Answer:
To compute the length of the curve f(x)=47(4−x2) over the interval 0≤x≤2, we need to use the formula for the arc length of a function:
L=∫ab1+(f′(x))2dx
where a and b are the endpoints of the interval. First, we need to find the derivative of f(x), which we can do by using the chain rule and the power rule:
f′(x)=4dxd7(4−x2)
f′(x)=427(4−x2)1dxd(7(4−x2))
f′(x)=427(4−x2)1(−14x)
f′(x)=−7(4−x2)28x
Next, we need to plug in f′(x) into the formula and simplify:
L=∫021+(−7(4−x2)28x)2dx
L=∫021+7(4−x2)784x2dx
L=∫027(4−x2)7(4−x2)+784x2dx
L=∫024−x228−21x2dx
Now, we need to evaluate the integral, which we can do by using a trigonometric substitution. Let x=2sinu, then dx=2cosudu and u=arcsin(x/2). The limits of integration change as follows:
x=0⟹u=0
x=2⟹u=2π
The integral becomes:
L=∫02π4−(2sinu)228−21(2sinu)2(2cosu)du
L=∫02π4−4sin2u28−84sin2u(2cosu)du
L=∫02π1−sin2u7−21sin2u(2cosu)du
L=∫02πcos2u7−21sin2u(2cosu)du
L=∫02π27−21sin2udu
Using a trigonometric identity, we can write:
L=∫02π4127−1221cos(2u)du
Using another trigonometric substitution, let v=2u, then dv=2du and u=v/2. The limits of integration change as follows:
u=0⟹v=0
u=2π⟹v=π
The integral becomes:
L=∫0π4127−1221cosv(21)dv
L=6∫0π