anonymous
  • anonymous
Theory: If f and g are continuous functions on the interval [a, b], and if f(x) ≥ g(x) for all x in [a, b], then the area of the region bounded above by y = f(x), below by y = g(x), on the left by the line x = a, and on the right by the line x = b is b A = ∫ [f(x) – g(x)] dx a Explain the theory and why it is true * 20 hours ago * - 3 days left to answer. Additional Details This question will be for a research paper. in essay format
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
http://tutorial.math.lamar.edu/cheat_table.aspx go to the complete cheat sheet for calculus, its explained there
anonymous
  • anonymous
i already checked that, i need better explaination
anonymous
  • anonymous
Well think of it this was if you have a function f(x) whose values are greater than a function g(x) then its only true that the function f(x) lies above g(x) for all values of that function. So in order to determine the area bounded by both function you would have to subtract greater function f(x) from the lesser function g(x) and integrate to determine the area contained within these two functions

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anonymous
  • anonymous
in essence these two functions function as boundaries for the area contained within them
anonymous
  • anonymous
well, what im trying to explain is, why would you intrgral or (antiderive it)? can you explain that aswell/.
anonymous
  • anonymous
aka = The Fundamental Theorems of the Calculus ( 1 and 2)
anonymous
  • anonymous
are you familiar with riemann sums?
anonymous
  • anonymous
Do you under stand the concept of this limit \[f'(a)=\lim_{h \rightarrow 0} [f(a+h)-f(a)]/h\]
anonymous
  • anonymous
Riemann sums are were integrals come from and goes hand in hand with the limit I posted above which coincides with the FTC 1 and 2
anonymous
  • anonymous
If you think about the limit I posted above it explains the the derivative but fails to exist at h=o, but the idea to get as close to 0 without ever being 0 which give you the derivative. Riemann sums work the same way but instead it uses smaller and smaller rectangles and approaches the function to determine the area under the function which is multiplied by the minute changes in x hence that's where you always notice a dx in integral: hence base x height which give you the area
anonymous
  • anonymous
yes i understand the riemann sum, it adds up all the boxes within the interval, to find the total area accordin to your limit
anonymous
  • anonymous
and the smaller you make the boxes the closer you get to the function
anonymous
  • anonymous
height = f(x) , base = dx , ami rite?
anonymous
  • anonymous
There you go
anonymous
  • anonymous
does it makes sense?
anonymous
  • anonymous
As f(x) changes to smaller and smaller intervals do does dx hence you have the change in y over the change in x which is a derivative....... multiplied over the entire function, which is the area
anonymous
  • anonymous
yea makes sense now, thank you, i will be sure to ask more questions if i need anymore help. THANK YOU!
anonymous
  • anonymous
no problem, I'm glad you understand
anonymous
  • anonymous
Are you in highschool?
anonymous
  • anonymous
I only ask bc I'm in college and I never had to prove such a thing for any of my cal classes
anonymous
  • anonymous
nope in uni. just a conus question for my midterm, i have to write an essay during my mid!! (counts only as bonus points towards mid marks)
anonymous
  • anonymous
cool
anonymous
  • anonymous
can u just explain the second FTC, how it works?
anonymous
  • anonymous
Part 1 says \[\int\limits_{a}^{b}f(x)dx= F(b)-F(a)\] Where F'=f Part 2 says \[d/dx(\int\limits_{a}^{x}f(t)dt)=f(x)\]
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus Wikipedia has an example of it under the Example section take a look at it and see if you can understand the concept
anonymous
  • anonymous
For part 2
anonymous
  • anonymous
Thank you.

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