`y = x^5/10 + 1/(6x^3) , [2, 5]` Find the arc length of the graph of the function over the indicated interval.

Expert Answers

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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 we have to differentiate the function,

`y=x^5/10+1/(6x^3)`

`dy/dx=1/10(5)x^(5-1)+1/6(-3)x^(-3-1)`

`dy/dx=x^4/2-1/2x^(-4)`

`dy/dx=x^4/2-1/(2x^4)`

`dy/dx=1/2(x^4-1/x^4)`

`dy/dx=1/2((x^8-1)/x^4)`

`L=int_2^5sqrt(1+((x^8-1)/(2x^4))^2)dx`

`L=int_2^5sqrt(1+(x^16-2x^8+1)/(4x^8))dx`

`L=int_2^5sqrt((4x^8+x^16-2x^8+1)/(4x^8))dx`

`L=int_2^5sqrt((x^16+2x^8+1)/(4x^8))dx`

`L=int_2^5sqrt(((x^8+1)/(2x^4))^2)dx`

`L=int_2^5(x^8+1)/(2x^4)dx`

`L=int_2^5(x^8/(2x^4)+1/(2x^4))dx`

`L=int_2^5(x^4/2+1/(2x^4))dx`

`L=[1/2(x^(4+1)/(4+1))+1/2(x^(-4+1)/(-4+1))]_2^5`

`L=[x^5/10-1/(6x^3)]_2^5`

`L=[5^5/10-1/(6(5)^3)]-[2^5/10-1/(6(2)^3)]`

`L=[3125/10-1/750]-[32/10-1/48]`

`L=[(234375-1)/750]-[(768-5)/240]`

`L=[234374/750]-[763/240]`

`L=(1874992-19075)/6000`

`L=1855917/6000`

`L=309.3195`

 

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