Question

Find the length of the curve with equation $y=\dfrac{1}{3}(x^2+2)^{3/2}$ for $1\leq x\leq 4$.

Answer 1

Answer:

Length of curve=$\int_{1}^{4}\sqrt{1+x^2(x^2+2)^2}$

Step-by-step explanation:

We are given that a curve

$y=\frac{1}{3}(x^2+2)^{\frac{3}{2}},1\leq x\leq 4$

We have to find the length of curve

Differentiate w.r.t x

$\frac{dy}{dx}=\frac{1}{3}\times \frac{3}{2}(x^2+2)(2x)=x(x^2+2)$

By using the formula:$\frac{dx^n}{dx}=nx^{n-1}$

Length of curve on the interval [a,b] is given by

$\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}$

By using the formula then , we get the length of given curve

$\int_{1}^{4}\sqrt{1+(x(x^2+2)^2}$

Length of curve=$\int_{1}^{4}\sqrt{1+x^2(x^2+2)^2}$