Question

calculus. someone who doesn't mind helping me with a "Sucky Integral"

Answer 1

\[y=e^5, 0\le x \le5\]

Answer 2

HI!!

Answer 3

hello!

Answer 4

that is not really a question is it?

Answer 5

\(e^5\) is a number

Answer 6

oops.. I meant \[e^x\]

Answer 7

still not a question

Answer 8

then o <= x <= 5

Answer 9

find the arc length

Answer 10

oooh
now it is a question

Answer 11

lol "now its a question"

Answer 12

\[\int _0^5\sqrt{1+e^{2x}}dx\]

Answer 13

\[e^x, 0\le x \le 5\] find the length of the curve

Answer 14

in general
\[\int_a^b\sqrt{1+(f'(x))^2}dx\]

Answer 15

i get lost at the second time they subsitute

Answer 16

\[u=5x, du= e^5 dx, \frac{ du }{ u }=dx\]
\[\int\limits_{1}^{e^5} \sqrt{1+u^2}\frac{ du }{ u }\]

Answer 17

\[\int\limits_{1}^{e^5} \ \frac{ \sqrt{1+u^2} }{ u^2 }du\]

Answer 18

oh you have a worked out solution? cuz this integral kind of sucks

Answer 19

\[\int\limits_{\sqrt2}^{\sqrt 1+e^{10}} \frac{ \sqrt (1+u^2) }{\sqrt( u^2) } du\]

Answer 20

yea. I was given an answer but I dont know how to work it really

Answer 21

apart from simplifying the square root with the u^2, I get lost here.

Answer 22

anyone who doesn't mind helping out with a sucky integral?

Answer 23

\[\int\limits_{0}^{5}\sqrt{1+(e^x)^2} dx \]
you can try a trib sub
let e^x=tan(u)

Answer 24

hmm i would say that makes it more complicated not easier

Answer 25

not @myininaya

Answer 26

they showed me an example with two different u subs

Answer 27

let me guess they did one sub for e^x as u
then they did u as tan(theta)

Answer 28

you can combine those so you only have to do one sub

Answer 29

no. They did u sub as i showed above and then made a v sub for the u sub and then applied partial fractions to that

Answer 30

they did v=u?

Answer 31

\[\int\limits_{0}^{5}\sqrt{1+(e^x)^2} dx \]
I don't understand your sub above...
maybe you mean to let u=e^x

Answer 32

if so we will do that sub first

Answer 33

yes \[v^2= 1 +u^2, ...u^2 = v^2-1,... v= \sqrt ( 1 u^2)\]

Answer 34

\[u=e^x \\ \frac{du}{dx}=e^x \\ du=e^x dx \\ \frac{du}{e^x}=dx \\ \frac{du}{u}=dx \text{ since } u=e^x \]

Answer 35

ok. BRB. sorry my phone

Answer 36

\[\int\limits_{x=0}^{x=5}\sqrt{1+(e^x)^2} dx \\ = \int\limits_{u=e^{0}}^{u=e^5} \sqrt{1+(u)^2} \frac{du}{u} \\ =\int\limits_{e^0}^{e^5} \frac{\sqrt{1+u^2}}{u} du \\ =\int\limits_1^{e^5} \frac{\sqrt{1+u^2}}{u} du\]
then you said you want to sub the inside of the square root...

Answer 37

before i go further
when you get back let me know if you understand what I have so far

Answer 38

yes, that is exactly what i typed above

Answer 39

accept i had a u^2 in denominator, so idk what i did there

Answer 40

\[v^2=1+u^2 \\ 2v dv=2u du \\ \text{ but if I divide both sides by 2 I have } v dv=u du \]

Answer 41

ok...

Answer 42

\[\frac{v}{u} dv=du \text{ by dividing } u \text{ on both sides } \\ \frac{v}{\sqrt{v^2-1}}dv= du \text{ since } v^2-1=u^2 \]

Answer 43

\[\int\limits\limits_{x=0}^{x=5}\sqrt{1+(e^x)^2} dx \\ = \int\limits\limits_{u=e^{0}}^{u=e^5} \sqrt{1+(u)^2} \frac{du}{u} \\ =\int\limits\limits_{e^0}^{e^5} \frac{\sqrt{1+u^2}}{u} du \\ =\int\limits\limits_{u=1}^{u=e^5} \frac{\sqrt{1+u^2}}{u} du \\ = \int\limits_{\sqrt{1+1^2}}^\sqrt{1+(e^5)^2} \frac{\sqrt{v^2}}{\sqrt{v^2-1}} \cdot \frac{v}{\sqrt{v^2-1}} dv \]

Answer 44

how do you feel about up to this point?

Answer 45

I am understanding you explanation better. they did partial fractions right about here

Answer 46

sqrt(a)*sqrt(a)=a

Answer 47

sqrt(a^2)=a when a>0

Answer 48

\[\int\limits_{\sqrt{1+1}}^{\sqrt{1+e^{10}}}\frac{v^2}{v^2-1} dv\]

Answer 49

\[\frac{v^2}{v^2-1}=\frac{v^2-1+1}{v^2-1}=\frac{v^2-1}{v^2-1}+\frac{1}{v^2-1}=1+\frac{1}{v^2-1}\]
so continuing what our integral looks like now:
\[\int\limits_{\sqrt{2}}^{\sqrt{1+e^{10}}}(1+\frac{1}{v^2-1}) dv\]

Answer 50

and yes partial fractions will be useful on the second term there

Answer 51

\[\frac{1}{v^2-1}=\frac{1}{(v-1)(v+1)}=\frac{A}{v-1}+\frac{B}{v+1}\]
do you know how to find A and B such that the equation I wrote is true.

Answer 52

I see what you are doing. It is so much better than \[1 + \frac{ 1 }{ 2 }/ (v-1)(v+1)\]

Answer 53

how does this compute \[\frac{ \sqrt{1+e^{10}}-1 }{ \sqrt{1+e^{10}}+1 }\]

Answer 54

there is an "ln" in front of that fraction that i forgot

Answer 55

ok you have went off to the end of the problem right?

Answer 56

yes. I see what the answer is, but when I am pluggin this in, I get confused with ln and the square roots

Answer 57

[ v + 1/2 ln (v-1) / (v+1) ]\[\sqrt (1+e^{10}) \to \sqrt2\]

Answer 58

that misty was right. this problem really sucks

Answer 59

Well let me get there too then because I don't know what the end is until I'm there.
I guess you don't have any problems finding A and B so I will do that...
\[\frac{1}{(v-1)(v+1)}=\frac{A}{v-1}+\frac{B}{v+1} \\ \text{ combining fractions } =\frac{A(v+1)+B(v-1)}{(v-1)(v+1)} \\ \text{ so we need \to find A and B such that } \\ 1=A(v+1)+B(v-1) \\ 1=(A+B)v+(A-B) \\ A+B+0 \text{ and } A-B=1 \\ A=-B \text{ so second equation \in terms of A is: } A+A=1 \\ A=\frac{1}{2} \text{ so } B=-\frac{1}{2}\]
So the integral now can be written as:
\[\int\limits\limits_{\sqrt{2}}^{\sqrt{1+e^{10}}}(1+\frac{1}{v^2-1}) dv \\ =\int\limits_{\sqrt{2}}^{\sqrt{1+e^{10}}}(1+\frac{1}{2} \frac{1}{v-1}-\frac{1}{2} \frac{1}{v+1} ) dv \\ =[v+\frac{1}{2} \ln|v-1|-\frac{1}{2}\ln|v+1| ]_\sqrt{2}^\sqrt{1+e^{10}}\]

Answer 60

i like the way you are setting this up better than what my course website showed me

Answer 61

\[=[v+\frac{1}{2}\ln|\frac{v-1}{v+1}|]_\sqrt{2}^\sqrt{1+e^{10}} \\ =(\sqrt{1+e^{10}}+\frac{1}{2} \ln|\frac{\sqrt{1+e^{10}}-1}{\sqrt{1+e^{10}}+1})-(\sqrt{2}+\frac{1}{2} \ln |\frac{\sqrt{2}-1}{\sqrt{2}+1}|)\]

Answer 62

so I guess they also transformed this answer a little too?

Answer 63

yes! this is where i stopped this time. this looks so confusing, my mind is having a hard time digesting all the rules here

Answer 64

the 1/2 (s) cancel on each other

Answer 65

move the sqrt2

Answer 66

I don't think the 1/2's cancel
but recall the power rule for log:
\[r \log(x)=\log(x^r)\]
maybe the 1/2 was moved

Answer 67

is it that the positive from the denominator moves to the top and the numerator cancels? \[\sqrt(1+e^{10}) -1 / \sqrt(1+e^{10}) +1 = \sqrt(1+e^{10}) +1 \]

Answer 68

but also I want to ask did they rationalize the bottoms in the ln( )'s ?

Answer 69

yes

Answer 70

that is what i am asking you how is done

Answer 71

\[\frac{\sqrt{a}-1}{\sqrt{a}+1} \cdot \frac{\sqrt{a}-1}{\sqrt{a}-1} =\frac{(\sqrt{a}-1)^2}{a-1}\]

Answer 72

do you understand that?

Answer 73

their final answer for this is \[L= \sqrt{1+e^{10}}- \sqrt{2}+ \ln (\sqrt{1+e^{10}}-1) -5 - \ln (\sqrt{2}-1)\]

Answer 74

yes i can see what you posted more clearly

Answer 75

\[=[v+\frac{1}{2}\ln|\frac{v-1}{v+1}|]_\sqrt{2}^\sqrt{1+e^{10}} \\ =(\sqrt{1+e^{10}}+\frac{1}{2} \ln|\frac{\sqrt{1+e^{10}}-1}{\sqrt{1+e^{10}}+1})-(\sqrt{2}+\frac{1}{2} \ln |\frac{\sqrt{2}-1}{\sqrt{2}+1}|) \\ \sqrt{1+e^{10}}-\sqrt{2}+\frac{1}{2}\ln|\frac{(\sqrt{1+e^{10}}-1)^2}{1+e^{10}-1}|-\frac{1}{2} \ln|\frac{(\sqrt{2}-1)^2}{2-1}|\]
how about this ?
is this cool so far

Answer 76

I just sorta used what I just said as like a formula

Answer 77

ok... it looks better on paper more spread out i guess

Answer 78

so where does the 5 come back into play?

Answer 79

\[=[v+\frac{1}{2}\ln|\frac{v-1}{v+1}|]_\sqrt{2}^\sqrt{1+e^{10}} \\ =(\sqrt{1+e^{10}}+\frac{1}{2} \ln|\frac{\sqrt{1+e^{10}}-1}{\sqrt{1+e^{10}}+1})-(\sqrt{2}+\frac{1}{2} \ln |\frac{\sqrt{2}-1}{\sqrt{2}+1}|) \\ =\sqrt{1+e^{10}}-\sqrt{2}+\frac{1}{2} \ln |(\sqrt{1+e^{10}}-1)^2)| \\ -\frac{1}{2} \ln|1+e^{10}-1|-\frac{1}{2} \ln|(\sqrt{2}-1)^2|+\frac{1}{2}\ln|2-1|\]
ran a little out of room those 2 bottom lines is where we are at
do you know which property of log I just used?
do you recall ln|a/b|=ln|a|-ln|b|

Answer 80

\[c \cdot \ln(\frac{a}{b})=c \cdot [\ln(a)-\ln(b)]=c \ln(a)- c \ln(b)\]

Answer 81

@jlockwo3 you there ?

Answer 82

We were almost to the end

Answer 83

just had to apply a couple more rules

Answer 84

\[\ln(1)=0 \\ a \ln(x)=\ln(x^a)\]

Answer 85

oh and ln(e)=1 :)