I'm self studying Stewart calculus and I'm working through the proof of FTC part 1. I'm stuck on a line in the proof that is preventing me from preceding.Given g(x) is the integral from a to x of f(t) dt, we were able to derive that g'(x) = 1/h integral from x to x+h of f(t) dt.By the extreme value theorem there exists a min m and max M on [x,x+h] such that f(u) = m and f(v) = M and f(u)h<= integral x to x+h f(t) dt <= f(v)h (by properties of integrals).Dividing by h we get f(u)<= (g(x+h)- g(x)) / h <= f(v) by substitution.Here is where I'm confused, the text says if we let h--> 0 then u --> x and v--> x since u and v lie in [x,x+h]. I'm not understanding how u--> x and v--> x. See 3rd image below, text is a red box. The Fundamental Theorem of Calculus, Part 1 If f is continuous on [a, b], then thefunction g defined byg(x) - S² (1) de a= x=bis continuous on [a, b] and differentiable on (a, b), and g'(x)Show Transcribed Text2that is,g(x + h)hM.Show Transcribed Text3g(x)=For now let's assume that h> 0. Since fis continuous on [x, x + h], the Extreme ValueTheorem says that there are numbers and in [x. x+h] such that f(u) m and(v)M, where and M are the absolute minimum and maximum values of fon[x, x + h]. (See Figure 6.)By Property 8 of integrals, we haveSince h> 0, we can divide this inequality by h:1 fathmh=f(t) dt = MhS*** (1) drf(u) =ATHf(u)h = ™" f(t) dt = f(0)hf(u) = = -1/2 √²+ h f(1) dt = f(v)hNow we use Equation 2 to replace the middle part of this inequality:g(x + h) - g(x)hgf(x).= f(v)ĴInequality 3 can be proved in a similar manner for the case where <0. (See Exercise 71.)Now we let h→0. Then u→x and v→ x, since u and v lie between x and x + h.

This is concept of limit , I will explain you numerically for better understanding .

Answered: Given g(x) is the integral from a to x…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

I'm self studying Stewart calculus and I'm working through the proof of FTC part 1. I'm stuck on a line in the proof that is preventing me from preceding.

Given g(x) is the integral from a to x of f(t) dt, we were able to derive that g'(x) = 1/h integral from x to x+h of f(t) dt.

By the extreme value theorem there exists a min m and max M on [x,x+h] such that f(u) = m and f(v) = M and f(u)h<= integral x to x+h f(t) dt <= f(v)h (by properties of integrals).

Dividing by h we get f(u)<= (g(x+h)- g(x)) / h <= f(v) by substitution.

Here is where I'm confused, the text says if we let h--> 0 then u --> x and v--> x since u and v lie in [x,x+h]. I'm not understanding how u--> x and v--> x. See 3rd image below, text is a red box.