009B Sample Midterm 2, Problem 1

Consider the region ${\displaystyle S}$ bounded by ${\displaystyle x=1,x=5,y={\frac {1}{x^{2}}}}$  and the ${\displaystyle x}$-axis.

a) Use four rectangles and a Riemann sum to approximate the area of the region ${\displaystyle S}$. Sketch the region ${\displaystyle S}$ and the rectangles and
indicate whether your rectangles overestimate or underestimate the area of ${\displaystyle S}$.
b) Find an expression for the area of the region ${\displaystyle S}$ as a limit. Do not evaluate the limit.

Approximation of integral with left endpoints is an overestimate.
Foundations:
Recall:
1. The height of each rectangle in the left-hand Riemann sum is given by
choosing the left endpoint of the interval.
2. The height of each rectangle in the right-hand Riemann sum is given by
choosing the right endpoint of the interval.
3. See the page on Riemann Sums for more information.

Solution:

(a)

Step 1:
Let ${\displaystyle f(x)={\frac {1}{x^{2}}}.}$ Since our interval is ${\displaystyle [1,5]}$ and we are using ${\displaystyle 4}$ rectangles, each rectangle has width ${\displaystyle 1.}$ Since the problem doesn't specify, we can choose either right- or left-endpoints. Choosing left-endpoints, the Riemann sum is
${\displaystyle 1\cdot (f(1)+f(2)+f(3)+f(4)).}$
Step 2:
Thus, the left-endpoint Riemann sum is
${\displaystyle {\begin{array}{rcl}\displaystyle {1\cdot (f(1)+f(2)+f(3)+f(4))}&=&\displaystyle {{\bigg (}1+{\frac {1}{4}}+{\frac {1}{9}}+{1}{16}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {205}{144}}.}\\\end{array}}}$
The left-endpoint Riemann sum overestimates the area of ${\displaystyle S.}$

(b)

Step 1:
Let ${\displaystyle n}$ be the number of rectangles used in the left-endpoint Riemann sum for ${\displaystyle f(x)={\frac {1}{x^{2}}}.}$
The width of each rectangle is
${\displaystyle \Delta x={\frac {5-1}{n}}={\frac {4}{n}}.}$
Step 2:
So, the left-endpoint Riemann sum is
${\displaystyle \Delta x{\bigg (}f(1)+f{\bigg (}1+{\frac {4}{n}}{\bigg )}+f{\bigg (}1+2{\frac {4}{n}}{\bigg )}+\ldots +f{\bigg (}1+(n-1){\frac {4}{n}}{\bigg )}{\bigg )}.}$
Now, we let ${\displaystyle n}$ go to infinity to get a limit.
So, the area of ${\displaystyle S}$ is equal to
${\displaystyle \lim _{n\to \infty }{\frac {4}{n}}\sum _{i=0}^{n-1}f{\bigg (}1+i{\frac {4}{n}}{\bigg )}\,=\,\lim _{n\to \infty }{\frac {4}{n}}\sum _{i=0}^{n-1}{\bigg (}1+i{\frac {4}{n}}{\bigg )}^{-2}.}$
(a) The left-endpoint Riemann sum is ${\displaystyle {\frac {205}{144}}}$, which overestimates the area of ${\displaystyle S}$.
${\displaystyle \lim _{n\to \infty }{\frac {4}{n}}\sum _{i=0}^{n-1}{\bigg (}1+i{\frac {4}{n}}{\bigg )}^{-2}}$