Answer:
Length of curve=[tex]\int_{1}^{4}\sqrt{1+x^2(x^2+2)^2}[/tex]
Step-by-step explanation:
We are given that a curve
[tex]y=\frac{1}{3}(x^2+2)^{\frac{3}{2}},1\leq x\leq 4[/tex]
We have to find the length of curve
Differentiate w.r.t x
[tex]\frac{dy}{dx}=\frac{1}{3}\times \frac{3}{2}(x^2+2)(2x)=x(x^2+2)[/tex]
By using the formula:[tex]\frac{dx^n}{dx}=nx^{n-1}[/tex]
Length of curve on the interval [a,b] is given by
[tex]\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}[/tex]
By using the formula then , we get the length of given curve
[tex]\int_{1}^{4}\sqrt{1+(x(x^2+2)^2}[/tex]
Length of curve=[tex]\int_{1}^{4}\sqrt{1+x^2(x^2+2)^2}[/tex]
