Help with solving integrals

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Help with solving integrals

Mathematics
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I basically would like to understand how my book solved thisintegral
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Like they totally ignored a negative
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du=-1
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whoops d(theta)=-1
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Sorry there is some kind of bug. It is really all the same document
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shldnt the answer be -(4-(theta))^-1
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no the book is right
http://letmegooglethat.com/?q=How+to+solve+integrals
so like how did they ignore the negative
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like if u check the answer you wont get the same integral
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Th eintegral will be negative instead of poitive
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Let \[u=4-\theta \] then \[\frac{du}{d\theta} = -1\] so \[d\theta = -du\] and hence our integral becomes, upon changing the limits, \[\lim_{a\rightarrow4^{+}}\left(-\displaystyle\int_{4-a}^{-2}u^{-2}du \right) = \lim_{a\rightarrow4^{+}} \left[-(-u^{-1})\right]^{-2}_{4-a} = \lim_{a\rightarrow4^{+}}u^{-1} \vert_{4-a}^{-2} = \lim_{a\rightarrow4^{+}}\left(\frac{1}{-2} - \frac{1}{4-a}\right)\]
wait i am just gonna process this
alright i get it but what i dont get is why u changed the bounds
I would have done it differently
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 \[u=4-\theta\]
ok Thanks :D I got it
Yup i got it thanks for ur help
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!
ya that is the way i did it

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