Question

Find the Length of the curve y = e^x 0<=x<=5

Answer 1

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

Answer 2

there is suppose to be a space after e^x

Answer 3

o to 5 is the limit

Answer 4

they have suggest u sub twice for me. but that has just confused me

Answer 5

\[\int\limits_{0}^{1} \sqrt{1+(e^ (2x))}\]

Answer 6

that is e^2x i dont know how to make the x up there with the 2 on the equation thing

Answer 7

```
e^{2x}
```
That'll do it =)

Answer 8

this is where they want me to do the first u sub

Answer 9

thanks

Answer 10

i guess the e has confused me

Answer 11

Sure, this is kind of a weird integral. What did they suggest to you that was confusing? It might be better to just explain what they're telling you to do, since they probably picked the best thing already.

Answer 12

ok, i will type the equation for you. give me a min

Answer 13

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

Answer 14

where u= e^x x= lnu then dx=du/u

Answer 15

but Idk how they got that

Answer 16

its like, they show me WHAT they got, but not HOW

Answer 17

then, they go right into the 2nd u sub here

Answer 18

v= \[\sqrt{1+u^2}\] v^2=\[1+u^2\] and vdv= udu

Answer 19

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

Answer 20

this is where I am COMPLETELY lost now! I cant figure out what they did

Answer 21

why would the limits become what they did?

Answer 22

Ok, also I think you might be wrong when you wrote these down, I think they should have been \[\int\limits_{1}^{e} \sqrt{\frac{ 1+u^2 }{ u^2 }}du\] and then \[\int\limits_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{ v }{ v^2-1 }dv\]
instead, without the extra u and v next to du and dv.

Answer 23

i that that was weird too

Answer 24

Wait, I'm also confused, I thought you said the limits were 0 to 5 and here you've written 0 to 1? \[\int\limits_{0}^{1} \sqrt{1+e^ {2x}}\]

Answer 25

cuz they made du, du/u

Answer 26

yes... im sorry... I meant 1! I am having a hard time with equation thing and I get confused

Answer 27

I have a 2nd problem where it is five.

Answer 28

this one is 1

Answer 29

That's fine, here I'll show you how I will do the first substitution step by step. =)

Answer 30

thank you

Answer 31

Before I get started, this is the correct integral, right? \[\Large \int\limits_{x=0}^{x=1}\sqrt{1+e^{2x}}dx\]

Answer 32

yes

Answer 33

First make the substitution\[\Large u = e^x \] Now take its derivative \[\Large \frac{du}{dx}=e^x\] Multiply both sides by dx \[\Large du = e^x dx\] Now we look and see we need to plug in e^x and dx from our original integral. So we need to solve for dx: \[\Large \frac{du}{e^x}=dx\] and we use the substitution to make this correct: \[\Large dx=\frac{du}{u}\] So we can plug these in: \[\Large \int\limits_{x=0}^{x=1} \sqrt{1+(u)^2} ( \frac{du}{u})\] but notice our bounds are still in terms of x! That's ok, we now just plug in our bounds to the substitution we made as well! \[\Large u=e^0=1 \\ \Large u=e^1=e\] just replace the bounds: \[\Large \int\limits\limits_{u=1}^{u=e} \frac{\sqrt{1+u^2}}{u}du \]We can of course square and square root the bottom to make it look similar to what they have: \[\Large \int\limits_{u=1}^{u=e}\sqrt{\frac{1+u^2}{u^2}}du\] and that's it for this part.

Answer 34

ok. let me look at this for a min plz

Answer 35

before we move on, could u tell me how the u becomes squared in the denominator?

Answer 36

oh wait... i see

Answer 37

because you took the root of it?

Answer 38

now...how do the limits become \[\int\limits_{\sqrt{2}}^{\sqrt{1+e^2}}\]

Answer 39

Try to do the same thing I just did on your own with the new substitution and I'll help you out. \[\Large v=\sqrt{1+u^2}\]

Answer 40

ok. because they are removing the square root and changing 1-e to 2-the function

Answer 41

ok, I have the final answer, but I was wondering if i could see how someone did their arithmetic.

Answer 42

i believe partial fractions was used

Answer 43

Yeah I think so too. Want to show me your work on attempting the partial fractions? It doesn't do either of us any good if I just do it, it's better for you to get some practice and I can help see where you need help. I know like two or three methods to do partial fractions so I might be able to give you a tip to make it easier for you.

Answer 44

sure. reminder... I am terrible at equation thing and will need some time to put all of this in

Answer 45

Haha believe me when I say I wasn't born knowing how to do partial fractions either. I practiced messing them up tons of times XD

Answer 46

\[\int\limits_{\sqrt{2}}^{\sqrt{1+e^2}} \frac{ v }{ v^2-1}dv\] =\[\int\limits_{\sqrt{2}}^{\sqrt{1+e^2}} 1+(\frac{ 1/2 }{ (v-1)(v+1) }dv\] =\[[v+\frac{ 1 }{ 2 }\ln(\frac{ v-1 }{ v+1 })]\]=\[(\sqrt{1+e^2}+\frac{ 1 }{ 2 }\ln (\frac{ \sqrt{1+e^2}-1 }{ \sqrt{1+e^2}+1 }))-(\sqrt{2}+\frac{ 1 }{ 2 }\ln(\sqrt{2}))\]

Answer 47

final answer is \[\sqrt{1+e^2}-\sqrt{2}+\ln (\sqrt{1+e^2}-1)-1-\ln(\sqrt{2}-1)\]

Answer 48

i still feel confused lol but that ok. you helped me a lot

Answer 49

L= of course

Answer 50

Hmm, sorry I'm having trouble concentrating, I stayed up all night working on stuff so I guess I'm finally braindead haha. I'll take a look tomorrow if you still want me to.

Answer 51

sure :) i truly appreciate the help i get. And..I understand. Somedays, im like "I GOT THIS" and the other days Its like I forget ALL THE RULES for EVERYTHING and don't know how to do it anymore

Answer 52

especially with e and lne and all that dumb stuff like square roots! My brain gets confused just looking at it

Answer 53

Haha yeah I know what you mean. It's a good feeling when you finally understand, actually I think one of the things that helped me as to teach other people. Somehow explaining things to other people is almost like I'm explaining it to myself and I end up learning at the same time somehow haha.