Now we explore why we are interested in the area under a graph. Consider an object moving in a straight line with constant velocity v (assumed positive). The distance traveled over a time interval [t1,t2] is equal to ____, where Δt = (t2-t1) is the _____ _____. State the appropriate distance formula
- time elapsed
- Distance traveled = velocity * time elapsed (vΔt)
Because v is constant, the graph of velocity is a _______ line (figure 1 pg 225) and vΔt is equal to the area of the _______ region under the graph of velocity over [t1,t2]. How do we rephrase the distance formula to better fit this concept?
- horizontal line
- rectangular region
- Distance traveled = area under the graph of velocity over [t1,t2]
There is, however, an important difference between these two equations: Eq. (1) makes sense only if velocity v is ______, whereas Eq. (2) is correct even if the velocity changes with time. Thus, the advantage of expressing distance traveled as an _____ is that it enables us to deal with much more general types of motion
To see why Eq (2) might be true in general, let's consider the case where velocity changes over time but is constant on intervals. In other words, we assume that the object's velocity _____ ______ from one interval to the next as in Figure 2 (pg 225). How would you find the distance traveled over each time interval? How would you find the total distance traveled?
- changes abruptly
- The distance traveled over each time interval is equal to the area of the rectangle above that interval
- The total distance traveled is the sum of the areas of the rectangles
What is our strategy when velocity changes continuously (Fig. 3 pg 226)? This idea leads to the concept of an ______
- To approximate the area under the graph by summing up areas of rectangles and then passing to a limit.
Distance traveled is equal to the _____ _____ the graph. It is approximated by the sum of the areas of the rectangles
- area under
- **See Fig 3 226
Our goal is to compute the area under the graph of a function f. In this section, we assume that f is _______ and _______, so that the graph of f lies above the x-axis (Fig 4 pg 226). The first step is to approximate the area using _______
- continuous and positive
To begin, choose a whole number N and divide [a,b] into N subintervals of _____ width as in Fig 4A (pg 226). The full interval [a,b] has width b - a, so each subinterval has width Δx = _______
- equal width
- Δx = (b - a)/N
State the right endpoints of the subintervals:
Hint, start with x1 =
x1 = a + Δx, x2 = a + 2Δx,..., xN-1 = a + (N - 1)Δx, xN = a + NΔx
Why is the last right endpoint xN = b?
Next, as in Fig 4B (pg 226), construct, above each subinterval, a rectangle whose height is the value of f(x) at the _____ _____ of the sub interval
- Because a + NΔx = a + N ((b - a)/N) = b
- right endpoint
The sum of the areas of these rectangles provides an approximation to the area under the graph. The first rectangle has base Δx and height f(x1), so its area is ______. Similarly, the second rectangle has height f(x2) and area _______, etc. The sum of the areas of the rectangles is denoted ____ and is called the ____ _____-______ _______ (State the initial formula).
Factoring out Δx, we obtain the formula:
- Nth right-endpoint approximation
- RN = f(x1)Δx + f(x2)Δx +...+f(xN)Δx
- RN = Δx(f(x1) + f(x2) +...+f(xN))
Explain the new formula for RN after Δx is factored out
RN is equal to Δx times the sum of the function values at the right endpoints of the subintervals
- a = left endpoint of interval [a,b]
- b = right endpoint of interval [a,b]
- N = number of subintervals in [a,b]
- Δx = (b - a)/N
Summation notation is a standard notation for writing _____ in compact form. How do we denote the sum of numbers am,..., an (m ≤ n)?
- nΣj=m aj = am + am + 1 +...+ an
The Greek letter Σ (capital sigma) stands for "sum," and the notation _____ tells us to start the summation at j = m and end it at j = n. For example 5Σj=1 j2 =
- 12 + 22 + 32 + 42 +52 = 55
In this summation, the jth term is aj = j2. We refer to j2 as the _____ term. The letter j is called the ______ _____. It is also referred to as a dummy variable (why?)
- general term
- summation index
- because any other letter can be used instead. For example, the j can be a k or an m see pg 227
6Σk=4 (k2 - 2k) =
9Σm=7 1 =
- (42 - 2(4)) + (52 - 2(5)) + (62 - 2(6)) = 47
- 1 + 1 + 1 = 3 (because a7 = a8 = a9 = 1)
The usual commutative, associative, and distributive laws of addition give us the following rules for manipulation summations:
- nΣj=m (aj + bj) = nΣj=m aj + nΣj=m bj
- nΣj=m Caj = C nΣj=m aj (C any constant)
- nΣj=1 C = nC (C any constant and n ≥ 1)
1) 5Σj=3 (j2 + j) =
2) 5Σj=3 j2 + 5Σj=3 j =
3) 100Σk=0 (7k2 - 4k +9) =
What can be noted about #1 and #2
- 1) (32 + 3) + (42 +4) + (52 +5) = 62
- 2) (32 + 42 + 52) + (3 + 4 + 5) = 62 (#1 and #2 are equivalent)
- 3) 100Σk=0 7k2 + 100Σk=0(-4k) + 100Σk=0 9 =
- 7 100Σk=0 k2 - 4 100Σk=0 k + 9 100Σk=0 1
It is convenient to use summation notation when working with area approximations. For example, RN is a sum with general term f(xj):
The summation extends from j = 1 to j = N, so we can write RN concisely as:
RN = Δx[f(x1) + f(x2) +...+f(xN)]
RN = Δx NΣj=1 f(xj)
We shall make use of two other rectangular approximations. Divide [a,b] into ___ subintervals as before. In the left endpoint approximation LN, the ______ of the rectangles are the values of f(x) at the left endpoints [Fig 6A pg 229]. What are these left endpoints? What are the sum areas of the left-endpoint rectangles?
- x0 = a, x1 = a +Δx, x2 = a + 2Δx,..., xN-1 = a + (N - 1) Δx
) + f(x1
) + f(x2
Note that both RN and LN have general term f(xj), but the sum for LN runs from _____ to ______ rather than from ______ to _____. Write in summation notation:
- j = 0 to j = N - 1
- j = 1 to j = N
- LN = Δx N-1Σj=0 f(xj)
In the midpoint approximation MN, the _____ of the rectangles are the values of f(x) at the midpoints of the subintervals rather than at the _______. As we see in Fig 6B (pg 229) what are the midpoints?
- (x0 + x1)/2, (x1 + x2)/2,..., (xN-1 + xN)/2
What is the sum of the areas of the midpoints rectangles?
How would you present this in summation notation?
- MN = Δx [f((x0 + x1)/2) + f((x1 + x2)/2) +...+ f((xN-1 + xN)/2)]
- MN = Δx N-1Σj=0 f((xj + xj+1)/2)
When f is increasing, the left-endpoint rectangles lie _____ the graph and right-endpoint rectangles lie _____ it
Figure 10 (pg 230) shows several right-endpoint approximations. Notice that the error in computing the area, corresponding to the yellow region above the graph, gets ______ as the number of rectangles increases. In fact, it appears that we can make the error as small as we please (How? and what would this entail?)
- By making the number N of rectangles large enough. This would entail considering the limit as N ⇾ ∞, as this should give us the exact area under the curve
The error _______ as we use more rectangles
Theorem 1: If f is continuous on [a,b], then the endpoint and midpoint approximations approach one and the same limit as N →
- N → ∞
- In other words, there is a value L such that:
- limN → ∞ RN = limN → ∞ LN = limN → ∞ MN = L
- If f(x) ≥ 0 on [a,b], we define the area under the graph over [a,b] to be L.
In Theorem 1, it is not assumed that f(x) ≥ 0. What if f(x) takes on negative values?
- The limit L no longer represents area under the graph, but we can interpret it as a "signed area,"
- **discussed in the next section
Theorem 1 can be illustrated using power sums (define)
The kth power sum is defined as the sum of the kth powers of the first N integers
**We shall use the power sum formulas for k = 1,2,3.
State 3 Power Sum examples
- NΣj=1 j = 1 + 2 +...+N = [N(N+1)/2] = N2/2 + N/2
- NΣj=1 j2 =12 +22 +...+ N2 = [N(N+1)(2N+1)/6] = N3/3 + N2/2 + N/6
- NΣj=1 j3 = 13 + 23 +...+N3 = [N2(N+1)2/4] = N4/4 + N3/2 + N2/4
- Bonus: NΣj=1 1 = N
Solve 6Σj=1 j2
12 + 22 + 32 +42 +52 + 62 = 63/3 + 62/2 + 6/6 = 91
**Notice the use of a Power Sum equation