# `y = 3/2 x^(2/3) , [1, 8]` Find the arc length of the graph of the function over the indicated interval.

Arc length (L) of the function y=f(x) on the interval [a,b] is given by the formula,

`L=int_a^bsqrt(1+(dy/dx)^2)` dx, if y=f(x) and  a `<=`  x `<=`  b,

Now let's differentiate the function,

`y=3/2x^(2/3)`

`dy/dx=3/2(2/3)x^(2/3-1)`

`dy/dx=1/x^(1/3)`

Now let's plug the derivative in the arc length formula,

`L=int_1^8sqrt(1+(1/x^(1/3))^2)dx`

`L=int_1^8sqrt(1+1/x^(2/3))dx`

`L=int_1^8sqrt((x^(2/3)+1)/x^(2/3))dx`

`L=int_1^8(1/x^(1/3))sqrt(x^(2/3)+1)dx`

Now let's evaluate...

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Arc length (L) of the function y=f(x) on the interval [a,b] is given by the formula,

`L=int_a^bsqrt(1+(dy/dx)^2)` dx, if y=f(x) and  a `<=`  x `<=`  b,

Now let's differentiate the function,

`y=3/2x^(2/3)`

`dy/dx=3/2(2/3)x^(2/3-1)`

`dy/dx=1/x^(1/3)`

Now let's plug the derivative in the arc length formula,

`L=int_1^8sqrt(1+(1/x^(1/3))^2)dx`

`L=int_1^8sqrt(1+1/x^(2/3))dx`

`L=int_1^8sqrt((x^(2/3)+1)/x^(2/3))dx`

`L=int_1^8(1/x^(1/3))sqrt(x^(2/3)+1)dx`

Now let's evaluate first the indefinite integral by using integral substitution,

Let `t=x^(2/3)+1`

`dt=2/3x^(2/3-1)dx`

`dt/dx=2/(3x^(1/3))`

`dx/x^(1/3)=3/2dt`

`intsqrt(x^(2/3)+1)(1/x^(1/3))dx=int3/2sqrt(t)dt`

`=3/2(t^(1/2+1)/(1/2+1))`

`=3/2(t^(3/2)/(3/2))`

`=t^(3/2)`

`=(x^(2/3)+1)^(3/2)`

`L=[(x^(2/3)+1)^(3/2)]_1^8`

`L=[(8^(2/3)+1)^(3/2)]-[(1^(2/3)+1)^(3/2)]`

`L=[5^(3/2)]-[2^(3/2)]`

`L=11.18033989-2.828427125`

`L=8.351912763`

Arc length (L) of the function over the given interval is `~~8.352`

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