`y = 1/2(e^x + e^(-x)) , [0, 2]` Find the arc length of the graph of the function over the indicated interval.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The arc length of the curve `y=f(x)` between `x=a` and `x=b,` `(a<b)` is given by 

`L=int_a^b sqrt(1+y'^2)dx`

Before we start using the above formula let us notice that `y=1/2(e^x+e^-x)=cosh x.` This should simplify our calculations.

`L=int_0^2sqrt(1+(cosh x)'^2)dx=`

`int_0^2sqrt(1+sinh^2x )dx=`

Now we use the formula `cosh^2 x=1+sinh^2x.`

`int_0^2cosh x dx=`

`sinh x|_0^2=`

`sinh 2-sinh 0=`

`sinh 2 approx3.62686`   

The arc length of the graph of the given function over interval `[0,2]` is `sinh 2` or approximately `3.62686.`

The graph of the function can be seen in the image below (part of the graph for which we calculated the length is blue).

` `

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Images:
Image (1 of 1)
Approved by eNotes Editorial