Respuesta :
Answer:
The arc length L is 21.1875.
Step-by-step explanation:
The function is
[tex]y=\frac{x^3}{3}+\frac{1}{4x}[/tex]
in the interval x=[1,4].
To find the arc length L of a curve, we need to solve this integral:
[tex]L=\int\limits^a_b {\sqrt{1+(dy/dx)^2} } \, dx[/tex]
So first, we calculate the first derivative
[tex]\frac{dy}{dx} =\frac{1}{3} (3x^2)+\frac{1}{4} (-x^{-2})=x^2-\frac{1}{4x^2}[/tex]
Then calculate the square of this derivative
[tex](\frac{dy}{dx})^2 =(x^2-\frac{1}{4x^2})^2=x^4-2(x^2*\frac{1}{4x^2})+\frac{1}{16x^4} \\\\(\frac{dy}{dx})^2 =x^4-\frac{1}{2} +\frac{1}{16x^4}[/tex]
Then, the arc length is
[tex]L=\int\limits^a_b {\sqrt{1+(dy/dx)^2} } \, dx\\\\L=\int\limits^4_1 {\sqrt{1+(x^4-1/2+1/16x^4)} } \, dx\\\\L=\int\limits^4_1 {\sqrt{x^4+1/2+1/16x^4} }\,dx\\\\L=\frac{1}{4} \int\limits^4_1 {\sqrt{16x^4+8+x^{-4}} }\,dx\\\\L=\frac{1}{4} \int\limits^4_1 {\sqrt{(4x^2+x^{-2)^2}} }\,dx\\\\L=\frac{1}{4} \int\limits^4_1 {(4x^2+x^{-2}) }\,dx\\\\L=\frac{1}{4} (4\frac{x^3}{3}+x^{-1}) }=\frac{x^3}{3}-\frac{1}{4x} \\\\[/tex]
[tex]L=(\frac{4^3}{3}-\frac{1}{4*4})-( \frac{1^3}{3}-\frac{1}{4*1})\\\\L=(\frac{64}{3} -\frac{1}{16})-(\frac{1}{3}-\frac{1}{4} )\\\\L=\frac{64*16-3}{48}-\frac{4-3}{12}= \frac{1021}{48} -\frac{1}{12} =\frac{1021-4}{48}=\frac{1017}{48} = 21.1875[/tex]
