定積分definite integral

For y= x^2 +3,
when x = 0
y = 3

For y = -x^2 +5
when y = 0
-x^2 + 5 = 0
x^2 = 5
x = ±√5
when x = 0
y = 5

y= x^2 +3 ...............(1)
y = -x^2 +5..............(2)
x^2 +3 = -x^2 +5
2x^2 -2 = 0
x^2 = 1
x =±√1
=±1

∫_(-1)^(1)▒((-x^2 +5)-(x^2+3))dx
=∫_(-1)^(1)▒(-2x^2 +2)dx
=[-2/3 * x^3 + 2x ](-1)^(1)
=[-2/3 * (1+1) +2(1+1)](-1)^(1)
=-2/3 *2 +4
=-4/3+4
=8/3 sq. units