Question

The next installment of the Integral Madness

Answer 1

\[\int\limits_ e^x(1-e^x)(1+e^x)^9dx\]

Answer 2

Try substitution :D :P d:

Answer 3

Wow actually

Answer 4

Um there isn't supposed to an x or an e on the integral sign.

Answer 5

I was wondering about that...lol

Answer 6

Yeah lol ok

Answer 7

I'd recommend the substitution \(u=1+e^x\).

Answer 8

u=1+e^x but you'll still get some long answer I'm sure

Answer 9

If anyone would like to know the answer I can post it.

Answer 10

But I don't know how to get to the answer

Answer 11

that substitution works nicely,
u wil get ideas if u start

Answer 12

maybe not, i see it wont be that smooth :O

Answer 13

du(u+2)(u^9) right?

Answer 14

I mean du(-u+2)(u^9)

Answer 15

LOL haha

Answer 16

looks you're assuming dx = du

Answer 17

I'm saying du= e^x(dx)

Answer 18

\[u = 1+e^x \implies dx = \dfrac{du}{u-1}\]

Answer 19

I think I got it.

Answer 20

You can actually also expand it and find the integral that might be easier :d

Answer 21

The u substitution that batman suggested works.

Answer 22

yeah that's what did

Answer 23

thanks for all the help

Answer 24

\[\int\limits(1-e^x)(1+e^x)^9dx = \int (1+e^x)^9dx - \int e^x(1+e^x)^9 dx \]

evaluating second integral is trivial with u substitution u = 1+ e^x

for first integral you could try below :
\[\int (1+e^x)^9 dx = \int \sum \limits_{k=0}^9e^{kx} ~dx = \sum \limits_{k=0}^9\dfrac{e^{kx}}{k} \]

Answer 25

Oh damn

Answer 26

I don't know if he knows power series though, if he's doing integrals right now haha. But that's nice xD
I think it would be easiest to just expand and integrate.

Answer 27

thats binomial theorem, not power series batty

Answer 28

Oh my

Answer 29

I just read about binomial a few days back to bleh lol

Answer 30

Well some of it haha

Answer 31

\[(a+b)^n = \sum\limits_{k=0}^n\binom{n}{k}a^kb^{n-k}\]

Answer 32

Oh yeah I asked kai what that notation was (n k) one he didn't know either

Answer 33

What does it mean

Answer 34

there is a mistake in previous post, i think thats the reason thought it was power series.. leme correct it

Answer 35

corrected : \[\int (1+e^x)^9 dx = \int \sum \limits_{k=0}^9 \binom{9}{k}e^{kx} ~dx = \sum \limits_{k=0}^9\binom{9}{k}\dfrac{e^{kx}}{k}\]

Answer 36

Nice :P

Answer 37

\[\binom{n}{k} = ^nC_k = nCk = \dfrac{n!}{k!(n-k)!}\]

Answer 38

\[f(x) = \sqrt{x}, ~~~a = 16\] I couldn't figure that part out

Answer 39

Ah that makes much more sense

Answer 40

they all represent same : number of ways of choosing k things from n things

Answer 41

if we assume n.k are naturals

Answer 42

I was doing a taylor series that was it and I couldn't figure some parts of it I needed the binomial theorem

Answer 43

Very nice

Answer 44

you're working taylor series of sqrt(x) centered at 16 ?

Answer 45

Yeah haha

Answer 46

okay got you :) taylor series doesn't require u to work binomial coefficients directly right ?

Answer 47

you will be messing with factorials instead

Answer 48

Like I found the pattern to be \[\frac{ 1 }{ 4^{2n-1} }(x-16)^n\] but then I got stuck because I guess I need binomial coefficients I should say

Answer 49

Ohk.. is that what you get for nth derivative of sqrt(x) at x=16 ?

Answer 50

Yup

Answer 51

I see, binomial thm makes ur life simple here yeah

Answer 52

\[\sum_{n=0}^{\infty} \left(\begin{matrix}1/2 \\ n\end{matrix}\right) = 1 + \frac{ 1 }{ 2 }+\frac{ 1/2(1/2-1) }{ 2! }+\frac{ 1/2(1/2-1)(1/2-2) }{ 3! }+...\] so this right?

Answer 53

binomial thm is more general+findamental than any other thm and thats one of the reasons why 0^0 is defined to be 1 for most parts of math

Answer 54

Oh wow, that makes a lot of sense lol

Answer 55

In a way it fills in the "gaps" haha

Answer 56

you're using generalized binomial thm for rational powers

Answer 57

Hmmmm. Fascinating. I have no idea what you two are talking about, but seems awesome.

Answer 58

Ah, ok. Lol don't worry I'm still learning this stuff, I'll have to read up on it.

Answer 59

Hey so with power series and taylor/ mclaurin I can do nearly any integral now right? Haha

Answer 60

that looks good, just use the modified definition for binomial coefficient

Answer 61

Is power series used in derivatives?

Answer 62

\[\large \binom{n}{k} = \dfrac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\]

Answer 63

just use the regular binomial thm and that modified coefficient formula for rational powers

Answer 64

alright cool

Answer 65

I like to call derivatives pessimistic they don't have many rules not that I can think off haha. @ArkGoLucky

Answer 66

They like to get rid of stuff

Answer 67

the replies keep dancing on my screen :o

Answer 68

Every problem on OS can be fixed by a laggy refresh xD

Answer 69

Coool

Answer 70

I used to hate series, but I'm starting to like them, maybe because I did so many last weekend...

Answer 71

which series do u like the most ?

Answer 72

I like power series problems, they're fun :P

Answer 73

My most favorite series is \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}\). :p

Answer 74

And when it comes to the tests, ratio test is by far the best as it gets rid of factorials

Answer 75

@myusernameisfuntosay Get the derivatives and solve for the 0's

Answer 76

You are a funny man mr. funny usernane

Answer 77

Unless you're a girl. Then ignore everything I said

Answer 78

I learned series problems last year, but I forgot most everything. I'm pretty bummed about that.

Answer 79

Oh guess you haven't done calculus yet