5.1: Area Problem

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Throughout this module, if something does not exist, write DNE in the answer box.

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Problems

If $f(x)$ is a continuous function on an interval $[a,b]$ , then $L_n$ (resp. $R_n$ ) will denote the left-endpoint (resp. right-endpoint) approximation for the area under $f(x)$ on $[a,b]$ with $n$ rectangles.

Consider $f(x) = 2x$ on $[1,2]$ .

Calculate $L_2$ and $R_2$ .
We need to do base times height and add up the areas. Notice that since we are using $2$ rectangles, our base is $\Delta x = \frac {b-a}{n} = \answer {1/2}.$ For the heights, we can plug in the left endpoint of our broken intervals into $f(x)$ . Our two intervals are $[1,1.5]$ and $[1.5,2]$ , so the heights are $f(1) = \answer {2}$ and $f(1.5) = \answer {3}$ . Therefore, the approximation $L_2 = \Delta x[f(1)+f(1.5)] = \answer {5/2}.$ For $R_2$ , our heights are given by the function at the right endpoints, so $f(1.5) = \answer {3}$ and $f(2) = \answer {4}$ . Therefore, $R_2 = \Delta x[f(1.5)+f(2)] = \answer {7/2}.$
Calculate $L_4$ and $R_4$ .
We need to repeat the previous part’s steps, only with four rectangles instead of $2$ . Now, the base has length $\Delta x = \answer {1/4}$ . The heights for $L_4$ , in increasing order of $x$ values, are $f(\answer {1}) = \answer {2},\quad f(\answer {5/4}) = \answer {5/2},\quad f(\answer {3/2}) = \answer {3},\quad f(\answer {7/4}) = \answer {7/2}.$ This gives us an approximation of $L_4 = \answer {2.75}$ . Similarly, for $R_4$ , the heights, in increasing order of $x$ -values, are $f(\answer {5/4}) = \answer {5/2},\quad f(\answer {3/2}) = \answer {3},\quad f(\answer {7/4}) = \answer {7/2},\quad f(\answer {2}) = \answer {4}.$ This gives us an approximation of $R_4 = \answer {3.25}$ .
Calculate $R_n$ in terms of $n$ . You may find the identity $\sum _{j=1}^n j = 1+2+3+\cdots +n = \frac {n(n+1)}{2}$ useful.
We proceed as before. If we use $n$ rectangles, the length of the base will be $\Delta x = \answer {1/n}$ (it’s ok to have things in terms of $n$ ). Now the heights. The first subinterval is $[1,1+\Delta x]$ , which means the first height is $f(1+\Delta x) = f\left (1+\frac {1}{n}\right ) = \answer {2+\frac {2}{n}}.$ The second subinterval will be $[1+\Delta x,1+2\Delta x]$ , so the second height is $f(1+2\Delta x) = f(1+\frac {2}{n}) = \answer {2+\frac {4}{n}}.$ Similarly the third height will be $\answer {2+\frac {6}{n}}$ . We will continue until the last height, which is $f(2) = \answer {4}$ . However, it will be useful to recognize that the last height comes from starting at $1$ and moving over $n$ $\Delta x$ ’s, i.e. writing the last height as $f(1+n\frac {1}{n}) = 2+\frac {2n}{n},$ which would simplify to $4$ . The reason for writing it in this form will be apparent in a moment. Now, we get that
$\begin{eqnarray*} R_n &=& \Delta x[f(1+\Delta x)+f(1+2\Delta x)+\cdots + f(1+n\Delta x)] \\ &=& \frac{1}{n}\left[\left(2+\frac{2}{n}\right) + \left(2+\frac{4}{n}\right)+\left(2+\frac{6}{n}\right)+\cdots+\left(2+\frac{2n}{n}\right)\right]. \end{eqnarray*}$
How can we simplify this? Let’s simplify the inside of the brackets by grouping the $2$ ’s and the rest. How many $2$ ’s are there? $\answer {n}$ . Therefore, adding all the $2$ ’s gives $2n$ , and the sum above simplifies to $\frac {1}{n}\left [2n+\frac {2}{n} + \frac {4}{n}+\frac {6}{n}+\cdots +\frac {2n}{n}\right ].$ Out of the remaining terms, we can factor out a $\frac {1}{n}$ , i.e. $\frac {1}{n}\left [2n+\frac {2}{n} + \frac {4}{n}\frac {6}{n}+\cdots +\frac {2n}{n}\right ] = \frac {1}{n}\left [2n+\frac {2}{n}\left (1+2+3+\cdots +n\right )\right ].$ Now, using the formula provided above, we get that this sum is equal to $\frac {1}{n}\left [2n+\frac {2}{n}\frac {n(n+1)}{2}\right ] = \frac {1}{n}\left [2n+(n+1)\right ] = \frac {3n+1}{n}.$ This is the formula for $R_n$ .
Use the previous part to find the area under $f(x)$ on $[1,2]$ .
We found $R_n = \frac {3n+1}{n}$ . The area will be $\lim _{n \to \infty } R_n$ . Taking a limit of this expression gives $\answer {3}$ , which is the area.

Consider $f(x) = 1-x^2$ on $[0,1]$ .

The approximation $L_3 = \answer {22/27}$ , and $R_3 = \answer {13/27}$ .
Based on the graph of $f(x)$ , we see that $L_3$ will be an
approximation for the area and $R_3$ will be an
approximation for the area.
If $n$ is any number $\geq 1$ , then (in terms of $n$ ) $|R_n - L_n| = \answer {\frac {1}{n}}.$
For this part, notice that when you subtract the two quantities, all the heights inside the interval are going to cancel out, leaving only the left and right endpoints. Therefore, a formula is $|R_n - L_n| = \frac {b-a}{n}|f(b)-f(a)|.$
The smallest $n$ so that $|R_n-L_n|\leq .001$ is $n=\answer {1000}$ .
Use the identity $\displaystyle 1+4+9+\cdots +n^2 = \frac {n(n+1)(2n+1)}{6}$ to find an expression for $R_n$ in terms of $n$ . We get $R_n = \answer {\frac {1}{n}\left (n-\frac {1}{n^2}\cdot \frac {n(n+1)(2n+1)}{6}\right )}.$
The area under the curve on $[0,1]$ is $\lim _{n \to \infty } R_n$ , which in this case turns out to be $\answer {2/3}$ .

Consider $f(x)$ on an interval $[a,b]$ . In which of the following situations will $R_n$ definitely be an over-estimate? Select all that apply.

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