Question

Hello there!

Can I get some help with this calculus practice problem? I don't think my answer is right :)
Please show your work and details. Thanks a lot!

Answer 1

Answer: 144

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:

`\int\limits^(2)_(-4) [(-2x^2-8x+32)-(2x^2)]dx\\\\\\=\int\limits^(2)_(-4) (-4x^2-8x+32)dx\\\\\\=\bigg(-(4)/(3)x^3-(8)/(2)x^2+32x\bigg)\bigg|\limits^2_(-4)\\\\\\=\bigg(-(4)/(3)(2)^3-4(2)^2+32(2)\bigg)-\bigg(-(4)/(3)(-4)^3-4(-4)^2+32(-4)\bigg)\\\\\\=\bigg(-(32)/(3)-16+64\bigg)-\bigg((256)/(3)-64-128\bigg)\\\\\\=\bigg((112)/(3)\bigg)-\bigg(-(320)/(3)\bigg)\\\\\\=(432)/(3)\\\\\\=\large\boxed{144}`

Answer: 4.5

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:

`\int\limits^(2)_(-1) [(x+3)-(x^2+1)]dx\\\\\\=\int\limits^(2)_(-1) (-x^2+x+2)dx\\\\\\=\bigg(-(x^3)/(3)+(x^2)/(2)+2x\bigg)\bigg|\limits^2_(-1)\\\\\\=\bigg(-(2^3)/(3)+(2^2)/(2)+2(2)\bigg)-\bigg(-((-1)^3)/(3)+((-1)^2)/(2)+2(-1)\bigg)\\\\\\=\bigg(-(8)/(3)+2+4\bigg)-\bigg((1)/(3)+(1)/(2)-2\bigg)\\\\\\=\bigg((10)/(3)\bigg)-\bigg(-(7)/(6)\bigg)\\\\\\=(9)/(2)\\\\\\=\large\boxed{4.5}`