y = x^2
y = 6 - x
Find the area of the region by integrating (a) with respect to x and (b) with respect to y.
To find the area of the region bounded by these two curves, we need to determine the points of intersection and then integrate the appropriate function with respect to the variable.
(a) Finding the points of intersection by setting the two equations equal to each other:
x^2 = 6 - x
Rearranging the equation:
x^2 + x - 6 = 0
Factoring the quadratic equation:
(x + 3)(x - 2) = 0
Solving for x:
x = -3 or x = 2
So, the two curves intersect at x = -3 and x = 2.
To find the area with respect to x, we need to identify the integral bounds. The upper bound is x = 2, and the lower bound is x = -3. The height of the region at each x is given by the function y = x^2 - (6 - x).
(b) To find the area with respect to y, we need to express x in terms of y. Rearranging the second equation, we have:
x = 6 - y
Now, the upper and lower bounds for the integral are the y-values where the two curves intersect. From the previous calculations, we found that the curves intersect at x = -3 and x = 2, which correspond to y-values 5 and 4, respectively.
Therefore, the bounds for integrating with respect to y are y = 5 to y = 4. The width of the region at each y is given by the difference between the corresponding x-values.
(a) Integrating with respect to x:
The area with respect to x can be calculated using the integral:
A = ∫[x=-3 to x=2] (x^2 - (6 - x)) dx
Expanding and integrating:
A = ∫[x=-3 to x=2] (x^2 - 6 + x) dx
= ∫[x=-3 to x=2] (x^2 + x - 6) dx
Evaluating the integral:
A = [1/3 x^3 + 1/2 x^2 - 6x] [x=-3 to x=2]
= [1/3(2)^3 + 1/2(2)^2 - 6(2)] - [1/3(-3)^3 + 1/2(-3)^2 - 6(-3)]
= [8/3 + 2 - 12] - [-27/3 + 9/2 + 18]
= [8/3 - 6] - [-9 + 9/2 + 18]
= (8/3 - 6) - (-9/2 + 18)
= 8/3 - 18/3 + 9/2 - 36/2
= -10 + 9/2 - 18
= -10 - 36/2 + 9/2
= -10 - 27/2
= -20/2 - 27/2
= -47/2
Therefore, the area with respect to x is -47/2.
(b) Integrating with respect to y:
The area with respect to y can be calculated using the integral:
A = ∫[y=5 to y=4] (6 - y - (y - 6)) dy
Simplifying:
A = ∫[y=5 to y=4] (-2y + 12) dy
Evaluating the integral:
A = [-y^2 + 12y] [y=5 to y=4]
= [-(4)^2 + 12(4)] - [-(5)^2 + 12(5)]
= [-16 + 48] - [-25 + 60]
= 32 - 25
= 7
Therefore, the area with respect to y is 7.