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NISA [10]
3 years ago
13

Find the area of the region bounded by the curves

rmula1" title="y=sin^-1\frac{x}{2}" alt="y=sin^-1\frac{x}{2}" align="absmiddle" class="latex-formula">, y=0, and x=2 obtained by integrating with respect to y. Please include the definite integral and the antiderivative.
Mathematics
1 answer:
Umnica [9.8K]3 years ago
7 0

Answer:

π − 2

Step-by-step explanation:

Graph of the region:

desmos.com/calculator/pcascl0frf

If we integrate with respect to x:

∫₀² (y − 0) dx

∫₀² sin⁻¹(x/2) dx

But we want to integrate with respect to y.  Let's start by finding the new limits of integration.

y = sin⁻¹(x/2), so when x = 0, y = 0.  When x = 2, y = π/2.

Next, we need to find x in terms of y.

sin y = x/2

x = 2 sin y

So the integral with respect to y is:

∫₀ᵖⁱ² (2 − x) dy

∫₀ᵖⁱ² (2 − 2 sin y) dy

Integrating:

(2y + 2 cos y + C) |₀ᵖⁱ²

(π + 2 cos (π/2) + C) − (0 + 2 cos 0 + C)

(π + 0 + C) − (0 + 2 + C)

π − 2

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Question 11

The directrix is a horizontal line, which means the parabola opens either upward or downward. In this case, it opens downward. This is because all answer choices have a negative leading coefficient. Also, it's because the focus is below the directrix.

For vertically opening parabolas, we use this form

4p(y-k) = (x-h)^2

where (h,k) is the vertex and p is the focal distance, aka the distance from the vertex the focus. To find (h,k), we start at the focus (0,-4) and move directly up until we reach the directrix y = 4. We'll arrive at (0,4). The midpoint of (0,-4) and (0,4) is (0,0) which is the vertex's location. So (h,k) = (0,0).

Note that in moving from (0,-4) to (0,4) is a span of 4 units. So this is the value of p.

Plug h = 0, k = 0, p = 4 into the equation mentioned and solve for y

4p(y-k) = (x-h)^2

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16y = x^2

y = (1/16)x^2

The only adjustment we need to make is to change the 1/16 to -1/16 so that the parabola opens downward.

<h3>Answer:  Choice D.  y = -(1/16)x^2</h3>

===============================================

Question 3

The given equation is in the form y = ax^2+bx+c

In this case,

  • a = 2
  • b = 4
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Let's compute the x coordinate of the vertex h

h = -b/(2a)

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This h value is plugged into the original function to find k

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We find that h = -1 and k = 1 pair up together. In short, (h,k) = (-1,1) is the vertex.

<h3>Answer: Choice B.  (-1,1)</h3>
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