For which function does the integral integral 0^3 sqrt(1+16e^4x)dx give the length of the curve on the interval [0,3] ?

1 Answer
Timber Lin
Jun 6, 2018

#f(x)=2e^(2x)+C# or #f(x)=-2e^(2x)+C#

Explanation:

the arc length from #x=0# to #x=3# is #int_0^3sqrt(1+(f'(x))^2)dx#

if the arc length is also equivalent to #int_0^3sqrt(1+16e^(4x))dx#, you can see that #(f'(x))^2=16e^(4x)#

solving for #f(x)#:
#f'(x)=+-sqrt(16e^(4x))#
#f'(x)=+-4e^(2x)#

now there are #2# cases:
#1)# if #f'(x)=4e^(2x)#, then #f(x)=int(4e^(2x))dx=2e^(2x)+C#
#2)# if #f'(x)=-4e^(2x)#, then #f(x)=int(-4e^(2x))dx=-2e^(2x)+C#

there's no more information so there are #2# possible answers.

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