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Section 4.2 Approximating Definite Integrals (IN2)

Subsection 4.2.1 Activities

Activity 4.2.1.

Suppose that a person is taking a walk along a long straight path and walks at a constant rate of 3 miles per hour.
(a)
On the left-hand axes provided in FigureΒ 80, sketch a labeled graph of the velocity function \(v(t) = 3\text{.}\)
Figure 80. At left, axes for plotting \(y = v(t)\text{;}\) at right, for plotting \(y = s(t)\text{.}\)
Note that while the scale on the two sets of axes is the same, the units on the right-hand axes differ from those on the left. The right-hand axes will be used in question (d).
(b)
How far did the person travel during the two hours? How is this distance related to the area of a certain region under the graph of \(y = v(t)\text{?}\)
(c)
Find an algebraic formula, \(s(t)\text{,}\) for the position of the person at time \(t\text{,}\) assuming that \(s(0) = 0\text{.}\) Explain your thinking.
(e)
For what values of \(t\) is the position function \(s\) increasing? Explain why this is the case using relevant information about the velocity function \(v\text{.}\)

Activity 4.2.2.

Suppose that a person is walking in such a way that her velocity varies slightly according to the information given in TableΒ 81 and graph given in FigureΒ 82.
\(t\) \(v(t)\)
\(0.00\) \(1.500\)
\(0.25\) \(1.789\)
\(0.50\) \(1.938\)
\(0.75\) \(1.992\)
\(1.00\) \(2.000\)
\(1.25\) \(2.008\)
\(1.50\) \(2.063\)
\(1.75\) \(2.211\)
\(2.00\) \(2.500\)
Table 81. Velocity data for the person walking.
Figure 82. The graph of \(y = v(t)\text{.}\)
(a)
Using the grid, graph, and given data appropriately, estimate the distance traveled by the walker during the two hour interval from \(t = 0\) to \(t = 2\text{.}\) You should use time intervals of width \(\Delta t = 0.5\text{,}\) choosing a way to use the function consistently to determine the height of each rectangle in order to approximate distance traveled.
(b)
How could you get a better approximation of the distance traveled on \([0,2]\text{?}\) Explain, and then find this new estimate.
(c)
Now suppose that you know that \(v\) is given by \(v(t) = 0.5t^3-1.5t^2+1.5t+1.5\text{.}\) Remember that \(v\) is the derivative of the walker’s position function, \(s\text{.}\) Find a formula for \(s\) so that \(s' = v\text{.}\)
(d)
Based on your work in (c), what is the value of \(s(2) - s(0)\text{?}\) What is the meaning of this quantity?

Definition 4.2.3.

If \(f(x)\) is a function defined on the interval \([a,b]\text{,}\) a Riemann sum for \(f\) on \([a,b]\) is a sum of the form
\begin{equation*} \sum_{i=1}^{n} f(s_{i}) \cdot (x_i - x_{i-1})\text{,} \end{equation*}
where \(a = x_0 \lt x_1 \lt \dots \lt x_{n-1} \lt x_n = b\) and where \(s_{i}\) is a point in the \(i\)-th subinterval.

Remark 4.2.4.

The Riemann sum in DefinitionΒ 4.2.3 is almost a sum of the areas of rectangles. The height of the \(i\)-th rectangle is \(f(s_{i})\) and the width is \(x_i - x_{i-1}\text{.}\)
Figure 83. A generic Riemann sum.

Activity 4.2.6.

Why is the Riemann sum in FigureΒ 83 only almost a sum of the areas of rectangles?
  1. The subintervals have different widths.
  2. The function is not differentiable.
  3. Some of the values \(f(s_i)\) are negative.
  4. The \(x\)-coordinates \(s_i\) are not equally spaced.

Activity 4.2.7.

There are some special Riemann sums that are often easier to work with than the general Riemann sum of DefinitionΒ 4.2.3.
In a left Riemann sum, the point \(s_i\) in each subinterval is the left endpoint of the subinterval. That is,
\begin{equation*} s_i = x_{i-1}\text{.} \end{equation*}
Consider the left Riemann sum for \(f(x) = x^{2/3}\) on the interval \([2, 4]\) with 3 subintervals.
(e)
What do you notice about the subinterval widths \(x_{i} - x_{i-1}\text{?}\)

Activity 4.2.8.

The right Riemann sum is similar to the left Riemann sum, but the point \(s_i\) in each subinterval is the right endpoint of the subinterval instead of the left endpoint.

Activity 4.2.9.

The midpoint Riemann sum is similar to the left and right Riemann sums, but the point \(s_i\) in each subinterval is the midpoint of the subinterval.

Activity 4.2.10.

Explain how to approximate the area under the curve
\begin{equation*} f(x) = \frac{1}{5} (x-4)(x-10)(x-12) \end{equation*}
on the interval \([4,10]\) using a right Riemann sum with 3 subintervals.
described in detail following the image
The graph of the function \(f(x) = 1/5 (x-4)(x-10)(x-12)\) crosses the \(x\)-axis upward at \((4,0)\) and downward at \((10,0)\) with a maximum at about \((6.3, 9.7)\text{.}\)

Subsection 4.2.2 Videos

Figure 84. Video for IN2

Subsection 4.2.3 Exercises