# Connection between Improper Integrals

1. Jul 11, 2010

### dami

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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.

2. Jul 11, 2010

### Staff: Mentor

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?

3. Jul 12, 2010

### dami

I know the two functions are both inverse of each other. I graphed the areas of each region. But i still dont know what to write to answer the question.

4. Jul 12, 2010

### Dick

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?

5. Jul 12, 2010

### 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

6. Jul 12, 2010

### Staff: Mentor

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?

7. Jul 13, 2010

### dami

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