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

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

Mathematics
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use the exponent form of a square root... ^(1/2)
or..use substitution methods designed for radical problems...
got anything specific?

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Other answers:

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.
you cant divde addition up into seperate radical parts :)
ohh so you mean i have to put (36+3x)^1/2 and go from there?
\[\sqrt{36 + x} \neq \sqrt{36} + \sqrt{x}\]
yes, thats what I mean :) if we let u = 36+3x then we evaluate u^(1/2) du
u = 36 + 3x du = 3 dx du/3 = dx [S] (36 +3x)^(1/2) dx [S] u^(1/2) du/3
\[\int\limits_{} (36+3x)^{1/2} dx \rightarrow \int\limits_{} \frac{u^{1/2}}{3}du\]
it ate my fraction bar ...lol
any of this make sense?
i follow so far, yes. thank you so much :)
so now, i plug in the two #'s and subtract them from eachother right ?
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 :)
what do you mean higher function of F(u)?
\[\frac{1}{3} \int\limits_{a}^{b} u^{1/2}du \rightarrow \frac{1}{3} * \frac{2u^{3/2}}{3} = \frac{2u^{3/2}}{9}|_{u(a)}^{u(b)}\]
F(u) is a function of "u"
we can resubstitute the "value" of u back like this:\[\frac{2u^{3/2}}{9} \rightarrow \frac{2(36+3x)^{3/2}}{9}|_{a}^{b}\]
now it become F(x) since "x" is the variable again.... F(b) - F(a) gets your answer
okay i think i get it now. thank you so much
youre welcome :) it gets better with practice ;)

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