anonymous
  • anonymous
Help with solving integrals
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
I basically would like to understand how my book solved thisintegral
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anonymous
  • anonymous
Like they totally ignored a negative
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anonymous
  • anonymous
du=-1
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anonymous
  • anonymous
whoops d(theta)=-1
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anonymous
  • anonymous
Sorry there is some kind of bug. It is really all the same document
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anonymous
  • anonymous
shldnt the answer be -(4-(theta))^-1
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anonymous
  • anonymous
no the book is right
anonymous
  • anonymous
http://letmegooglethat.com/?q=How+to+solve+integrals
anonymous
  • anonymous
so like how did they ignore the negative
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anonymous
  • anonymous
like if u check the answer you wont get the same integral
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anonymous
  • anonymous
Th eintegral will be negative instead of poitive
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anonymous
  • anonymous
Let \[u=4-\theta \] then \[\frac{du}{d\theta} = -1\] so \[d\theta = -du\] and hence our integral becomes, upon changing the limits, \[\lim_{a\rightarrow4^{+}}\left(-\displaystyle\int_{4-a}^{-2}u^{-2}du \right) = \lim_{a\rightarrow4^{+}} \left[-(-u^{-1})\right]^{-2}_{4-a} = \lim_{a\rightarrow4^{+}}u^{-1} \vert_{4-a}^{-2} = \lim_{a\rightarrow4^{+}}\left(\frac{1}{-2} - \frac{1}{4-a}\right)\]
anonymous
  • anonymous
wait i am just gonna process this
anonymous
  • anonymous
alright i get it but what i dont get is why u changed the bounds
anonymous
  • anonymous
I would have done it differently
anonymous
  • anonymous
I used the substitution \[u=4-\theta\] and to find the new bounds, i.e. the ones for the integral in terms of u, one must substitute the old bounds into the theta position i.e. the first bound is \[4-(6) = -2\] and the second is \[4-(a) = 4-a\] in both cases, we use the equation \[u=4-\theta\]
anonymous
  • anonymous
ok Thanks :D I got it
anonymous
  • anonymous
Yup i got it thanks for ur help
anonymous
  • anonymous
In fact, substitution doesn't change anything about the integral really, it's still the same integral, just expressed differently. If we rearrange to get \[\theta = 4-u\] and then continue from here: \[...=\lim_{a\rightarrow 4^{+}}u^{-1}\vert_{4-a}^{-2}\] we see that the bounds may be changed via:\[4-(-2) = 6\] and \[4-(4-a) = a\] now we simply sub back \[u=4-\theta \] to arrive at the same thing your book had in the second to last step: \[=\lim_{a\rightarrow 4^{+}}(4-\theta )^{-1}\vert^{6}_{a}\] so really, the true 'identity' of this integral is unchaged under the substitution - you can uncover it back!
anonymous
  • anonymous
ya that is the way i did it

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