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anonymous

  • 4 years ago

Help with solving integrals

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    I basically would like to understand how my book solved thisintegral

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  3. anonymous
    • 4 years ago
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    Like they totally ignored a negative

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  4. anonymous
    • 4 years ago
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    du=-1

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  5. anonymous
    • 4 years ago
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    whoops d(theta)=-1

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  6. anonymous
    • 4 years ago
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    Sorry there is some kind of bug. It is really all the same document

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  7. anonymous
    • 4 years ago
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    shldnt the answer be -(4-(theta))^-1

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  8. anonymous
    • 4 years ago
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    no the book is right

  9. anonymous
    • 4 years ago
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    http://letmegooglethat.com/?q=How+to+solve+integrals

  10. anonymous
    • 4 years ago
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    so like how did they ignore the negative

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  11. anonymous
    • 4 years ago
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    like if u check the answer you wont get the same integral

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  12. anonymous
    • 4 years ago
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    Th eintegral will be negative instead of poitive

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  13. anonymous
    • 4 years ago
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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, \[\lim_{a\rightarrow4^{+}}\left(-\displaystyle\int_{4-a}^{-2}u^{-2}du \right) = \lim_{a\rightarrow4^{+}} \left[-(-u^{-1})\right]^{-2}_{4-a} = \lim_{a\rightarrow4^{+}}u^{-1} \vert_{4-a}^{-2} = \lim_{a\rightarrow4^{+}}\left(\frac{1}{-2} - \frac{1}{4-a}\right)\]

  14. anonymous
    • 4 years ago
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    wait i am just gonna process this

  15. anonymous
    • 4 years ago
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    alright i get it but what i dont get is why u changed the bounds

  16. anonymous
    • 4 years ago
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    I would have done it differently

  17. anonymous
    • 4 years ago
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    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 \[u=4-\theta\]

  18. anonymous
    • 4 years ago
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    ok Thanks :D I got it

  19. anonymous
    • 4 years ago
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    Yup i got it thanks for ur help

  20. anonymous
    • 4 years ago
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    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!

  21. anonymous
    • 4 years ago
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    ya that is the way i did it

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