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anonymous
 5 years ago
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.
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This question will be for a research paper. in essay format
anonymous
 5 years ago
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/cheat_table.aspx go to the complete cheat sheet for calculus, its explained there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i already checked that, i need better explaination

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in essence these two functions function as boundaries for the area contained within them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, what im trying to explain is, why would you intrgral or (antiderive it)? can you explain that aswell/.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0aka = The Fundamental Theorems of the Calculus ( 1 and 2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you familiar with riemann sums?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you under stand the concept of this limit \[f'(a)=\lim_{h \rightarrow 0} [f(a+h)f(a)]/h\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Riemann 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
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes i understand the riemann sum, it adds up all the boxes within the interval, to find the total area accordin to your limit

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the smaller you make the boxes the closer you get to the function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0height = f(x) , base = dx , ami rite?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As 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
 5 years ago
Best ResponseYou've already chosen the best response.0yea makes sense now, thank you, i will be sure to ask more questions if i need anymore help. THANK YOU!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no problem, I'm glad you understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you in highschool?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I only ask bc I'm in college and I never had to prove such a thing for any of my cal classes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nope 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
 5 years ago
Best ResponseYou've already chosen the best response.0can u just explain the second FTC, how it works?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Part 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
 5 years ago
Best ResponseYou've already chosen the best response.0http://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
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