Checking Answer: Area between 2 Curves

[SeekerOfDragons]

Disclaimer: I tried to make the following as readable as possible and hopefully I wrote the Integrals correctly... Also I realized in advance that Part 1 and Part 3 were symmetrical and could have just multiplied by 2 to save time, but just to be totally sure, I solved each part individually...

I'm trying to verify that my answer is correct IRT the following problem:

Find the area between the curves Y = 12 - X^2 and Y = X^2 - 6 [-5, 5]

12 - X^2 = X^2 - 6
18 = 2X^2
9 = X^2

X = -3, 3

functions cross at -3 and 3

Area =
(Integral(-5, -3) [(X^2 - 6) - (12 - X^2)] dx) +
(Integral(-3, 3) [(12 - X^2) - (X^2 - 6)] dx) +
(Integral(3, 5) [(X^2 - 6) - (12 - X^2)] dx)

Part 1:
(Integral(-5, -3) [(X^2 - 6) - (12 - X^2)] dx)
(Integral(-5, -3) [(X^2 - 6 - 12 + X^2)] dx)
(Integral(-5, -3) (2X^2 - 18) dx
(Integral(-5, -3) (2X^2 - 18) dx
2X^3/3 - 18X ] (-5, -3)

[2(-3^3)/3 - 18(-3)] - [2(-5^3)/3 - 18(-5)]
[2(-27)/3 - (-54)] - [2(-125)/3 - (-90)]
(-54/3 + 54) - (-250/3 + 90)
(-18 + 54) - (20/3)
36 - 20/3

*** 88/3 Units^2

Part 2:
(Integral(-3, 3) [(12 - X^2) - (X^2 - 6)] dx)
(Integral(-3, 3) [12 - X^2 - X^2 + 6] dx)
(Integral(-3, 3) [18 - 2X^2] dx)
18X - 2X^3/3 ] (-3, 3)

[18(3) - 2(3^3)/3] - [18(-3) - 2(-3^3)/3]
[54 - 2(27)/3] - [-54 - 2(-27)/3]
[54 - 54/3] - [-54 - (-54)/3]
[54 - 18] - [-54 + 18]
36 - (-36)
36 + 36

*** 72 Units ^2

Part 3:
(Integral(3, 5) [(X^2 - 6) - (12 - X^2)] dx)
(Integral(3, 5) [(X^2 - 6 - 12 + X^2)] dx)
(Integral(3, 5) (2X^2 - 18) dx
(Integral(3, 5) (2X^2 - 18) dx
2X^3/3 - 18X ] (3, 5)

[2(5^3)/3 - 18(5)] - [2(3^3)/3 - 18(3)]
[2(125)/3 - 90] - [2(27)/3 - 54]
(250/3 - 90) - (54/3 - 54)
(-20/3) - (18 - 54)
(-20/3) - (-36)

*** 88/3 Units^2


88/3 + 72 + 88/3
176/3 + 216/3

**** 392/3 units^2 or 130 2/3 units^2 or ~130.6666 units^2

Pretty sure I'm right, but wanted to verify just to be safe. thanks in advance for your time.

r/

SoD

 

[galactus]

\(\displaystyle 2\int_{-5}^{-3}(-2x^{2}+18)dx=\frac{176}{3}\)

Then, the area under the parabola is \(\displaystyle \int_{-3}^{3}(-2x^{2}+18)dx=72\)

\(\displaystyle 72+\frac{176}{3}=\frac{392}{3}\approx 130.\overline{6}\)

Good job