Consider the functionf (x) = 11 – 3x on the interval, math homework help
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Upper Sum and Integral
Consider the functionf (x) = 11 – 3x on the interval [1, 3]. Let P6 be the partition of [1, 3] consisting of 7 equally spaced points: 1, 4/3, 5/3, 2, 7/3, 8/3, 3. The points divide [1, 3] into 6 subintervals of equal width.
The first graph below depicts the rectangles obtained by evaluating f at the left endpoints of the subintervals, and the second graph depicts the rectangles obtained by evaluating f at the right endpoints of the subintervals.
#4 (a) Fill in the blank: The width of each of the six subintervals is ______.
#4 (b) Find the value of U(f, P6), the upper sum of f with respect to P6.Show some work. Note that the upper sum is the sum of the areas of the rectangles depicted in the first graph.
#4 (c) Find the value of L(f, P6), the lower sum of f with respect to P6.Show some work. Note that the lower sum is the sum of the areas of the rectangles depicted in the second graph.
REMARK: It is a good idea to look Part 3 of http://sandsduchon.org/sands/MATH301/Videos/Chapter7/MATH301Chapter7.html, which is similar in approach, but considers a different function y = x2 on the interval [0, 1]. (I havevideos on this topic. For an additional example, see https://www.youtube.com/watch?v=UMIm2haAGYY)
Let Pn be the partition of [1, 3] consisting of n + 1 equally spaced points, so, these points divide [1, 3] into n subintervals of equal width.
#4(d)Fill in the blanks:
For the partition of [1, 3], the width of each of the subintervals is _________ for all i, 1 £ i £ n.
Then + 1 points in the partition Pn of [1, 3] are_________________________________________________.
(Write in an analogous fashion to Pn shown on http://sandsduchon.org/sands/MATH301/Videos/Chapter7/Section7.1Part3.html . Note that the points in your partition must be ³ 1. When n = 6, your partition should agree with P6 on the previous page. )
The supremum of f (x) = 11 – 3x on the i th subinterval = Mi = f (___________)= ______________________
U(f, Pn)= the upper sum of f with respect to partition Pn.Fill in the blank below.
____________________
(just substitute the appropriate quantities from the previous blanks.)
#4 (e) Show how to simplify the sum for U(f, Pn) in part (d) so that n is the only variable involved, in a similar fashion to problem #3 and the webpage example and videos. Make use of appropriate summation formulas.
We say that a bounded functionf on the interval[a, b] is integrable if L(f)= U(f), where
L(f)= sup {L(f, P) : P is a partition of [a, b]} and U(f)= inf {U(f, P) : P is a partition of [a, b]}.
L(f)is called the lower (Darboux) integral and U(f) is called the upper (Darboux) integral.
In this case we say that and the integral of f on [a, b] = the number U(f).
Note that we need the bounded hypothesis so that the lower sums L(f, P) and the upper sums U(f, P) are guaranteed to exist.
#4 (f) Find where U(f, Pn) is the expression you simplified in part e.
It turns out that . (You are not asked to show this.)
Say we call this limit I.
Since and L(f) is the supremum of the lower sums and U(f) is the infimum of the upper sums, we can conclude thatL(f)³ I and U(f) £I and since L(f)£ U(f), we conclude that L(f)= U(f)= I , so f (x) = 11 – 3x is integrable on interval [1, 3], and we write
.
#4 (g)State a function F such that F’ (x) = 11 – 3x. In other words, you are stating an antiderivative of f.(no work expected; use what you know from calculus.)
A version of the Fundamental Theorem of Calculus can be stated this way:
Fundamental Theorem of Calculus:
If f is integrable on [a, b] and F is differentiable on [a, b], with F‘ = f(i.e., F is an antiderivative of f),then
#4 (h).In #4 part(f), we have determined that f (x) = 11 – 3x is integrableon [1, 3] and in part #4(g) we have found a suitable function F whose derivative is f.Apply the Fundamental Theorem of Calculus as stated above to find Thus, you will verify that the value of the integral is in fact I, the limit you got in part (f).
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