Question

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

Answer 1

use the exponent form of a square root... ^(1/2)

Answer 2

or..use substitution methods designed for radical problems...

Answer 3

got anything specific?

Answer 4

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.

Answer 5

you cant divde addition up into seperate radical parts :)

Answer 6

ohh so you mean i have to put (36+3x)^1/2 and go from there?

Answer 7

\[\sqrt{36 + x} \neq \sqrt{36} + \sqrt{x}\]

Answer 8

yes, thats what I mean :) if we let u = 36+3x then we evaluate u^(1/2) du

Answer 9

u = 36 + 3x

du = 3 dx

du/3 = dx

[S] (36 +3x)^(1/2) dx

[S] u^(1/2) du/3

Answer 10

\[\int\limits_{} (36+3x)^{1/2} dx \rightarrow \int\limits_{} \frac{u^{1/2}}{3}du\]

Answer 11

it ate my fraction bar ...lol

Answer 12

any of this make sense?

Answer 13

i follow so far, yes. thank you so much :)

Answer 14

so now, i plug in the two #'s and subtract them from eachother right ?

Answer 15

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 :)

Answer 16

what do you mean higher function of F(u)?

Answer 17

\[\frac{1}{3} \int\limits_{a}^{b} u^{1/2}du \rightarrow \frac{1}{3} * \frac{2u^{3/2}}{3} = \frac{2u^{3/2}}{9}|_{u(a)}^{u(b)}\]

Answer 18

F(u) is a function of "u"

Answer 19

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

Answer 20

now it become F(x) since "x" is the variable again....

F(b) - F(a) gets your answer

Answer 21

okay i think i get it now. thank you so much

Answer 22

youre welcome :) it gets better with practice ;)