anonymous
  • anonymous
re-evaluating the integrals limits
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\int\limits_{0}^{10}\frac{ dx }{ 1+x }\]
anonymous
  • anonymous
I can't remember how to re-evaluate the limits after u-substitution
anonymous
  • anonymous
|dw:1370747567196:dw|

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anonymous
  • anonymous
$$\int_0^{10}\frac{dx}{1+x}=\left[\log(1+x)\right]_0^{10}=\log11-\log1=\log11$$
anonymous
  • anonymous
You simply plug in the values as x. If u = x+1 then for the bottom limit you have 0+1 = 1 and for the top you have 10+1 = 11 giving: \[\int\limits_1^{11} \frac{du}{u}\]
anonymous
  • anonymous
so, u=1+x then plug in the x valvues to get the new limits?
anonymous
  • anonymous
|dw:1370747587106:dw|
anonymous
  • anonymous
memory jogged... thx everyone
anonymous
  • anonymous
Which upon evaluation gives @oldrin.bataku's answer even though they are too worried about solving the integral to read the question @ChmE :P
anonymous
  • anonymous
Medal? :D
anonymous
  • anonymous
And medals...
anonymous
  • anonymous
@ChmE your bounds are \(x\) values; you "pull" them "back" by passing through our change of variables. Given \(u=1+x\) and \(x=0\) then \(u=1+0=1\); if \(x=10\), \(u=11\).
anonymous
  • anonymous
@ChmE or you can change your U back to X+1 and keep your old limits
anonymous
  • anonymous
after you integrate. Gotcha
anonymous
  • anonymous
Undoing our pullback is only really needed if you're doing an indefinite intgeral.

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