Question

Shown is the graph of traffic on an Internet service provider's\({\bf{T1}}\)data line from midnight to 8:00 AM. \(D\)is the data throughput, measured in megabits per second. Use Simpson's Rule to estimate the total amount of data transmitted during that time period.

Short answer

Amount of data transmitted is\( \approx 15636\)megabits.

Step-by-step solution

Simpson's Rule

By the formula\(\int_a^b f (x)dx \approx {S_n} = \frac{{\Delta x}}{3}\left( {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 4f\left( {{x_3}} \right) + L + 2f\left( {{x_{n - 2}}} \right) + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right)\)

Where\({\bf{x}}\)is even and\(\Delta x = \frac{{b - a}}{n}\).

If\({E_s}\)is the error involved in the Simpson's rule then\(\left| {{E_s}} \right| \le \frac{{k{{(b - a)}^5}}}{{180{n^4}}}\).

Apply rule

By the graph time interval is\((0,8)\). Divide this interval in to 8 subintervals then\(\Delta t = 1\), then the total amount of data transmitted during the time interval\((0,8)\) can be estimate, by Simpson's rule as follows

Amount \( = \int_0^8 D (t)dt \times 3600\) (Since time is in hours and 1 hour\( = 3600\)seconds)

\(\begin{aligned}{c}\int_0^8 D (t)dt \approx {S_8} = \frac{{\Delta t}}{3}(D(0) +\( = 4.34\bar 3 \times 3600\) 4D(1) + 2D(2) + 4D(3) + 2D(4) + 4D(5) + 2D(6) + 4D(7) + D(8))\\ = \frac{1}{3}((0.35) + 4(0.32) + 2(0.41) + 4(0.5) + 2(0.51) + 4(0.56) + 2(0.56) + 4(0.83) + 0.88)\\ = \frac{{13.03}}{3}\\ = 4.34\bar 3\end{aligned}\)

Therefore, the total amount of data transmitted megabits\( \approx 15636\) mega bits.