# Evaluate the integral

1. Dec 10, 2007

### ggcheck

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

integral from 0 to 7 (49-x^2)^(1/2)dx

I just need help getting started; not sure how to find the anti-derivative of a function that is under a radical

the way to do this is by using the FTC once I find the anti-D, right?

3. The attempt at a solution
I started messing around with it and the closest I could get was this:

(2/3)(49-x^2)^(3/2)

any help would be great

2. Dec 10, 2007

### rock.freak667

Try a trig substition like say, x=7sin$\theta$

3. Dec 10, 2007

### quasar987

sin²+cos²=1 ==> 1-cos²=sin

4. Dec 10, 2007

### ggcheck

how did you guys know to do that?

5. Dec 10, 2007

### quasar987

When you see something of the form a-bx², this cries for a x=cos or x=sin substitution. Even more so when a-bx² is under a radical!

6. Dec 10, 2007

### ggcheck

Is there any other way to approach this problem? because we weren't taught that technique--we just learned the FTC last week

7. Dec 10, 2007

### quasar987

Well, the function is unbounded on (0,7) [look what happens near 7]... so it's not integrable automatically.

8. Dec 10, 2007

### ggcheck

trig substitute? we haven't learned any substitution methods yet... just the ftc...

9. Dec 10, 2007

### Avodyne

Maybe you're supposed to notice that this integral gives the area of a quarter-circle of radius 7.

10. Dec 10, 2007

### ggcheck

O_O

wat

11. Dec 10, 2007

### quasar987

Nevermind that, I thought the function was 1/(...)^1/2

12. Dec 10, 2007

### quasar987

Maybe Avodyne has it.

If you have a function y(x), what does the integral of y(x) from a to b represent? The area under the curve, I'm sure you know. Now what's the equation of a circle of radius 7? x²+y²=7². Isolate y and take the square root and keep the positive sign (this means, keep the part of the curve that is above the x axis), and you get (49-x²)^1/2. Therefor, the integral of (49-x²)^1/2 is the area of a quarter circle of radius 7.

13. Dec 10, 2007

### unplebeian

If you don't want to use trig subst. then one way is by expanding
(49-x^2)^(1/2) using the binomial theorem.

I dont know what you mean by FTC.

The physical interpretation of this problem is exactly as said above, it is the area of a quarter-circle of radius 7 i.e. in the first quadrant.

14. Dec 10, 2007

### ggcheck

binomial theorem?

ftc= fundamental theorem of calculus

15. Dec 10, 2007

### Avodyne

You don't want to know.

16. Dec 10, 2007

### ggcheck

since I haven't learned the trig sub. or the binomial theorem, either I recognize that is a half circle or I'm fu(ked?

17. Dec 10, 2007

Pretty much!