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NISA [10]
4 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]4 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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Answer:

f(x) = x + 2x^\frac{3}{2} + 5

Step-by-step explanation:

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The first step would be to find the antiderivative of f'(x)

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Now substitute into the antiderivative

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