# Single Variable Calculus, Chapter 5, 5.1, Section 5.1, Problem 6

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a.) Graph the function $\displaystyle f(x) = \frac{1}{(1 + x^2)}, -2 \leq x \leq 2$

b.) Estimate the area under the graph of $f$ using four approximating rectangles at sample points.

The width of the rectangle is..

$\displaystyle \Delta x = \frac{2 - (-2)}{4} = 1$

(i) Right endpoints

At right endpoint,

\begin{aligned} R_4 =& \sum \limits_{i = 1}^4 f(xi) \Delta x \\ \\ R_4 =& 1 [f(-1) + f(0) + f(1) + f(2)] \\ \\ R_4 =& [0.5 + 1 + 0.5 + 0.20] \\ \\ R_4 =& 2.2 \end{aligned}

We can see in the graph that at $x < 0$,

(ii) Midpoints

In each case, sketch the curve and the rectangles in part (b)

\begin{aligned} M_4 =& \sum \limits_{i = 1}^4 f(xi) \Delta x \\ \\ M_4 =& 1 [f(-1.5) + f(-0.5) + f(0.5) + f(1.5)] \\ \\ M_4 =& \left[ \frac{4}{13} + \frac{4}{5} + \frac{4}{5} + \frac{4}{13} \right] \\ \\ M_4 =& 2.22 \end{aligned}

We can see the graph that our estimates is a mixed of overestimate and underestimate.

By using eight rectangles, the new width of the rectangle will be.

$\displaystyle \Delta x = \frac{2 - (-2)}{8} = 0.5$

(i) at right endpoint,

\begin{aligned} R_8 =& \sum \limits_{0 = 1}^8 f(xi) \Delta x \\ \\ R_8 =& 0.5 [f(-1.5) + f(-1) + f(-0.5) + f(0) + f(0.5) + f(1) + f(1.5) + f(2)] \\ \\ R_8 =& 0.5 \left[ \frac{4}{13} + \frac{1}{2} + \frac{4}{5} + 1 + \frac{4}{3} + \frac{1}{2} + \frac{4}{3} + \frac{1}{3} \right] \\ \\ R_8 =& 2.2077 \end{aligned}

(ii) at midpoint

\begin{aligned} M_8 =& \sum \limits_{i = 1}^8 (xi) \Delta x \\ \\ M_8 =& 0.5 [f(-1.75) + f(-1.25) + f(-0.75) + f(-0.25) + f(0.25) + f(0.75) + f(1.25) + f(1.75)] \\ \\ M_8 =& 0.5 \left[ \frac{16}{65} + \frac{16}{41} + \frac{16}{25} + \frac{16}{17} + \frac{16}{17} + \frac{16}{25} + \frac{16}{41} + \frac{16}{65} \right] \\ \\ M_8 =& 2.2176 \end{aligned}