Question

Find the arc length of the curve below on the given interval. y equals one third (x squared plus 2 )Superscript 3 divided by 2y= 1 3x2+23/2 on ​[00​,66​]

Answer 1

`\int_(0)^(6)√(1+12x^4+8x^2)dx`

Step-by-step explanation:

We are given that

`y=(1)/(3)(3x^2+2)^{(3)/(2)}`

Interval=[0,6]

a=0 and b=6

Differentiate w.r. t x

`(dy)/(dx)=(1)/(3)(3x^2+2)^{(1)/(2)}* 6x=2x(3x^2+2)^{(1)/(2)}`

By using the formula ;
`(dx^n)/(dx)=nx^(n-1)`

We know that arc length of curve

`s=\int_(a)^(b)\sqrt{1+((dy)/(dx))^2}dx`

Substitute the values

`s=\int_(0)^(6)\sqrt{1+(2x(3x^2+2)^{(1)/(2)})^2}dx`

`s=\int_(0)^(6)√(1+4x^2(3x^2+2))dx`

`s=\int_(0)^(6)√(1+12x^4+8x^2)dx`

Length of curve,=
`s=\int_(0)^(6)√(1+12x^4+8x^2)dx`