need help on the attachment @Calculus1

- anonymous

need help on the attachment @Calculus1

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- anonymous

##### 1 Attachment

- anonymous

hey you know on the other problem you were helping me with lagrange?

- anonymous

what did you get for the answer because I got -4 and they counted it wrong

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## More answers

- anonymous

i got 2

- anonymous

are u sure you evaluated correctly

- anonymous

oh, made a stupid sign error

- anonymous

could you help me with this new problem on the attachment

- anonymous

yeah let me look at it

- anonymous

okay so this is what you wanna do, set u=x+4, then you have to then rewrite x in terms of u. You can do this simply by saying x=u-4

- anonymous

ok

- anonymous

|dw:1321490017407:dw|

- anonymous

|dw:1321490044699:dw|

- anonymous

|dw:1321490067161:dw|

- anonymous

do you replug the original function into u after integrating

- anonymous

yes after you, integrate, you must replace the orginal function

- anonymous

did you get u^2/2-3x*2U^(1/2) after integrating

- anonymous

give me a sec

- anonymous

i led you wrong, my mistake. What you want to do, is let u=sqrt(x+4), now we rewrite x interm of u. To do this simply:|dw:1321490856188:dw|

- anonymous

okay, now you should get:|dw:1321490948435:dw|

- anonymous

now let fix up that integral:|dw:1321491019469:dw|

- anonymous

|dw:1321491091199:dw|

- anonymous

now recall that u=sqrt(x+4), thus we can divide out the u in the denmominator and the sqrt(x+4) as so:|dw:1321491198348:dw|
The two can be put outside the integral since it is a constant

- anonymous

now simply we have:|dw:1321491263984:dw|

- anonymous

next you have to integrate that, and then replace the u's. Then evaluate over your limits

- anonymous

or, we could simply change the limits interms of u. Which would actaully be much eaier in this case. But i leave that up to you to decide on

- anonymous

sqrt(x+4) did both cancel out and when you replace the u on the bottom back to x+4 don't you have to do the same thing to the top?

- anonymous

on top we had u^2-3, cause we replaced the x remeber

- anonymous

oh, ok also you forgot to label the endpoints on the integral

- anonymous

well,i didnt forget, for me it is simpler to first leave the limits while i do the integration, since they really dont matter untill we get the antiderivative, which we then evlauate over the limits

- anonymous

I got -44

- anonymous

is that right?

- anonymous

???

- anonymous

yes i did,1 is the correct answer

- anonymous

Thanks could you help me with one more of these problem?

- anonymous

sure

- anonymous

did you post it already?

- anonymous

##### 1 Attachment

- anonymous

i cant open it

- anonymous

##### 1 Attachment

- anonymous

could you open it?

- anonymous

yeah, okay lets do this, can we multiply the x into the brackets, then seperate the integral into two spereate integrals:|dw:1321500013159:dw|

- anonymous

now lets take out some constants:
|dw:1321500075388:dw|

- anonymous

now let try this, can we for the first integral, let x^2=u, then du=2xdx, which further means du/2x=dx

- anonymous

now it should look something like this:|dw:1321500219046:dw|

- anonymous

now we see that the x will divide out, and the 1/2 from the du can be taken outside the integral, so now we will get :|dw:1321500319063:dw|

- anonymous

now 2*1/2 is simply 1, so we now have:|dw:1321500384165:dw|

- anonymous

now we can integrate

- anonymous

for the first integral, what i the antiderivative of f'? well its simply f, the orginal function , and the second integral is very simpyle to integrate

- anonymous

|dw:1321500486597:dw|

- anonymous

|dw:1321500522223:dw|

- anonymous

but lets replace the u:|dw:1321500561883:dw|

- anonymous

now you can evaluate this over you limits but remeber that you were given certain conditions, namely f(1)=3 and f(0)=1.so we have|dw:1321500688385:dw|

- anonymous

furthermore:|dw:1321500784828:dw|

- anonymous

Got it

- anonymous

yes, I was just was reviewing all the work, thanks!

- anonymous

anytime:)soon you will be an expert on calculus and you will be helping others as well

- anonymous

I hope so, thanks again

- anonymous

your welcome:)

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