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anonymous
 5 years ago
How do you evalute definite integrals with a radical in it?
anonymous
 5 years ago
How do you evalute definite integrals with a radical in it?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0use the exponent form of a square root... ^(1/2)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or..use substitution methods designed for radical problems...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0got anything specific?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its the square root of 36+3x i tried putting 36^1/2 + 3x^1/2 but i get lost on what to do next.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you cant divde addition up into seperate radical parts :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh so you mean i have to put (36+3x)^1/2 and go from there?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{36 + x} \neq \sqrt{36} + \sqrt{x}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yes, thats what I mean :) if we let u = 36+3x then we evaluate u^(1/2) du

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0u = 36 + 3x du = 3 dx du/3 = dx [S] (36 +3x)^(1/2) dx [S] u^(1/2) du/3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} (36+3x)^{1/2} dx \rightarrow \int\limits_{} \frac{u^{1/2}}{3}du\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it ate my fraction bar ...lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0any of this make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i follow so far, yes. thank you so much :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now, i plug in the two #'s and subtract them from eachother right ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if 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 :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean higher function of F(u)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{3} \int\limits_{a}^{b} u^{1/2}du \rightarrow \frac{1}{3} * \frac{2u^{3/2}}{3} = \frac{2u^{3/2}}{9}_{u(a)}^{u(b)}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0F(u) is a function of "u"

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we can resubstitute the "value" of u back like this:\[\frac{2u^{3/2}}{9} \rightarrow \frac{2(36+3x)^{3/2}}{9}_{a}^{b}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now it become F(x) since "x" is the variable again.... F(b)  F(a) gets your answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay i think i get it now. thank you so much

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0youre welcome :) it gets better with practice ;)
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