Definite Integrals with inverse of function

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Homework Help Overview

The problem involves evaluating the integral of the inverse function \( f^{-1}(x) \) over a specified interval, given certain properties of the function \( f(x) \). The context is within the subject area of calculus, specifically focusing on definite integrals and inverse functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to approach the problem, noting the relationship \( f^{-1}(f(x)) = x \). Some participants suggest using a substitution method, while others introduce a specific formula relating the integrals of a function and its inverse.

Discussion Status

Participants are exploring different methods to tackle the problem, including substitution and the application of a specific formula. There is no explicit consensus on a single approach, but several lines of reasoning are being discussed.

Contextual Notes

The problem is constrained by the properties of the function \( f(x) \), which is continuous and decreasing on the specified interval, and the values of the function at the endpoints are provided. The original poster's lack of familiarity with the topic is also noted.

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Homework Statement


Suppose f(x) is continuous and decreasing on the closed interval 5 <= x < 13, that f(5) = 9, f(13) = 5 and that the

integral of f(x) from 5 to 13 is 70.64758.

Then the integral of f^-1(x) from 5 to 9 is equal to what?

(Note: f^-1(x) is the inverse of f(x))


Homework Equations






The Attempt at a Solution



I really don't know how to solve this problem. I know that f^-1(f(x)) = x. Any ideas would be great. Thanks.
 
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You want to find [tex]\int_{u=5}^9 f^{-1}(u)\,du[/tex]


Try the substitution u=f(x)
 
f is a positive monotone function on [a,b] where 0 < a < b and f has an inverse [tex]f^-1[/tex]. Set [tex]\alpha = f(a), \beta = f(b)[/tex] and then use this formula:[tex]\int_{\alpha}^{\beta} f^-1(y) dy = b \beta - a \alpha - \int_a^b f(x) dx[/tex].
 
JG89 said:
f is a positive monotone function on [a,b] where 0 < a < b and f has an inverse [tex]f^-1[/tex]. Set [tex]\alpha = f(a), \beta = f(b)[/tex] and then use this formula:[tex]\int_{\alpha}^{\beta} f^-1(y) dy = b \beta - a \alpha - \int_a^b f(x) dx[/tex].

This formula can be seen by drawing an example graph. Along the x-axis, you are integrating f(x)dx. ALong the y-axis you are integrating f-1(y)dy. Draw some rectangles and add/subtract areas to get the formula.
 

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