## QRAwarrior Group Title Evaluate the definite integral from u = 0 to u = -4 of: \sqrt{1 + u^2} dx As you may have noticed, I have already done U-sub. 2 years ago 2 years ago

1. Mimi_x3

Then where do you needhelp with?

2. QRAwarrior

Evaluating

3. QRAwarrior

Where do you people get your bike avatars from?

4. QRAwarrior

5. lgbasallote

looks trig sub...have you learned that yet?

6. QRAwarrior

Yes

7. lgbasallote

but we really did gust google this! i swear! :(

8. Mimi_x3

What's the whole question?

9. lgbasallote

just*

10. lgbasallote

|dw:1337521244254:dw|

11. QRAwarrior

|dw:1337521266266:dw|

12. QRAwarrior

Yes I know that trig sub.

13. lgbasallote

|dw:1337521291094:dw|

14. QRAwarrior

But ultimately, you have to put a -4 into a trigonometric function's argument. How will that work?

15. Mimi_x3

the problem is subbing in the limits???

16. LagrangeSon678

convert the limits when using the trig sub

17. lgbasallote

dont forget to change back to x before evaluating

18. Mimi_x3

No, you don't need to convert back

19. lgbasallote

otherwise you'll have to convert the limits in terms of theta

20. QRAwarrior

Ok let me see what happens: - tan(θ) = u/1 - sec(θ) = \sqrt{1 + u^2} Then, sec^2(θ)dθ = du

21. lgbasallote

yep.. $\int \sec^3 \theta d\theta$

22. QRAwarrior

Then what is limits of integration wrt θ?

23. QRAwarrior

What is arctan(-4)?

24. QRAwarrior

How did you get that?

25. QRAwarrior

I got sec^3(θ)

26. lgbasallote

it was sec^3 theta...ohh woops

27. lgbasallote

sec^3 theta = (sec^2 theta)sec theta (1+ tan^2 theta) sec theta $\int \sec \theta + \tan^2 \theta \sec \theta$ lol got me stuck now

28. QRAwarrior

|dw:1337523905957:dw| @LagrangeSon678 @shivam_bhalla

29. QRAwarrior

@experimentX

30. experimentX

evaluate it individually .. $\int \sec \theta d\theta = \ln |\sec\theta + \tan\theta|$

31. experimentX

this is a lot better http://answers.yahoo.com/question/index?qid=20110524035942AAXxTdM

32. shivam_bhalla

@QRAwarrior , i just remember this $\int\limits_{}^{}\sqrt{x^2+a^2} = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log{|x+\sqrt{x^2+a^2}|}+C$

33. shivam_bhalla

You can go ahead and derive this

34. shivam_bhalla

I forgot the dx in the question :P

35. QRAwarrior

Ok, skrew that question and instead would you mind looking at this: You have the curves x = (y-7)^2, and x = 4 that enclose a region. You must rotate this region about y = 5. |dw:1337525316103:dw| I need to use the shell method here, but it looks like as if I will get two cylinders here. Thanks for the help on the opening post question...

36. experimentX

I think that's formula for standard integral ... what do we call it ... i have bunch of them in my books ... but they are derived in the same way!!

37. experimentX

brb

38. shivam_bhalla

slight Correction: $\int\limits\limits_{}^{}\sqrt{x^2+a^2} \space \space dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log{|x+\sqrt{x^2+a^2}|}+C$

39. QRAwarrior

@shivam_bhalla look at my recent post above

40. QRAwarrior

Those are two functions!

41. QRAwarrior

I just realized! I would have to use the washer.

42. shivam_bhalla

@QRAwarrior , I am not so good at this. @TuringTest can surely help you in this :)

43. QRAwarrior

Alright thanks your help for the above.

44. QRAwarrior

@TuringTest help please @amistre64 @Hero @asnaseer

45. QRAwarrior

@FoolForMath would you mind helping me here?

46. QRAwarrior

Someone is bound to come!

47. QRAwarrior

http://www.wolframalpha.com/input/?i=x+%3D+%28y-7%29^2%2C++x%3D4 $\int_{5}^{9}2\pi (y-7)(y-7)^2 dy$