Consider the area between the graphs x + 3y = 36 and x + 4 = y^2. this area can
ID: 2891396 • Letter: C
Question
Consider the area between the graphs x + 3y = 36 and x + 4 = y^2. this area can be computed in two different ways using integrals
(1 point) Consider the area between the graphs x + 3y = 36 and x + 4 = y. This area can be computed in two different ways using integrals First of all it can be computed as a sum of two integrals fx) dx +/g(a) dx b = | 21 where a = f(x) = g(x) Alternatively this area can be computed as a single integral ,c=160 and y)dy where 1-8 h(y) = Either way we find that the area is 5 andExplanation / Answer
A = 2(x+4) dx [-4,21] + (36-x)/3 + (x+4) dx [21,60] = 2197/6 units²
Solve the equations so that you have two functions x(y) instead of y(x):
x(y) = 36 - 3y
x(y) = y² - 4
Upwards parabola is on the bottom, line is on the top.
Set the functions equal to find the intersection points:
y² - 4 = 36 - 3y
y² + 3y - 40 = 0
(y + 8)(y - 5) = 0
y = -8, 5
x(-8) = 36 - 3(-8) = 60
x(5) = 36 - 3(5) = 21
Integrating dy, so you only care about the y-limits.
= -8
ß = 5.
H(y) = x_top(y) - x_bot(y)
H(y) = (36 - 3y) - (y² - 4)
H(y) = -(y² + 3y - 40)
You should get the negation of the parabola you found when solving for the limits (because we want positive area and -ay² parabolas enclose area above the axis).