Question

Find the exact length of the curve y = (x^6 +8)/(16x^2) from x =2 to x= 3.

Answer 1

`y=(x^6+8)/(16x^2)\implies y'=(x^6-4)/(4x^3)`

The length of the curve over the given integral is


`\displaystyle\int_2^3√(1+(y')^2)\,\mathrm dx`

`\displaystyle\int_2^3\sqrt{1+((x^6-4)^2)/(16x^6)}\,\mathrm dx`

`\displaystyle\int_2^3\sqrt{1+(x^(12)-8x^6+16)/(16x^6)}\,\mathrm dx`

`\displaystyle\int_2^3\sqrt{(x^(12)+8x^6+16)/(16x^6)}\,\mathrm dx`

`\displaystyle\int_2^3\sqrt{((x^6+4)^2)/(16x^6)}\,\mathrm dx`

`\displaystyle\frac14\int_2^3(x^6+4)/(x^3)\,\mathrm dx`

`\displaystyle\frac14\int_2^3\left(x^3+\frac4{x^3}\right)\,\mathrm dx`

`\displaystyle\frac14\left(\frac{x^4}4-\frac2{x^2}\right)\bigg|_(x=2)^(x=3)=(595)/(144)`