Question

find the arc length of f over [2,6] for f(x) =4/3x^3/2

Answer 1

We have function $f(x) = 4/3 *$ x^3/2.and interval [2;6]. We can use one formula for finding arc length: $l = \int_{a}^{b} \sqrt{1+(f'(x))^2} \,dx$

where a, b are ends of the interval, f'(x) is derivative of our function.

So let's find f'(x):

f'(x) = 4/3*3/2*x^1/2= 2*x^1/2=$2*\sqrt{x}$

And find our integral:

$l = \int_{2}^{6} \sqrt{1+(2*\sqrt{x} )^2} \,dx = \int_{2}^{6} \sqrt{1+4x} \,dx=$

$= 1/4*2/3*(1+4x)^{3/2} |_{2}^{6}= 1/6*(1+4x)^{3/2} |_{2}^{6} =$

$= 1/6*(25^{3/2}-9^{3/2})=1/6*(125-27)= 98/6 = 49/3$

Finally, we have our answer: 49/3 is arc length of f(x) over [2;6], where f(x) =4/3*x^3/2