∫[-4,-1] 4-(-y) dy
+∫[-1,1] 4-(1) dy
+∫[1,2] 4-(y^2) dy
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
2 answers
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)
1 answer