathrm{~m}$ wide. The depth of river is found in metres from one side of the embankment to the other with the results shown in the table below.$$
\begin{array}{|l|llllll|}
\hline \text { Dist. (m) } & 0 & 1.5 & 3.0 & 4.5 & 6.0 & 7.5 \\
\text { Depth }(\mathrm{m}) & 0 & 1.04 & 1.65 & 3.10 & 4.66 & 4.12 \\
\hline \text { Dist. }(\mathrm{m}) & 9.0 & 10.5 & 12.0 & 13.5 & 15.0 & \\
\text { Depth }(\mathrm{m}) & 3.21 & 2.33 & 1.78 & 0.76 & 0 & \\
\hline
\end{array}
$$Given that the velocity of water is $2.05 \mathrm{~m} / \mathrm{s}$, find the approximate number of cubic metres of water flowing down the river per second.
[Hint: Volume of flow per second = cross-sectional area $\times$ velocity.]

Averell Hause

Carnegie Mellon University

01:44

Problem 6

the values at intervals of $0.1 \mathrm{~s}$ shown in the table below.$$
\begin{array}{|c|ccccccc}
\hline t(\mathrm{~s}) & 0 & 0.1 & 0.2 & 0.3 & 0.4 & 0.5 & 0.6 \\
\hline v(\text { volts }) & 4 & 3.92 & 3.86 & 3.77 & 3.61 & 3.52 & 3.41 \\
\hline
\end{array}
$$

Erika Bustos

Numerade Educator