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IceJOKER [234]
1 year ago
7

NEED FUNNY PPL TO JOIN https://www.remind.com/join/hellae

Advanced Placement (AP)
1 answer:
KATRIN_1 [288]1 year ago
6 0

Answer:

I joined

Explanation:

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Answer-B.
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(1) The integral is straightforward; <em>x</em> ranges between two constants, and <em>y</em> ranges between two functions of <em>x</em> that don't intersect.

\displaystyle\int_{-2}^1\int_{-x}^{x^2+2}\mathrm dy\,\mathrm dx

(2) First find where the two curves intersect:

<em>y</em> ² - 4 = -3<em>y</em>

<em>y</em> ² + 3<em>y</em> - 4 = 0

(<em>y</em> + 4) (<em>y</em> - 1) = 0

<em>y</em> = -4, <em>y</em> = 1   →   <em>x</em> = 12, <em>x</em> = -3

That is, they intersect at the points (-3, 1) and (12, -4). Since <em>x</em> ranges between two explicit functions of <em>y</em>, you can capture the area with one integral if you integrate with respect to <em>x</em> first:

\displaystyle\int_{-4}^1\int_{y^2-4}^{-3y}\mathrm dx\,\mathrm dy

(3) No special tricks here, <em>x</em> is again bounded between two constants and <em>y</em> between two explicit functions of <em>x</em>.

\displaystyle\int_1^5\int_0^{\frac1{x^2}}\mathrm dy\,\mathrm dx

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