Asked by dario

set up sums of integrals that can be used to find the area of the region bounded by the graphs of the equations by integrating with respect to y
y=square root of x; y= -x, x=1, x=4

Answers

Answered by Steve
∫[-4,-1] 4-(-y) dy
+∫[-1,1] 4-(1) dy
+∫[1,2] 4-(y^2) dy
Answered by Reiny
The two verticals x=1 and x=4 intersect
y =√x at A(1,1) and B(4,2) and they intersect
y = -x at D(1,-1) and C(4,-4)

from A draw a horizontal to hit BC at (4,1)
So we have a trapezoid, for which we don't need Calculus to find its area
AD = 2, EC = 5 , and the height between them is 3
Area of trap AECD = (1/2)(2+5)(3) = 21/2 square units
the area of the region outlined by ABE is
∫(4 - y^2) dy from 1 to 2
= [4y - (1/3)y^3] from 1 to 2
= (8 - 8/3) - (4 - 1/3)
= 5/3
total area = 21/2 + 5/3 = 73/6

I don't understand why they would ask that it be done by integrating with respect to y, when integrating it with respect to x would have been so much easier.

Area = ∫(x^(1/2) + x) dx from 1 to 4
= [(2/3)x^(3/2) + (1/2)x^2] from 1 to 4
= (16/3 + 8) - (2/3 + 1/2)
= 73/6

(that took 4 lines)
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