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Mathematics 20 Online
OpenStudy (anonymous):

how to find anti-derivative for (x^1/3-3)/(x^2/3) and (e^x +3)/(e^x)

OpenStudy (loser66):

1/ = x^(1/3)(x ^(-2/3) - 3 x^(-2/3) take int to get the answer 2/ =1 + 3e^(-x) take int to get the answer

OpenStudy (anonymous):

what is int?

OpenStudy (anonymous):

i did everything you did but I am confused for finding the anti-derivitive

OpenStudy (loser66):

antiderivative is integral

OpenStudy (anonymous):

can you go through it step by step with me

OpenStudy (loser66):

@zepdrix

OpenStudy (anonymous):

i am confused because of the product rule and quotient rule

zepdrix (zepdrix):

\[\Large\bf\sf \frac{x^{1/3}-3}{x^{2/3}}\quad=\quad \frac{x^{1/3}}{x^{2/3}}-\frac{3}{x^{2/3}}\quad=\quad x^{-1/3}-3x^{-2/3}\]Simplify before looking for the anti-derivative. Understand those steps?

zepdrix (zepdrix):

We don't have a nice product rule or quotient rule for anti-differentiation. So simplifying first is necessary.

OpenStudy (anonymous):

yes i already did that but i am confused for what to do next

zepdrix (zepdrix):

Next we apply the `Power Rule for Anti-Differentation`: The anti-derivative of \(\Large\bf\sf x^n\) will be \(\Large\bf\sf \dfrac{1}{n+1}x^{n+1}\) 2 steps: ~Increase the power by 1, ~Divide by the new power. This looks familiar yes?

zepdrix (zepdrix):

Our exponents are fractions, so it will be a little bit messier.

OpenStudy (anonymous):

yes i understand that

OpenStudy (anonymous):

I am having trouble understanding your abbreviations though

zepdrix (zepdrix):

abbreviations? You shouldn't see abbreviations. Is the math code not loading correctly?

OpenStudy (anonymous):

its not loading

OpenStudy (anonymous):

\(\Large\bf\sf x^n\) will be \(\Large\bf\sf \dfrac{1}{n+1}x^{n+1}\)

zepdrix (zepdrix):

Ugh, the LaTeX plugin has been acting weird today :( Are you using Internet explorer?

OpenStudy (anonymous):

google chrome

OpenStudy (anonymous):

i could try safari?

zepdrix (zepdrix):

It should work fine on Chrome.. that really stinks.

zepdrix (zepdrix):

The alternative is the drawing tool, which I know doesn't work.. Hmm :p

OpenStudy (anonymous):

i tried inserting it on wolfram

zepdrix (zepdrix):

antiderivative of x^n is (1/(n+1)) * x^(n+1)

zepdrix (zepdrix):

I can do it like that I guess.. it's just a lot uglier XD lol

OpenStudy (anonymous):

lol its fine. Thanks for your help. I really appreciate it.

zepdrix (zepdrix):

So for the first term, The -1/3 .... adding 1 will change it to 2/3. Then we divide by that new exponent, (which is the same as multiplying by the reciprocal, yes?

zepdrix (zepdrix):

We would divide by 2/3, or multiply by 3/2

OpenStudy (anonymous):

okay i follow you there

OpenStudy (anonymous):

so what would it look like expressed out? 2/3x^4/3

zepdrix (zepdrix):

So antiderivative of the first term: (3/2)x^(2/3)

OpenStudy (anonymous):

okay so we are going to find all antiderivities in 1 step? Should i use the anti-derivitive blank integral sign?

zepdrix (zepdrix):

woops, i had simplified things down earlier.. maybe the code didn't show up though. We're antidifferenting these two terms: x^(-1/3)-3x^(-2/3)

zepdrix (zepdrix):

blank integral sign? :o you mean the curly S thing?

OpenStudy (anonymous):

yeah sorry

OpenStudy (anonymous):

and there is no x^-1/3??

OpenStudy (anonymous):

can you just do it and take a snap shot and upload it?

zepdrix (zepdrix):

Mmm lemme see if I can pull that off. One sec lemme test the drawing tool

zepdrix (zepdrix):

M:mm darn no drawing tool.. one sec

OpenStudy (anonymous):

okay thanks

zepdrix (zepdrix):

Okay let's see if this works or not.

OpenStudy (anonymous):

it works. Thanks!!!

zepdrix (zepdrix):

For the next problem, is the 3 supposed to be in the exponent position or no? e^(x-3) or e^x - 3

zepdrix (zepdrix):

Err it's +3 but hopefully you know what I'm asking :x

OpenStudy (anonymous):

wolfram alpha and copy and paste it.

zepdrix (zepdrix):

The next one is very similar.

zepdrix (zepdrix):

If you get confused on the anti-derivative of e^(-x), hmm I'm not really sure if I can explain that in a way that you'll understand. I wonder if you've learned enough to be able to do that. Have you learned U-substitution for integration yet?

OpenStudy (anonymous):

no.

OpenStudy (anonymous):

I can tell my teacher it was undoable and he will understand

OpenStudy (anonymous):

thanks for all of your help

zepdrix (zepdrix):

np c:

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