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anonymous

  • 5 years ago

How do you evalute definite integrals with a radical in it?

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  1. amistre64
    • 5 years ago
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    use the exponent form of a square root... ^(1/2)

  2. amistre64
    • 5 years ago
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    or..use substitution methods designed for radical problems...

  3. amistre64
    • 5 years ago
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    got anything specific?

  4. anonymous
    • 5 years ago
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    its the square root of 36+3x i tried putting 36^1/2 + 3x^1/2 but i get lost on what to do next.

  5. amistre64
    • 5 years ago
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    you cant divde addition up into seperate radical parts :)

  6. anonymous
    • 5 years ago
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    ohh so you mean i have to put (36+3x)^1/2 and go from there?

  7. amistre64
    • 5 years ago
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    \[\sqrt{36 + x} \neq \sqrt{36} + \sqrt{x}\]

  8. amistre64
    • 5 years ago
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    yes, thats what I mean :) if we let u = 36+3x then we evaluate u^(1/2) du

  9. amistre64
    • 5 years ago
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    u = 36 + 3x du = 3 dx du/3 = dx [S] (36 +3x)^(1/2) dx [S] u^(1/2) du/3

  10. amistre64
    • 5 years ago
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    \[\int\limits_{} (36+3x)^{1/2} dx \rightarrow \int\limits_{} \frac{u^{1/2}}{3}du\]

  11. amistre64
    • 5 years ago
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    it ate my fraction bar ...lol

  12. amistre64
    • 5 years ago
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    any of this make sense?

  13. anonymous
    • 5 years ago
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    i follow so far, yes. thank you so much :)

  14. anonymous
    • 5 years ago
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    so now, i plug in the two #'s and subtract them from eachother right ?

  15. amistre64
    • 5 years ago
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    if it had an interval set; then we integrate this new integral to get the higher function of F(u); then resubstitute back in the "x" parts from the "u" parts to get the F(x) OR... modify the original interval to work for u instead if x :)

  16. anonymous
    • 5 years ago
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    what do you mean higher function of F(u)?

  17. amistre64
    • 5 years ago
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    \[\frac{1}{3} \int\limits_{a}^{b} u^{1/2}du \rightarrow \frac{1}{3} * \frac{2u^{3/2}}{3} = \frac{2u^{3/2}}{9}|_{u(a)}^{u(b)}\]

  18. amistre64
    • 5 years ago
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    F(u) is a function of "u"

  19. amistre64
    • 5 years ago
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    we can resubstitute the "value" of u back like this:\[\frac{2u^{3/2}}{9} \rightarrow \frac{2(36+3x)^{3/2}}{9}|_{a}^{b}\]

  20. amistre64
    • 5 years ago
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    now it become F(x) since "x" is the variable again.... F(b) - F(a) gets your answer

  21. anonymous
    • 5 years ago
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    okay i think i get it now. thank you so much

  22. amistre64
    • 5 years ago
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    youre welcome :) it gets better with practice ;)

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