Integration Problem?

Question

Could you please tell me what the answer is and how to figure out this problem.

Find the length of the curve y=e^x
0<=x=<1

Answer 1

I'll work out the hardest part of this and let you finish it.
As was pointed out, the arclength of this curve is
given by the integral
Int(sqrt(1 + e^(2x) ) dx) between 0 and 1.
So the question is, how does one evaluate this
integral? I'll work it without limits for now, just
to illustrate the techniques.
First, let u = e ^x, x = ln u, dx = du/u.
Then the integral becomes

int( sqrt(1 + u^2)/u du).

I tried several ideas on this one, but the
only one that worked was to rationalise
the numerator! Doing this, and splitting
the integral in 2 gives

int( du/(u*sqrt(1 + u^2)) + int (u du/sqrt(1 + u^2)).
The second integral immediately evaluates
(by a simple substitution) to sqrt(1 + u^2).
So, how to do the first one?
Here a trig substitution works nicely.
Try letting u = tan t and see what happens!
I know this is only an outline of the work,
but I hope it's enough for you to finish the problem!

Answer 2

you need the formula for arc length
You need the function and its derivative to be continuous (no prob for this particular case)
The length s of the part of the graph of the function between x = 0 and x = 1 is found by the formula
s= the integral between 0 and 1 of:
square root (1+(f'(x))^2)
as in this case f'(x)=e^x you have

the integral between 0 and 1 of:
square root (1+(e)^(2x))

Hope its clear, not being able to include symbols here makes it complicated

Answer 3

i forgot how to do this but look up line integral. the first two answers here are not right.

Answer 4

You take the integral from 0 to 1 of (e^x)dx.

so the integral of (e^x)dx is e^x.

so your answer is e^1 - e^0 = e - 1 which is about 1.71

Answer 5

You integrate e^x between the endpoints you listed. The exponential function is unique because it is its own integral and derivative. Therefore, the answer to you problem is e^1-e^0=1.718.