Question

1. If, use a graph to find an upper bound for
2. Use Simpson’s Rule withto approximateand use part(a) to estimate the error.
3. How large should n be to guarantee that the size of the error in usingis less than 0.00001?

Short answer

1. The fourth derivative ofis\({f^{(4)}}(x) = \sin x\left( {6{{\cos }^2}x + 1} \right)\cos (\sin x) + \sin (\sin x)\left( { - 3{{\sin }^2}x + {{\cos }^4}x + 4{{\cos }^2}x} \right)\)
2. to approximate the integral is\(\int_0^\pi {\sin } (\sin x)dx \approx 1.786721\)

The error estimation is

1. The large error size is

Step-by-step solution

Step 1: Solution for an upper bound for

By using online algebra system, the fourth derivative of

\({f^{(4)}}(x) = \sin x\left( {6{{\cos }^2}x + 1} \right)\cos (\sin x) + \sin (\sin x)\left( { - 3{{\sin }^2}x + {{\cos }^4}x + 4{{\cos }^2}x} \right)\)

The graph of the system is

Step 2: Solution of  to approximate  by using Simpson’s rule

Let, consider the interval, there are 10 subintervals to consider

The length of each interval

\(\begin{aligned}{l}\Delta x = \frac{{b - a}}{n}\\ = \frac{{\pi - 0}}{{10}}\\ = \frac{\pi }{{10}}\end{aligned}\)

to approximate the integral is

\(\begin{aligned}{l}\int_0^\pi {\sin } (\sin x)dx \approx {S_{10}}\\ \approx 1.786721\end{aligned}\)

The error estimation is

Step 3: Explanation of error size in using  is less than 0.00001

The errorby using Simpson’s rule\(\left| {{E_S}} \right| \le \frac{{K{{(b - a)}^5}}}{{180{n^4}}}\)

\(\begin{aligned}{l}\left| {{E_S}} \right| < 0.00001\\\frac{{3.8{{(\pi - 0)}^5}}}{{180{n^4}}} < 0.00001\\{n^4} > \frac{{3.8{\pi ^5}}}{{180(0.00001)}}\\ \approx 646,041.557\\n > 28.35\end{aligned}\)

The large error size is