Question

need help with integrating w/ radicals
will post eq w/ editor

Answer 1

\[\int\limits_{1}^{\sqrt[3]{2}} (s^2+\sqrt[3]{s})/s^2 ds\]

Answer 2

I suggest to simplify the integrand. Divide both terms by s^2

Answer 3

Then, you power rule for the remaining 's' term. If you get what I'm saying.

Answer 4

= 1-3/2s^-2/3 + C

Answer 5

try 1 + s^1/3 / s^2 ds

Answer 6

how you get that

Answer 7

\[\int_1^{\sqrt[3]{2}} \frac{s^2+\sqrt[3]{s}}{s^2} \, ds = \int_1^{\sqrt[3]{2}} ds + \int_1^{\sqrt[3]{2}} \frac{s^{1/3}}{s^2}\,ds = \int_1^{\sqrt[3]{2}} ds + \int_1^{\sqrt[3]{2}} {s^{-5/3}}\,ds \]

Answer 8

yeh like that ^

Answer 9

sorry crashed

Answer 10

@whpalmer4 how did you get s^-5/3

Answer 11

@iPwnBunnies little help

Answer 12

Well, Palmer did what I would be. Split the fraction into two integrals, and simplify.
s^(1/3)/s^2 simplifies to s^(-5/3); that's a property of powers when it comes to variables

Answer 13

so the 1-3/2s^-2/3 isnt right

Answer 14

same thing?

Answer 15

Omg, I just wrote something out but it didn't post.

Answer 16

No, look at the second integral Palmer wrote out.
s^(1/3) / s^2 simplies to s^(-5/3), a property of powers, right?

Answer 17

show me please you just add right 1/3 + -3 ?

Answer 18

You'll get this now:
\[\int\limits_{1}^{\sqrt[3]{2}} 1 + s^{-5/2} ds\]

Answer 19

Err, I brought the s^2 up to the numerator:
s^(1/3)*s(-2)
Then yes, I added the powers.
(1/3) + (-6/3) = -5/3

Answer 20

Oh I messed up my equation, that it's supposed to be s^(-5/3)

Answer 21

oh yea you dont increase exp until you drop it back down, ok.

Answer 22

Eh, well, you change the sign of the power if you have to switch sides in a fraction. :3
Well now, you have a pretty simple integral. Use power rule now.

Answer 23

for the -5/3?
or just plug in limits?

Answer 24

s^-2/3 / -2/3

Answer 25

Yeah, we still have to integrate the integrand. :3

Answer 26

Right, you forgot to integrate the '1' as well doh xp

Answer 27

so back up

Answer 28

are you saying if there is a cube root, you have to change sides?

Answer 29

dont know where a 3 came from except cube root

Answer 30

No, we did that just to simplify the original problem. There was an s^2 in the denominator, right? And there were 's' terms in the numerator. The problem would be easier if we simplified that fraction.

Answer 31

Well, you can do that with any exponent and radical. Rewrite it as a fractional exponent.

Answer 32

ok which is 1 + s^-5/3 yes?

Answer 33

Right, we found the integrand simplifies to that. So, find the integral and evaluate it at the limits.

Answer 34

ok which is
1/2s^2 - 3/2 s^-2/3?

Answer 35

Whoa, the integral of a constant would just be the variable times that constant.

Answer 36

ok

Answer 37

s-3/2(s^(-2/3))

Answer 38

Yeah :3
\[\int\limits_{?}^{?} 1 ds = s + C\]

Answer 39

So we have [s - (3/2)s^(-2/3)] evaluated as s goes from 1 to cube root of 2, I believe.

Answer 40

yes, however the coefficient -3/2 is that always the reciprocal of the exponent?

Answer 41

but the same sign

Answer 42

Um, yes, so, here's the general form:
\[\int\limits_{?}^{?} x^n dx = \frac{x^{n}}{n} + C\]

When using the power rule, you are dividing by the raised power, which is the same as multiplying by the reciprocal of that raised power.
Like in this case, it's much easier to just right the fraction as a reciprocal.

Answer 43

Omg I messed that up T.T
\[\int\limits_{?}^{?} x^n dx = \frac{x^{n+1}}{n+1}\]

Answer 44

yea i got that but that just gets you s^-5/3 yes?

Answer 45

We had s^(-5/3) in the integrand, right. The power rule still applies to negative exponents. Except if it's -1 I guess. You'll use natural log.

Answer 46

ok so s^-2/3

Answer 47

Right, times the reciprocal of the raised power. (-3/2)*s^(-2/3), as I wrote a few posts ago.

Answer 48

so the coefficient is always the reciprocal yes?

Answer 49

Yes.

Answer 50

and keep the sign?

Answer 51

Yes.

Answer 52

ok cool now just plug in limits?

Answer 53

Yep, Fundamental Theorem of Calculus now.

Answer 54

f(b)-f(a)

Answer 55

Yessir.

Answer 56

Remember the integral was s - (3/2)s^(-2/3)