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How do I find the area enclosed by #x=5y-5y^2# and #x=0#?

Answer 1

# "Area" = 5/6 #

Here is a graph of the function: graph{x=5y-5y^2[-2, 2, -1, 2]}

When # x = 0 =>5y-5y^2 = 0 # # " " => 5y(1-y) = 0 # # " " :. y = 0,1 #

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 #

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Answer 2

Scarlett Allison

To find the area enclosed by the curves (x = 5y - 5y^2) and (x = 0), you need to first find the points of intersection between the two curves by setting them equal to each other and solving for (y). Then integrate the difference between the upper and lower curves with respect to (y) within the interval where they intersect. The integral will give you the area enclosed by the curves.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.