Establishing a definite integral is btw 12&24

In summary: Mto give your bounds on the areas below the graph of f^(-1).In summary, the problem asks to establish that the integral of f^-1(x) from 1 to 7 must be between 12 and 24. This can be shown by considering the graphical interpretation of an integral as the area under a graph. Since f(2)=7, f^-1(7)=2, and f'(x)<0 for all x, the graph of f^-1(x) between 1 and 7 can be plotted with points (1,2), (1,4), (7,
  • #1
Calcgeek123
20
0

Homework Statement


Suppose f(2)=7, f(4)=1, and f'(x)< 0 for all x. Assuming f^(-1) is differentiable everywhere, establish that

12 < Integral from 1 to 7, of f^(-1)(x)dx < 24


Homework Equations


N/A


The Attempt at a Solution


I do not know where to begin... =/
 
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  • #2
For this problem, you need to think about the graphical interpretation of an integral as the area under a graph. Note also that since f(2)=7, f^-1 (7) =2, and ditto for the other equation.

Don't get caught up with the 12 and 24. Think about what you would answer if I asked you what values the integral should be between.
 
  • #3
Start with a graph of y = f-1(x) between 1 and 7. You know that f(2) = 7 and that f(4) = 1. Can you get some points on the graph of f-1? You also know that f'(x) < 0 for all x.
 
  • #4
I made a sort of rectangle when I graphed. I have the points (1,2), (1,4), (7,4) and (7,2).
From this, i see that the area below 2 is 12, and the area below 4 is 24..
 
  • #5
Good! You also know that f'(x) < 0, which means f^-1(x) decreases as x increases, and that means...
 
  • #6
That means.. that f^-1 will eventually be 3. To the left of that point though, it will have to be less than 4, and to the right of that point, it will hve to be greater than 2. ANd... this is why f^-1 can't be equal to the areas of the rectangle that its in..?
 
  • #7
Or is this last part i did wrong?
 
  • #8
Your post isn't quite clear, but I think you got it. If you draw out the rectangle bounded by y=2, then you draw out the rectangle bounded by y=4, you'll see that the area under the graph has to be larger than the former and smaller than the latter. There's no way you can twist, turn, bend, or morph the graph so that the area under it is larger than 24 (the larger rectangle's area) or smaller than 12 (the smaller rectangle's area).
 
  • #9
use the fact that if
m<f<M when a<x<b
then
[tex]m (b-a)<\int_a^b f dx<M(b-a)[/tex]
along with some other basic facts
in particular find m and M so that
m<f^(-1)(x)<M when 1<x<7
 

Related to Establishing a definite integral is btw 12&24

1. What is a definite integral?

A definite integral is a mathematical concept used to calculate the area under a curve or between two points on a graph. It is represented by the symbol ∫ and has a lower and upper limit of integration.

2. How do you establish a definite integral?

To establish a definite integral, you need to follow the steps of integration, which include identifying the limits of integration, finding the anti-derivative of the function, and evaluating the integral using the limits of integration.

3. Why is it important to establish a definite integral?

Establishing a definite integral allows us to find the exact area under a curve, which has many real-world applications such as calculating volumes, work done, and average values.

4. What are the differences between definite and indefinite integrals?

A definite integral has specific limits of integration and gives a numeric value as the result, while an indefinite integral has no limits and gives a general solution in the form of a function.

5. Can you explain the concept of Riemann sums in relation to establishing a definite integral?

Riemann sums are an approximation method used to calculate the area under a curve. By dividing the area into smaller rectangles, we can find the sum of their areas, which gives us an estimate of the definite integral. The more rectangles we use, the more accurate the approximation becomes.

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