Single Variable Calculus, Chapter 5, 5.2, Section 5.2, Problem 4

a.) Suppose that $\displaystyle f(x) = \sin x, 0 \leq x \leq 3 \frac{\pi}{2}$, find the Riemann sum with $n = 6$, taking the sample points to be right end points. What does Riemann sum represents? Illustrate with a diagram.

With $n = 6$, we divide the interval $\displaystyle (0, 3 \frac{\pi}{2})$ into 6 rectangles with widths

$\displaystyle \Delta x = \frac{\displaystyle \frac{3 \pi}{2} - 0}{6} = \frac{\pi}{4}$ at $\displaystyle x = 0, x = \frac{\pi}{4}, x = \frac{\pi}{2}, x = \frac{3 \pi}{4}, x = \pi, x = \frac{5 \pi}{4}$ and $\displaystyle x = \frac{3 \pi}{2}$.

Evaluating $f(x)$ on the right end points (starting from $\displaystyle x = \frac{\pi}{4}$)


$
\begin{array}{|c|c|}
\hline\\
x & f(x) = x^2 - 2x \\
\hline\\
\displaystyle \frac{\pi}{4} & \displaystyle \frac{\sqrt{2}}{2} \\
\hline\\
\displaystyle \frac{\pi}{2} & 1 \\
\hline\\
\displaystyle \frac{3 \pi}{4} & \displaystyle \frac{\sqrt{2}}{2} \\
\hline\\
\pi & 0 \\
\hline\\
\displaystyle \frac{5 \pi}{4} & \displaystyle - \frac{\sqrt{2}}{2} \\
\hline\\
\displaystyle \frac{3 \pi}{2} & -1\\
\hline
\end{array}
$


Now, the total area of the rectangle is..

$\displaystyle \frac{\pi}{4} \left[ \frac{\sqrt{2}}{2} + 1 + \frac{\sqrt{2}}{2} + 0 - \frac{\sqrt{2}}{2} - 1 \right] = 0.5554 \text{ units}^2$

The Riemann sum represents ab estimate of the area between the curve and the $x$-axis. Although in some cases, some areas result to a negative value because some rectangles are located below the $x$-axis. With this, you have to take the absolute values of such areas to get the actual area.

b.) Repeat part (a) with midpoints as sample points.

By using midpoints,

Evaluating $f(x)$ at midpoints


$
\begin{array}{|c|c|}
\hline\\
x & f(x) = \sin x \\
\hline\\
\displaystyle \frac{\displaystyle 0 + \frac{\pi}{4}}{2} = \frac{\pi}{8}
& 0.38327 \\ \hline\\
\displaystyle \frac{\displaystyle \frac{\pi }{4} + \frac{\pi}{2} }{2} = \frac{3 \pi}{8}
& 0.9239 \\
\hline\\
\displaystyle \frac{\displaystyle \frac{\pi}{2} + \frac{3 \pi}{4} }{2} = \frac{5 \pi }{8}
& 0.9239 \\
\hline\\
\displaystyle \frac{\displaystyle \frac{3 \pi}{4} + \pi}{2} = \frac{7 \pi}{8}
& 0.3827 \\
\hline\\
\displaystyle \frac{\displaystyle \pi + \frac{5 \pi}{4}}{2} = \frac{9 \pi}{8}
& -0.3827 \\
\hline\\
\displaystyle \frac{\displaystyle \frac{5 \pi}{4} + \frac{3 \pi}{2}}{2} = \frac{11 \pi}{8}
& -0.9239\\
\hline
\end{array}
$


Now, the total area of the rectangle is ..

$\displaystyle \frac{\pi}{4} [0.3827 + 0.9239 + 0.9239 + 0.3827 - 0.3827 - 0.9239] = 1.0262 \text{ units}^2$