y = 2 Sqrt x from o to 1 (Curve , and Area)

Hi guys you can see my problem in the attachment

I would be happy if someone could solve the last part for me, and show it in detail
[attachment=0:4amlwj7g]IMG_0003.jpg[/attachment:4amlwj7g]

Thank you very much

Riazy said:

Hi guys you can see my problem in the attachment

I would be happy if someone could solve the last part for me, and show it in detail
[attachment=0:298a3mdd]IMG_0003.jpg[/attachment:298a3mdd]

Thank you very much

Click to expand...

Substitute

u = sec(?)

Don't forget to change the limits

our professors generally don't use sec,, any other way?

No I am not familiar with sec at all, I dont know how to evaluate it, haven't learnt that here in sweden.

so if we have sec^2 how is it evaluated if i try ur method

i am serious, they don't use it in scandinavia.

Well - then I don't know how to do it.

After substitution and back substitution the answer would be

1/2*[x?(x[sup:36k0m72k]2[/sup:36k0m72k]-1)-log{2[?(x[sup:36k0m72k]2[/sup:36k0m72k]-1)+x]}] + C

Strange - scandanavian countries score very high in math competitions.

For integrals of the form \(\displaystyle \int\sqrt{u^{2}-a^{2}}du\)

use the hyperbolic sub \(\displaystyle u=a\cdot cosh(t), \;\ du=a\cdot sinh(t)dt\)

In your case, a=1.

They come in handy for integrals of this form.

Makes for an easier integration than with the sec sub anyway. At least, I always thought so.

Continuing:

\(\displaystyle \frac{\pi}{2}\int_{1}^{3}\sqrt{u^{2}-1}du\)

Make the sub \(\displaystyle u=cosh(t), \;\ du=sinh(t)dt\)

\(\displaystyle \frac{\pi}{2}\int_{0}^{cosh^{-1}(3)}sinh^{2}(t)dt\)

\(\displaystyle \frac{\pi}{2}\int_{0}^{cosh^{-1}(3)}\frac{1}{2}(cosh(2t)-1)dt\)

\(\displaystyle \frac{\pi}{2}\left(\frac{1}{4}sinh(2t)-\frac{t}{2}\right)\left|_{0}^{cosh^{-1}(3)}\right\)

\(\displaystyle \approx 5.28\)

Do y'all use hyperbolic trig in Sweden?.