Calculus: Integration Problems?

Question

I believe I did these two problems right, but I just want to make sure I got the right solutions’, as there was quite a lot of algebra so I want to make sure I did not make any mistakes anywhere.

For these problems, we are asked to find the area enclosed by the lines and curves of the two functions’.

1. y = sin(pi*x/2) and y = x

The first equation reads as the sin of pi times x divided by 2.

For my solution, I got that the area of the region is (4/pi) -1.

2. x – y^2 = 0 and x + 2y^2 = 3


(the first equation reads as x minus y squared equals 0, and the second equation reads as x plus 2y squared equals 3).

For my solution, I got that the area of the region is 4.

Answer 1

1)
y = sin(pi*x/2) and y = x intercept at (0,0) and (1,1)
so,
area enclosed by y = sin(pi*x/2) and y = x
= ∫ [sin(pi*x/2) - x ] dx
= [-2/pi * cos(pi*x/2) - 1/2 x^2] (from x = 0 to x = 1)
= -2/pi * cos(pi/2) - 1/2(1)^2 + 2/pi *cos(0) + 1/2(0)^2
= -2/pi*(0) - 1/2 + 2/pi (1) + 0
= 2/pi - 1/2

2)
x-y^2 = 0 and x+2y^2=3 intercept at (1,1) and (1,-1)
so,

x-y^2 = 0
x = y^2

x+2y^2 = 3
x= -2y^2+3

area enclosed by x-y^2=0 and x+2y^2=3
= ∫ [(-2y^2+3) - (y^2)] dy
= ∫[-3y^2+3] dy
= [-y^3 + 3y] (from y=1 to y=-1)
= -(1)^3 + 3(1) + (-1)^3 - 3(-1)
= 4

Answer 2

1.
y = sin(pi*x/2) and y = x intersect at (-1,-1), (0,0), and (1,1). Both functions are symmetrical about 0, so
. . . . 1
A = 2∫[sin(πx/2) - x]dx =
. . . . 0
2[- (2/π)cos(πx/2) - (1/2)x^2] from 0 to 1 =
2[- 0 - 1/2 + 2/π + 0] =
4/π -1

2.
x – y^2 = 0 and x + 2y^2 = 3
Finding intersections,
3x = 3, x = 1
y = -1, 1
x = y^2, x = 3 - 2y^2
dA = (3 - 3y^2)dy
A = 3y - y^3 from - 1 to 1
A = 3 + 3 - 1 - 1 = 4