Find the area bound by x = y3 and the y-axis from y = 0 to y = 2
A = ∫02 y3 dy
A = [(y4)/4] |02
A = [(24)/4] − 0
A = 4
Find the area bound by y = sinx and the x-axis from x = 0 to x = 2π
A = ∫02π sinx dx
A = − cosx |02π
A = −cos2π− (− cos0)
A = −1 + 1 = 0
An area of zero doesn't quite make sense here, so using the straight integral is insufficient. What we can do is treat this as two separate integrals, one where the area is above the x-axis and one where it is below and add their effective area.
sinx = 0, x = 0, π, 2π in this interval.
A = ∫0π sinx dx + (−∫π2π sinx dx)
The negative sign is there to compensate for the "negative area"
A = −cosx |0π + cosx |π2π
A = − cosπ+ cos0 + cos2π− cosπ
A = 1 + 1 + 1 + 1
A = 4
Find the area bound by x = y2 and y = x − 2
We could find the integral in terms of dx, but the we would have to split it up into at least two separate integrals. Besides, x = y2 is not a function of x (there are two outputs for every x input).
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Area Between Curves
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