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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 4.1 Areas & Distance

Subsection 4.1.1 Before Class

https://mymedia.ou.edu/media/4.1-1/1_3gm43l27
Figure 30. Pre-Class Video 1
https://mymedia.ou.edu/media/4.1-2/1_2sv8x0jk
Figure 31. Pre-Class Video 2

Subsubsection 4.1.1.1 The Area Problem

Example 4.1.1.
Consider the function \(y=x^2\) on the interval \([0,1]\text{.}\)
  1. Use four rectangles to approximate the area under the parabola \(y=x^2\) from \(x=0\) to \(x=1\text{;}\) start from the left endpoint, \(x=0\text{.}\)
    Figure 32. Left-rectangle approximiation for \(y=x^2\text{.}\) Slide \(n\) to see different approximations (hide the menu by pressing the arrows).
  2. Now start from the right endpoint, \(x=1\text{,}\) and use four rectangles to approximate the area under the parabola.
    Figure 33. Right-rectangle approximiation for \(y=x^2\text{.}\) Slide \(n\) to see different approximations (hide the menu by pressing the arrows).
Solution.
  1. \(\displaystyle A\approx 0.21875\)
  2. \(\displaystyle A\approx 0.46875\)
Example 4.1.2.
Now use eight rectangles to approximate the area under \(y=x^2\) from \(x=0\) to \(x=1\text{,}\) using both left and right endpoints. Compare your estimates to the previous two calculations.
Solution.
\(L_8 = 0.2734375\) and \(R_8 = 0.3984375\)
Question 4.1.3.
What do you think would happen to our approximation of area if we used more rectangles?
Solution.
We would get a better approximation.
Generically, the process we used above has the following steps:
Approximating Area Under a Curve.
  1. Determine the length of the interval
  2. Divide the interval into \(n\) equal subintervals
  3. Find the area of each subinterval’s rectangle
  4. Sum the areas of the rectangles
Question 4.1.4.
  1. If the interval we are concerned about is \([a,b]\text{,}\) and we are using \(n\) rectangles to approximate the area, how could we write the base length of each individual rectangle? Denote the length by \(\Delta x\)
  2. Let \(a=x_0\text{.}\) How could we express \(x_1\text{?}\) How about \(x_2\text{?}\) What about a generic \(x_i\text{,}\) where \(1\leq i\leq n\text{?}\)
  3. Let \(f(x_i)\) be the function value at \(x_i\text{.}\) Using the previous two exercises, write an expression for \(S_n\text{,}\) the sum of areas of our approximating rectangles.
Solution.
  1. \(\displaystyle \Delta x = \dfrac{b-a}{n}\)
  2. \begin{align*} x_1 &= a+\Delta x\\ x_2 &= a+2\Delta x\\ x_i &= a+i\Delta x \end{align*}
  3. \(\displaystyle S_n\approx f(a+i\Delta x)\cdot \Delta x\)
From this, we can generalize to two special situations.
Left Rectangle Approximation.
The left rectangle approximation for the area under the curve \(f(x)\) on the interval \([a,b]\) (using \(n\) rectangles) is given by
\begin{equation*} f(x_0)\Delta x + f(x_1)\Delta x + \cdots + f(x_{n-2})\Delta x + f(x_{n-1})\Delta x \end{equation*}
or, in sigma notation,
\begin{equation*} \ds \sum_{n=0}^{n-1} f(x_i)\Delta x \end{equation*}
The \(n\)th left rectangle approximation is denoted \(L_n\text{.}\)
Right Rectangle Approximation.
The right rectangle approximation for the area under the curve \(f(x)\) on the interval \([a,b]\) (using \(n\) rectangles) is given by
\begin{equation*} f(x_1)\Delta x + f(x_2)\Delta x + \cdots + f(x_{n-1})\Delta x + f(x_n)\Delta x \end{equation*}
or, in sigma notation,
\begin{equation*} \ds \sum_{n=1}^{n} f(x_i)\Delta x \end{equation*}
The \(n\)th right rectangle approximation is denoted \(R_n\text{.}\)

Subsubsection 4.1.1.2 Sigma Notation

Sigma notation is a shorthand way to write sums. For example, for the left-rectangle approximation \(L_4\text{,}\) instead of writing
\begin{equation*} L_4 = f(x_0)\Delta x + f(x_1)\Delta x + f(x_2)\Delta x + f(x_3)\Delta x \end{equation*}
we would write
\begin{equation*} \ds L_4 = \sum_{n=0}^3 f(x_i)\Delta x \end{equation*}
Here, \(i=0\) refers to the starting index, and \(n=3\) is called the ending index.

Subsection 4.1.2 Pre-Class Activities

Example 4.1.5.

Consider \(f(x) = \dfrac{1}{x}\text{.}\)
  1. Carefully sketch the graph of \(f(x)\) on the interval \([1,4]\text{.}\)
  2. Find \(R_3\text{,}\) and draw the rectangles on your sketch.
  3. Express the sum from (b) in sigma notation.
  4. Find \(L_3\text{,}\) and draw the rectangles on your sketch.
  5. Express the sum from (d) in sigma notation.
Solution.
  1. The graph of \(f(x) = \dfrac{1}{x}\) on the interval \([1,4]\)
  2. Figure 34. Right-rectangle approximiation for \(y=\dfrac{1}{x}\text{.}\) Slide \(n\) to see different approximations (hide the menu by pressing the arrows).
    \(R_3 = 1.083\)
  3. \(\displaystyle R_3=\ds \sum_{i=1}^3 f(x_i)\Delta x\)
  4. Figure 35. Left-rectangle approximiation for \(y=\dfrac{1}{x}\text{.}\) Slide \(n\) to see different approximations (hide the menu by pressing the arrows).
    \(L_3 = 1.833\)
  5. \(\displaystyle L_3=\ds \sum_{i=0}^2 f(x_i)\Delta x\)

Example 4.1.6.

Again, consider \(f(x) = \dfrac{1}{x}\) on \([1,4]\)
  1. Resketch the graph of \(f(x)\text{.}\)
  2. Find \(R_6\) and include the rectangles on your sketch from (a). How do these rectangles compare to the \(R_3\) rectangles?
  3. Find \(L_6\) and include the rectangles on your sketch from (a). How do these rectangles compare to the \(L_3\) rectangles?
Solution.
  1. The graph of \(f(x) = \dfrac{1}{x}\) on the interval \([1,4]\)
  2. Figure 36. Right-rectangle approximiation for \(y=\dfrac{1}{x}\text{.}\) Slide \(n\) to see different approximations (hide the menu by pressing the arrows).
    \(R_6 = 1.218\)
  3. Figure 37. Left-rectangle approximiation for \(y=\dfrac{1}{x}\text{.}\) Slide \(n\) to see different approximations (hide the menu by pressing the arrows).
    \(L_6 = 1.593\)

Subsection 4.1.3 In Class

Example 4.1.7.

For \(f(x) = x^2\) on \([0,1]\text{,}\) show that the sum of the areas of the upper approximating rectangles approaches \(\dfrac{1}{3}\text{,}\) i.e. \(\ds \lim_{n\to\infty} R_n = \dfrac{1}{3}\text{.}\)
Solution.
First,
\begin{equation*} \Delta x = \dfrac{b-a}{n}\dfrac{1}{n} \end{equation*}
Now, to find \(f(x_i)\text{,}\) write \(x_i\) as
\begin{equation*} x_i= i\Delta x = \dfrac{i}{n} \end{equation*}
This means that
\begin{equation*} f(x_i) = \lrpar{\dfrac{i}{n}}^2 = \dfrac{i^2}{n^2} \end{equation*}
So, the area of a single rectangle can be expressed as
\begin{equation*} f(x_i)\Delta x = \lrpar{\dfrac{i}{n}}^2\cdot \dfrac{i}{n} = \dfrac{i^2}{n^3} \end{equation*}
To find the area of a finite number of rectangles, we add these single rectangles:
\begin{equation*} R_n = \ds \sum_{i=1}^n f(x_i)\Delta x = \sum_{i=1}^n \dfrac{i^2}{n^3} \end{equation*}
Since the summation is over the variable \(i\text{,}\) we can use the formula \(\ds \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\) to get
\begin{equation*} R_n = \ds \sum_{i=1}^n \dfrac{i^2}{n^3} = \dfrac{1}{n^3}\cdot \dfrac{n(n+1)(2n+1)}{6} \end{equation*}
Now, the area under the curve is found by taking infinitely many rectangles in the approximation. The way we can do that is by taking a limit:
\begin{equation*} A = \ds \lim_{n\to \infty} R_n = \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x = \lim_{n\to\infty} \dfrac{1}{n^3}\cdot \dfrac{n(n+1)(2n+1)}{6} = \dfrac{1}{3} \end{equation*}
So we conclude the area is precisely \(\dfrac{1}{3}\)

Definition 4.1.8. Area Under a Curve.

The area of the region \(S\) that lies under the graph of the continuous function \(f\) is the limit of the sum of the areas of approximating rectangles:
\begin{equation*} A = \lim_{n\to\infty} L_n = \lim_{n\to\infty} \sum_{i=0}^{n-1} f(x_i)\Delta x \end{equation*}
or
\begin{equation*} A = \lim_{n\to\infty} R_n = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\Delta x \end{equation*}

Example 4.1.9.

Let \(A\) be the area of the region that lies under the graph of \(f(x) = x^3\) between \(x = 1\) and \(x=b\text{,}\) where \(1\leq b\leq 3\)
  1. Using left endpoints, find an expression for \(A\) as a limit; do not evaluate the limit.
  2. Estimate the area when \(b=3\) by taking sample points to be midpoints and using four subintervals.
Solution.
  1. \(\displaystyle A = \ds \lim_{n\to \infty} L_n = \lim_{n\to\infty} \sum_{i=0}^{n-1} \lrpar{1 + \dfrac{(b-1)i}{n}}^3\cdot \dfrac{b-1}{n}\)
  2. \(\displaystyle L_4 = 14\)

Example 4.1.10.

The expression \(\ds \lim_{n\to \infty} \sum_{i=1}^n \dfrac{3}{n} \sqrt{1+\dfrac{3i}{n}}\) gives the area of a region; describe such a region. Do not try to evaluate the limit.
Solution.
One possible description is the function \(f(x) = \sqrt{1+x}\) on the interval \([0,3]\)

Subsubsection 4.1.3.1 The Distance Problem

Given constant velocity, we can use the formula distance\(=\)velocity\(\times\) time to calculate the distance an object travels over a certain period of time.
Example 4.1.11.
Consider a moving car, with increasing velocity. The velocity was measured every two seconds, and the results collected in the table below.
Time (sec) 0 2 4 6 8 10
Velocity (ft/s) 20 30 38 44 48 50
  1. Find an upper estimate for the distance the car traveled in 10 seconds.
  2. Find a lower estimate for the distance the car traveled in 10 seconds.
Solution.
  1. 420 feet
  2. 360 feet
Example 4.1.12.
Roger is training for a marathon. His friend Jeff rides behind him on a bicycle and clocks his speed every 15 minutes. Roger starts out strong, but stops after an hour and a half.
Time elapsed(min) 0 15 30 45 60 75 90
Speed (mph) 12 11 10 10 8 7 0
Give upper and lower rectangle estimates for the distance Roger ran.
Solution.
The upper distance is 14.5 miles and the lower distance is 11.5 miles.

Subsection 4.1.4 After Class Activities

Example 4.1.13.

We showed that, for \(f(x) =x^2\) on \([0,1]\text{,}\) \(\ds \lim_{n\to\infty} R_n = \dfrac{1}{3}\) was the area under the curve. Show that \(\ds \lim_{n\to\infty} L_n = \dfrac{1}{3}\) as well. Use the fact that \(\ds \sum_{i=0}^{n-1} f(x_i)\Delta x = \sum_{i=1}^{n} f(x_{i-1})\Delta x\) and that \(\ds \sum_{i=1}^n i =\dfrac{n(n+1)}{2}\text{.}\)
Solution.
First,
\begin{equation*} \Delta x = \dfrac{b-a}{n}\dfrac{1}{n} \end{equation*}
Now, to find \(f(x_i)\text{,}\) write \(x_i\) as
\begin{equation*} x_i= i\Delta x = \dfrac{i}{n} \end{equation*}
This means that
\begin{equation*} f(x_i) = \lrpar{\dfrac{i}{n}}^2 = \dfrac{i^2}{n^2} \end{equation*}
So, the area of a single rectangle can be expressed as
\begin{equation*} f(x_i)\Delta x = \lrpar{\dfrac{i}{n}}^2\cdot \dfrac{i}{n} = \dfrac{i^2}{n^3} \end{equation*}
To find the area of a finite number of rectangles, we add these single rectangles:
\begin{equation*} L_n = \ds \sum_{i=0}^{n-1} f(x_i)\Delta x = \sum_{i=1}^n \dfrac{(i-1)^2}{n^3} \end{equation*}
Since the summation is over the variable \(i\text{,}\) we can use the formula \(\ds \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\) to get
\begin{equation*} L_n = \ds \sum_{i=1}^n \dfrac{(i-1)^2}{n^3} = \dfrac{1}{n^3}\sum_{i=1}^n i^2-2i+1 = \dfrac{1}{n^3}\cdot \lrpar{\dfrac{n(n+1)(2n+1)}{6}- n(n+1) + 1} \end{equation*}
Now, the area under the curve is found by taking infinitely many rectangles in the approximation. The way we can do that is by taking a limit:
\begin{equation*} A = \ds \lim_{n\to \infty} L_n = \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\Delta x = \lim_{n\to\infty} \dfrac{1}{n^3}\cdot \lrpar{\dfrac{n(n+1)(2n+1)}{6}- n(n+1) + 1} = \dfrac{1}{3} \end{equation*}
So we conclude the area is precisely \(\dfrac{1}{3}\)

Example 4.1.14.

A stock car prototype is being tested. Over the course of one minute, its speed was recorded in ten-second intervals.
Time (sec) 0 10 20 30 40 50 60
Velocity (mi/h) 182.9 168 106.6 99.8 124.5 176.1 175.6
Find upper and lower estimates for the distance traveled by the car in one minute.
Solution.
The upper estimate is 2.59 miles and the lower estimate is 2.15 miles

Example 4.1.15.

If a curve measures the rate of change of some function, what is the explicit connection between the area under the curve on a given interval and the total change of the function on that interval?
Solution.
The net area between the rate of change curve and the axis gives the total change of the function on the interval.
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