anonymous
  • anonymous
Anyone want to help with the integral of x/(x+2)^1/2 from -1 to 2?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I just don't know how to set it up, what should u be?
anonymous
  • anonymous
say u = 1/(sqrt(x-2) then you have integral of u/u^2-2. use partial fractions to solve.
anonymous
  • anonymous
sorry,u = 1/sqrt(x+2)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
lemme know how it works out.
anonymous
  • anonymous
ok thx I'll try it out
anonymous
  • anonymous
did you get du=-1/2(x+2)^3/2? or did I do something wrong
anonymous
  • anonymous
thats right, just checked it
anonymous
  • anonymous
ok great! . i'd actually not done it. I just looked at the problem and did some calculation in my head. sorry.
anonymous
  • anonymous
np, wait though not done yet
anonymous
  • anonymous
so next du=-1(x+2)/2(x+2)^1/2 right?
anonymous
  • anonymous
umm, you can solve it entirely in u and plug in the values of u, since this is a definite integral.
anonymous
  • anonymous
no need to deal with x at all
anonymous
  • anonymous
wait how come? I thought we had to make du look like the remaining terms in the integral, sorry I've just learned integrals today
anonymous
  • anonymous
Okay. let me get back to you with this. I will take out my pen and paper and write it down. Do ask someone else too.
anonymous
  • anonymous
ok
anonymous
  • anonymous
Ok, found it
anonymous
  • anonymous
hey!
anonymous
  • anonymous
Ok, so first Let \(u = (x+2)^{1/2} \implies du = (1/2)(x+2)^{-1/2}dx\) \[ \implies dx = 2(x+2)^{1/2} du\]
anonymous
  • anonymous
So then \[\int_{-1}^2\frac{x}{(x+2)^{1/2}}dx = \int_1^2[u^2-2]du\]
anonymous
  • anonymous
Whoops forgot the 2
anonymous
  • anonymous
wow, ok let me try that out thanks, do u know of any rules to find these? or to you just have to guess the u sometimes?
anonymous
  • anonymous
and does the integral change from -1 to 2, to 1 to 2, or is that a typo?
anonymous
  • anonymous
The integral changes because \(u = (x+2)^{1/2}\) so when x = -1, u = 1 and when x = 2, u = 2
anonymous
  • anonymous
right right, forgot about that, when you said you forgot the 2, is it ]2u^2-2]
anonymous
  • anonymous
\(2[u^2 - 2]\)
anonymous
  • anonymous
ah right, that would make more sense, well thx again
anonymous
  • anonymous
Do you see how I arrived at \(2[u^2-2]\)?
anonymous
  • anonymous
Cause I kinda skipped some steps there.
anonymous
  • anonymous
well no, but I understand that it's the right equation to the integral
anonymous
  • anonymous
I can show the steps.
anonymous
  • anonymous
there are steps to it?
anonymous
  • anonymous
Yes certainly.
anonymous
  • anonymous
awesome
anonymous
  • anonymous
Since \( u = (x+2)^{1/2} \implies x = u^2 -2\) \(\implies dx = 2u\ du\) \[\implies \frac{x}{(x+2)^{1/2}}dx = \frac{u^2-2}{u} *2u\ du\] \[=2[u^2-2]\ du\]
anonymous
  • anonymous
awesome thanks a lot, I don't think we even learned that yet
anonymous
  • anonymous
do you always implicitly differentiate x? (x'=dx and not 1?)
anonymous
  • anonymous
I found that out nvm

Looking for something else?

Not the answer you are looking for? Search for more explanations.