What is the Anti-Derivative of a Radical Function?

[ggcheck]

Homework Statement

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?

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

 

[rock.freak667]

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

 

[quasar987]

sin²+cos²=1 ==> 1-cos²=sin

 

[ggcheck]

how did you guys know to do that?

 

[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!

 

[ggcheck]

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

 

[quasar987]

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

 

[ggcheck]

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

 

[Avodyne]

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

 

[ggcheck]

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

O_O

wat

 

[quasar987]

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

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

 

[quasar987]

O_O

wat

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.

 

[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 don't 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.

 

[ggcheck]

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 don't 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.

binomial theorem?

ftc= fundamental theorem of calculus

 

[Avodyne]

binomial theorem?

You don't want to know.

 

[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?

 

[quasar987]

Pretty much!