Question

length of curve y = e^x from 0 to 1

Answer 1

Do it the same as before

Answer 2

Use the same formula we just used in your last problem and see how far you can get. Ask when you're stuck, but I think it's fair that we don't work this problem for you from start to finish.

Answer 3

I'm at \[\int\limits_{0}^{1}\sqrt{1 + e ^{2x}}\] and I'm stuck

Answer 4

Use the most banal substitution you can imagine:

u = sqrt(1 + e^(2x) )

and work this through patiently.

Answer 5

how would i take the integral of it

Answer 6

I remember this integral very well: it was the last problem in my HS textbook. It's not hard, but you need to go through several steps.

Answer 7

find du/dx

Answer 8

okay so u = \[\sqrt{1 + e ^{2x}} \] so du = \[e ^{2x}/\sqrt{1+e ^{2x}}\]

Answer 9

You dropped a 2

Answer 10

but the 1/2 and 2 cancel out

Answer 11

my bad

Answer 12

okay but where do i go from there?

Answer 13

You want an integral in u, so express that derivative in u

Answer 14

but there is no du in the equation

Answer 15

Remember an integral looks like Sf(x)dx

Answer 16

We're using a substition

Answer 17

yeah but i set u as the whole integral, so du isn't in the integral

Answer 18

no, u is only sqrt(1 + e^(2x) )

Answer 19

there is still a dx hanging there...

Answer 20

which is the whole equation

Answer 21

okay so how do i put du back in the equation?

Answer 22

yeah i know how to do u substitution i just don't understand how to do it int his problem

Answer 23

We've used the simpliest, dumbest substitution possible,
f(x) = \sqrt(1 + e^(2x) ) = u, so g(u) = u

Answer 24

So what now is du/dx in terms of u? You need to calculate this and I really think if it takes you 5 minutes to figure it out, good, do it.

Answer 25

I know ti took me a while the first time as well.

Answer 26

what if i set u = e^x then did \[\int\limits_{0}^{1}\sqrt{1 + u ^{2}} \] which is just arctan u

Answer 27

No, it is NOT arctan u.

Answer 28

Use the substitution we gave you. It works.

Answer 29

i did a trig substitution
and now i'm using partial fractions lol
it does seem to be long

Answer 30

Yes it is long.

Answer 31

okay so right now i have integral u, how do i put in a du?

Answer 32

You need to find an expression for du/dx in terms of u and then use the substitution rule for integrals I wrote above.

Answer 33

So try!

Answer 34

You have almost got it, mate

Answer 35

I've got to go; conference call.

Answer 36

Have fun conference calling

Answer 37

i just don't understand how to put a du in, i'd have to put in the whole derivative somehow

Answer 38

Just substitute it in James' formula

Answer 39

I.e., integral of f(x) dx = u . dx/du du

So actually you need an expression for dx/du in terms of u; just the reciprocal of du/dx

Answer 40

i think james's way is easiest
so he said let u=(something)^(1/2)
then u^2=something
now taking derivative of both sides we get...?

Answer 41

Here's an example problem, so you can the method.

Evaluate
\[I = \int\limits \sqrt{4 + e^x} \ dx\]

Let
\[u = \sqrt{4 + e^x}\]
then
\[e^x = u^2 - 4\]
and
\[\frac{du}{dx} = \frac{e^x}{2 \sqrt{4+e^x}} = \frac{u^2 - 4}{2u}\]

Thus
\[I = \int\limits u(x) dx = \int\limits u(x) \frac{dx}{du} du = \int\limits \frac{2u^2}{u^2 - 4} du\]

Answer 42

Now,
\[\frac{2u^2}{u^2 - 4} = 2 + \frac{8}{u^2 - 4} = 2 + 2 \left( \frac{1}{u- 2} - \frac{1}{u+2} \right)\]
Thus
\[I = 2u + \ln(u-2) - \ln(u+2) + C\]

Answer 43

Corrections: I dropped a couple of 2s
\[I = 2u + 2\ln(u - 2) - 2 \ln(u + 2)\]

Answer 44

Now substitute back in u(x) = sqrt(4 + e^x)

Answer 45

\[I = \int\limits\limits \sqrt{4 + e^x} \ dx \]
\[u=\sqrt{4+e^x} => u^2=4+e^x =>2 u du=e^x dx\]
\[I=\int\limits_{}^{} \frac{e^x}{e^x} \sqrt{4+e^x} dx=\int\limits_{}^{} \frac{u \cdot 2udu}{u^2-4}\]
\[=\int\limits_{}^{}\frac{2(u^2-4)+8}{u^2-4} du=\int\limits_{}^{}(2+\frac{8}{u^2-4}) du\]
\[2u+8\int\limits_{}^{} \frac{1}{(u-2)(u+2)} du =2 \sqrt{4+e^x} +8\int\limits_{}^{}(\frac{A}{u-2}+\frac{B}{u+2}) du \]
....
and blah blah is the way i would have done it
my computer is freaking out so i can't finish

Answer 46

Yes, we have exactly the same thing.

Like I said, this was the last problem in my high school mathematics text book. I was very happy when I solved it the first time! :-)

Answer 47

i mean is the samething james has basically