Step 1

Interval \([0,\ 3\ \frac{\pi}{4}]\) is to be devided into 3 sub-intervals whos length is given by

\(h=\ \frac{b\ -\ a}{n}=\frac{\frac{3\pi}{4}\ -\ 0}{3}=\frac{\pi}{4}\)

Sep 2

Therefore, the three sub-intervals are

\(\left[0,\ \frac{\pi}{4}\right],\ \left[\frac{\pi}{4},\ \frac{2\pi}{4}\right]\left[\frac{2\pi}{4},\ \frac{3\pi}{4}\right]\)

Step 3

Area bounded by the graph of f(x) and the axis on the interval using a left end point Riemann sum id given by

\(A_{L}=h\left[|f(0)|\ +\ |f\left(\frac{\pi}{4}\right)|\ +\ |f\left(\frac{2\pi}{4}\right)|\right]\)

\(=\frac{\pi}{4}\left[|\cos\ 0|\ +\ |\frac{\cos\ \pi}{2}|\ +\ |\cos\ \pi|\right]=\frac{\pi}{4}[1\ +\ 0\ +\ 1]=\frac{\pi}{2}\)

Step 4

Area bounded by the graph of f(x) and the axis on the interval using a right end point Riemann sum is given by

\(A_{r}=h\left[|f\left(\frac{\pi}{4}\right)|\ +\ |f\left(\frac{2\pi}{4}\right)|\ +\ |f\left(\frac{3\pi}{4}\right)|\right]\)

\(=\frac{\pi}{4}\left[|\cos\ \times\ \frac{\pi}{2}|\ +\ |\cos\ \pi|\ +\ |\cos\ \times\ \frac{3\pi}{2}|\right]=\frac{\pi}{4}[0\ +\ 1\ +\ 0]=\frac{\pi}{4}\)

Step 5

Area bounded by the graph of f(x) and the axis on the interval using a mid point Riemann sum is given by

\(A_{m}=\left[\left|f\left(\frac{\pi}{8}\right)\right|\ +\ \left|f\left(\frac{3\pi}{8}\right)\right|\ +\ \left|f\left(\frac{5\pi}{8}\right)\right|\right]\)

\(=\frac{\pi}{4}\left[\left|\cos\ \times\ \frac{\pi}{4}\right|\ +\ \left|\cos\ \times\ \frac{3\pi}{4}\right|\ +\ \left|\cos\ \times\ \frac{5\pi}{4}\right|\right]=\frac{\pi}{4}\left[\frac{1}{\sqrt{2}}\ +\ \frac{1}{\sqrt{2}}\ +\ \frac{1}{\sqrt{2}}\right]=\frac{3\pi}{4\sqrt{2}}\)