Connection between Improper Integrals

[dami]

Homework Statement

Explain geometrically why the integral of dx/sqrt(x) for (x, 0,1)= [the integral of dx/x^2 for (x, 1, infinity)] + 1
Including graphs as well.

Homework Equations

The Attempt at a Solution

I graphed both functions and they met at a point (1,1) but i don't really understand how to show it geometrically.

 

[Mark44]

Homework Statement

Explain geometrically why the integral of dx/sqrt(x) for (x, 0,1)= [the integral of dx/x^2 for (x, 1, infinity)] + 1
Including graphs as well.

Homework Equations

The Attempt at a Solution

I graphed both functions and they met at a point (1,1) but i don't really understand how to show it geometrically.

Graph the regions whose areas are represented by the two integrals. That might get you going in the right direction.

Also, there is a relationship between y = 1/sqrt(x) and y = 1/x^2. Can you figure out what it is?

 

[dami]

I know the two functions are both inverse of each other. I graphed the areas of each region. But i still don't know what to write to answer the question.

 

[Dick]

I know the two functions are both inverse of each other. I graphed the areas of each region. But i still don't know what to write to answer the question.

Do you know that if two functions are inverses of each other then their graphs are reflections of each other across the line x=y?

 

[dami]

How does this affect their area on the intervals i wrote in the initial question i asked.Why do need to add 1 to make both areas equal

 

[Mark44]

The area represented by the integral \int_0^1 \frac{dx}{\sqrt{x}} is a bit larger (1 unit) than the area represented by the integral \int_1^{\infty}\frac{dx}{x^2}}.

For the first integral, the area is calculated using vertical strips of width dx and height 1/sqrt(x). What would the integral be if you used horizontal strips?

 

[dami]

Yeah, finally solve it. I used the inverses and reflection of the two graphs at x=y.