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.

- anonymous

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- Mimi_x3

Then where do you needhelp with?

- anonymous

Evaluating

- anonymous

Where do you people get your bike avatars from?

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## More answers

- anonymous

I do not want "Google" as an answer... :O

- lgbasallote

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

- anonymous

Yes

- lgbasallote

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

- Mimi_x3

What's the whole question?

- lgbasallote

just*

- lgbasallote

|dw:1337521244254:dw|

- anonymous

|dw:1337521266266:dw|

- anonymous

Yes I know that trig sub.

- lgbasallote

|dw:1337521291094:dw|

- anonymous

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

- Mimi_x3

the problem is subbing in the limits???

- anonymous

convert the limits when using the trig sub

- lgbasallote

dont forget to change back to x before evaluating

- Mimi_x3

No, you don't need to convert back

- lgbasallote

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

- anonymous

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

- lgbasallote

yep.. \[\int \sec^3 \theta d\theta\]

- anonymous

Then what is limits of integration wrt Î¸?

- anonymous

What is arctan(-4)?

- anonymous

How did you get that?

- anonymous

I got sec^3(Î¸)

- lgbasallote

it was sec^3 theta...ohh woops

- 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

- anonymous

|dw:1337523905957:dw|
@LagrangeSon678
@shivam_bhalla

- anonymous

@experimentX

- experimentX

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

- experimentX

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

- anonymous

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

- anonymous

You can go ahead and derive this

- anonymous

I forgot the dx in the question :P

- anonymous

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...

- 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!!

- experimentX

brb

- anonymous

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

- anonymous

@shivam_bhalla look at my recent post above

- anonymous

Those are two functions!

- anonymous

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

- anonymous

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

- anonymous

Alright thanks your help for the above.

- anonymous

@TuringTest help please
@amistre64
@Hero
@asnaseer

- anonymous

@FoolForMath would you mind helping me here?

- anonymous

Someone is bound to come!

- anonymous

@apoorvk help please?

- anonymous

Please look at the most recent question (it is the one with the sketch just above, NOT THE OPENING POST)

- experimentX

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

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