## anonymous 4 years ago sketch and then calculate the area of the region bounded by the graphs of x=1, y=x+1 and y=1/sqrt(x^2+1)

1. anonymous

Is it possible to sketch the graph $$y = \frac{1}{\sqrt{x^2 + 1}}$$? 

2. anonymous

I think I would use wolfram alpha to sketch the graph.

3. anonymous

Such a complicated Graph to sketch all by yourself.

4. anonymous

Okay I got it We need to fins the intersection of $$\frac{1}{\sqrt{x^2+1}}=y$$ and y = x + 1 $x + 1 = \frac{1}{\sqrt{x^2+1}} \implies (x+1)^2 (x^2+1) = 1$ To do it quickly I will use wolfram alpha but you should solve on your own. BTW from the graph it is clearly visible that the point is x =0 and y=1 So here is the equation you should use to get the area of the region required $\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx$

5. anonymous

I did all the talking and solving :-/

6. anonymous

sorry somthing came up

7. anonymous

8. anonymous

thats what i tried to do

9. anonymous

So, did you get it? I mean the answer did you get the answer?

10. anonymous

i got (x^2/2)+x - Lnsec theta+ tan theta

11. anonymous

.618626 area?

12. anonymous

$\int_0^{1} (x+1)dx-\int_0^1 \frac{1}{\sqrt{x^2+1}}dx = \left[ \frac{x^2}{2} + x \right]_0^1 - \left[ \tan^{-1}x \right]_0^1$ I think this is what you are supposed to get.

13. anonymous

$\frac{1}{2} + 1 - \frac{\pi}{4}$

14. anonymous

I am getting .714... something

15. anonymous

the second part i got LN(sec theta + tan theta) for integral

16. anonymous

did i do it wrong?

17. anonymous

Yeah I don't seem to get ln(sec theta) anywhere. You should recheck your solution and I will check mine. Maybe I am wrong.

18. anonymous

i let x = tan theta and du = sec^2 theta

19. anonymous
20. anonymous

Umm Lammy I think $\frac{1}{\sqrt{x^2+1}} = \tan^{-1}x$ I am not sure. I am gonna apply same substitution as yours $$dx = d\theta \sec^2\theta$$ $\frac{1}{\sqrt{\tan^2\theta + 1}}\times \sec^2\theta d\theta = \sec \theta d\theta$ I think you are right I was wrong, sorry for misleading you

21. anonymous

so integral of sec theta = LN abs(sec theta + tan theta) right?

22. anonymous

no no you were helping =)

23. anonymous

then i just subtract and will get the answer right?

24. anonymous

yeah

25. anonymous

do you think you have time to help me with on more? befor you go to sleep?

26. anonymous

hahah

27. anonymous

yeah sure, post it :-D

28. anonymous

just posted it

29. anonymous

=)