The value in the part (c) is 1/(b-a) if the graph of any continuous function y = f(x) for a ≤ x ≤ b is rotated about the horizontal line y = L, the volume obtained depends on L.
It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.
We have a graph of any continuous function y = f(x) for a ≤ x ≤ b is rotated about the horizontal line y = L.
For part(c)
[tex]\rm V(L) = \int\limits^b_a {\pi(f(x) -L)^2} \, dx\\\\\rm V(L) = \int\limits^b_a {\pi((f(x))^2 -2f(x)+L^2)} \, dx\\\\\rm V(L) = \pi \int\limits^b_a {(f(x))^2} \, dx-2\pi L \int\limits^b_a {f(x)} \, dx +L^2\pi\int\limits^b_a {} \, dx[/tex]
Now finding V'(L)
[tex]\rm V'(L) = -2\pi \int\limits^b_a {f(x)} \, dx+2\pi L \int\limits^b_a {} \, dx[/tex]
[tex]\rm V'(L) = -2\pi \int\limits^b_a {f(x)} \, dx+2\pi L (b-a)[/tex]
Now equation V'(L) to zero; V'(L) = 0
[tex]\rm -2\pi \int\limits^b_a {f(x)} \, dx+2\pi L (b-a)=0[/tex]
[tex]\rm \int\limits^b_a {f(x)} \, dx= L (b-a)[/tex]
[tex]\rm L = \frac{1}{b-a} \int\limits^b_a {f(x)} \, dx[/tex]
The above expression is representing the critical point.
Now finding V''(L)
V''(L) = 2π(b-a) >0 at L
[tex]\rm L = \frac{1}{b-a} \int\limits^b_a {f(x)} \, dx[/tex] (minimizines the volume)
Thus, the value in the part (c) is 1/(b-a) if the graph of any continuous function y = f(x) for a ≤ x ≤ b is rotated about the horizontal line y = L, the volume obtained depends on L.
Learn more about integration here:
brainly.com/question/18125359
#SPJ1
