- #1
iggybaseball
- 57
- 0
This is my last homework problem and I feel that I almost have it solved. The problem is as followed:
[tex] f(x) = \sqrt{4-x^2} [/tex]
Find the arc length of the given function from x=0 to x=2.
I know that I am supposed to use this formula to solve for arclength:
[tex] \int_{0}^{2} \sqrt{1 + (f\prime(x))^2} dx[/tex]
Ok so I take the derivative of
[tex] f(x) = \sqrt{4-x^2} [/tex]
and I get:
[tex] f\prime(x) =\frac{-x}{\sqrt{4-x^2}} [/tex]
I then plug this into the equation get:
[tex] \int_{0}^{2} \sqrt{1 + (\frac{-x}{\sqrt{4-x^2}}) ^2} [/tex]
[tex] \int_{0}^{2} \sqrt{1 + \frac{x^2}{4-x^2}} [/tex]
I get stuck here because I can't integrate this. I can't use any integral tables and I even try some algebraic manipulation inside of the square root and get:
[tex] \int_{0}^{2} \sqrt{\frac{4}{4-x^2}} [/tex]
Any help? Thank you.
[tex] f(x) = \sqrt{4-x^2} [/tex]
Find the arc length of the given function from x=0 to x=2.
I know that I am supposed to use this formula to solve for arclength:
[tex] \int_{0}^{2} \sqrt{1 + (f\prime(x))^2} dx[/tex]
Ok so I take the derivative of
[tex] f(x) = \sqrt{4-x^2} [/tex]
and I get:
[tex] f\prime(x) =\frac{-x}{\sqrt{4-x^2}} [/tex]
I then plug this into the equation get:
[tex] \int_{0}^{2} \sqrt{1 + (\frac{-x}{\sqrt{4-x^2}}) ^2} [/tex]
[tex] \int_{0}^{2} \sqrt{1 + \frac{x^2}{4-x^2}} [/tex]
I get stuck here because I can't integrate this. I can't use any integral tables and I even try some algebraic manipulation inside of the square root and get:
[tex] \int_{0}^{2} \sqrt{\frac{4}{4-x^2}} [/tex]
Any help? Thank you.