Arc Length on an Interval?

Question

Using a graphing utility, approximate the arc length of the graph of f(x)=e^x on the interval (0,20)

I think I'm supposed to use the formula (Integral) Squareroot of (1+f(x)^2), but I'm not sure. Detailed explanations are welcome, thank you.

Answer 1

Not quite. The integral formula is

∫ √ [1 + f'(x)²] dx,

where f'(x) is the derivative of f at x. Since f(x) = e^x here, this becomes

∫ √ [1 + e^2x] dx.

You should proceed from there.

Answer 2

I agree with first answer, but how to evaluate this?
√ [1 + e^2x] hasn't a primitive, as far as I know, so that's why you've been told to use a graphing utility. And not being familiar with graphing utilities, I can't suggest what to do with one so I'm not much help there.

Using old-fashioned Simpson's rule and a calculator or spreadsheet, I'd probably evaluate
F(x) = 1 + e^2x for x = 0, 1, 2, 3, ... 20
and then find
(1/3)[F0 + F20 +4(F1 + F3 + F5 + ... + F19) + 2(F2 + F4 + ... + F18)]

Answer 3

y=2t-3 dy/dt =2 x=t^2+a million dx/dt=2t dy/dx = dy/dt / dx/dt = 2 / 2t = a million/t (dy/dx)^2 = a million/t^2 a million+(dy/dx)^2 = a million+a million/t^2 sqrt (a million+(dy/dx)^2 ) = sqrt [ a million+ a million/t^2] 0?t?a million t^2=x-a million at the same time as t=0, x=a million at the same time as t=a million, x=2 ? sqrt [ a million+ a million / (x-a million) ] dx from a million to 2 ? sqrt [ x / (x-a million) ] dx from a million to 2 This essential seems to diverge at the same time as x=a million