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anonymous
 4 years ago
Help with solving integrals
anonymous
 4 years ago
Help with solving integrals

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I basically would like to understand how my book solved thisintegral

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Like they totally ignored a negative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry there is some kind of bug. It is really all the same document

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0shldnt the answer be (4(theta))^1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so like how did they ignore the negative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0like if u check the answer you wont get the same integral

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Th eintegral will be negative instead of poitive

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let \[u=4\theta \] then \[\frac{du}{d\theta} = 1\] so \[d\theta = du\] and hence our integral becomes, upon changing the limits, \[\lim_{a\rightarrow4^{+}}\left(\displaystyle\int_{4a}^{2}u^{2}du \right) = \lim_{a\rightarrow4^{+}} \left[(u^{1})\right]^{2}_{4a} = \lim_{a\rightarrow4^{+}}u^{1} \vert_{4a}^{2} = \lim_{a\rightarrow4^{+}}\left(\frac{1}{2}  \frac{1}{4a}\right)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait i am just gonna process this

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright i get it but what i dont get is why u changed the bounds

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I would have done it differently

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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 \[4(a) = 4a\] in both cases, we use the equation \[u=4\theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok Thanks :D I got it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yup i got it thanks for ur help

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In fact, substitution doesn't change anything about the integral really, it's still the same integral, just expressed differently. If we rearrange to get \[\theta = 4u\] and then continue from here: \[...=\lim_{a\rightarrow 4^{+}}u^{1}\vert_{4a}^{2}\] we see that the bounds may be changed via:\[4(2) = 6\] and \[4(4a) = a\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0ya that is the way i did it
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