How do I find the area enclosed by #x=5y-5y^2# and #x=0#?
1 Answer
Apr 10, 2017
# "Area" = 5/6 #
Explanation:
Here is a graph of the function:
graph{x=5y-5y^2[-2, 2, -1, 2]}
When
So the bounded area is given by:
# A = int_0^1 f(y) \ dy #
# \ \ \ = int_0^1 \ 5y-5y^2 \ dy #
# \ \ \ = 5 \ int_0^1 \ y-y^2 \ dy #
# \ \ \ = 5 \ [ \ y^2/2-y^3/3 \ ]_0^1 #
# \ \ \ = 5 \ {(1/2-1/3) - (0-0)} #
# \ \ \ = 5 \ {1/6} #
# \ \ \ = 5/6 #
