Definite Integrals with inverse of function

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The problem involves finding the integral of the inverse function f^-1(x) from 5 to 9, given that f(x) is continuous and decreasing on the interval [5, 13]. The integral of f(x) from 5 to 13 is provided as 70.64758. A suggested approach is to use the substitution u = f(x) and apply the formula that relates the integral of a function and its inverse: ∫ f^-1(y) dy = bβ - aα - ∫ f(x) dx. This method relies on the properties of monotonic functions and their inverses. The discussion emphasizes visualizing the relationship through graphing to understand the areas involved in the integration.
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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 \int_{u=5}^9 f^{-1}(u)\,du


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

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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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