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
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
+∫[-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)
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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