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

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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).

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