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sergeinik [125]
2 years ago
8

Help please I really need help :)

Mathematics
1 answer:
Firdavs [7]2 years ago
6 0

Answer:

Solution given:

For 1st

x²=71

For x=±\sqrt{71}

and

For 2nd

x³=71

For x=±\sqrt[3]{71}

So answer is:

In first row

<u>Solution to x²=71</u>

In second row

<u>Solution to </u><u>x³</u><u>=71</u>

In third row

<u>neither</u>

In forth row

<u>Solution to x³=71</u>

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Answer:

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Calculus Problem
Roman55 [17]

The two parabolas intersect for

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and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

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\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

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b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

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\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

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\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

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\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

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