Answer
[tex]\int_{0}^{6}\sqrt{1+12x^4+8x^2}dx[/tex]
Step-by-step explanation:
We are given that
[tex]y=\frac{1}{3}(3x^2+2)^{\frac{3}{2}}[/tex]
Interval=[0,6]
a=0 and b=6
Differentiate w.r. t x
[tex]\frac{dy}{dx}=\frac{1}{3}(3x^2+2)^{\frac{1}{2}}\times 6x=2x(3x^2+2)^{\frac{1}{2}}[/tex]
By using the formula ;[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
We know that arc length of curve
[tex]s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx[/tex]
Substitute the values
[tex]s=\int_{0}^{6}\sqrt{1+(2x(3x^2+2)^{\frac{1}{2}})^2}dx[/tex]
[tex]s=\int_{0}^{6}\sqrt{1+4x^2(3x^2+2)}dx[/tex]
[tex]s=\int_{0}^{6}\sqrt{1+12x^4+8x^2}dx[/tex]
Length of curve,=[tex]s=\int_{0}^{6}\sqrt{1+12x^4+8x^2}dx[/tex]
