Answer: 144
Step-by-step explanation:
First, find the x-values where the equations intersect:
2x² = -2x² - 8x + 32
4x² + 8x - 32 = 0
4(x² + 2x - 8) = 0
(x + 4)(x - 2) = 0
x = -4 and x = 2 (see graph to confirm these values)
Now, find the area under the curve:
[tex]\int\limits^{2}_{-4} [(-2x^2-8x+32)-(2x^2)]dx\\\\\\=\int\limits^{2}_{-4} (-4x^2-8x+32)dx\\\\\\=\bigg(-\dfrac{4}{3}x^3-\dfrac{8}{2}x^2+32x\bigg)\bigg|\limits^2_{-4}\\\\\\=\bigg(-\dfrac{4}{3}(2)^3-4(2)^2+32(2)\bigg)-\bigg(-\dfrac{4}{3}(-4)^3-4(-4)^2+32(-4)\bigg)\\\\\\=\bigg(-\dfrac{32}{3}-16+64\bigg)-\bigg(\dfrac{256}{3}-64-128\bigg)\\\\\\=\bigg(\dfrac{112}{3}\bigg)-\bigg(-\dfrac{320}{3}\bigg)\\\\\\=\dfrac{432}{3}\\\\\\=\large\boxed{144}[/tex]
Answer: 4.5
Step-by-step explanation:
First, find the x-values where the equations intersect:
x+3 = x² + 1
0 = x² - x - 2
0 = (x - 2)(x + 1)
x = 2 and x = -1 (see graph to confirm these values)
Now find the area under the curve:
[tex]\int\limits^{2}_{-1} [(x+3)-(x^2+1)]dx\\\\\\=\int\limits^{2}_{-1} (-x^2+x+2)dx\\\\\\=\bigg(-\dfrac{x^3}{3}+\dfrac{x^2}{2}+2x\bigg)\bigg|\limits^2_{-1}\\\\\\=\bigg(-\dfrac{2^3}{3}+\dfrac{2^2}{2}+2(2)\bigg)-\bigg(-\dfrac{(-1)^3}{3}+\dfrac{(-1)^2}{2}+2(-1)\bigg)\\\\\\=\bigg(-\dfrac{8}{3}+2+4\bigg)-\bigg(\dfrac{1}{3}+\dfrac{1}{2}-2\bigg)\\\\\\=\bigg(\dfrac{10}{3}\bigg)-\bigg(-\dfrac{7}{6}\bigg)\\\\\\=\dfrac{9}{2}\\\\\\=\large\boxed{4.5}[/tex]
