Find the area of the region bounded by the curves y = sin^-1(x/2), y = 0, and x = 2 obtained by integrating with respect to y. Please include the definite integral and the antiderivative.

To find the area of the region bounded by the given curves, we can integrate with respect to y.

Let's start by sketching the curves on a coordinate system. The first curve is y = sin^-1(x/2), which is the inverse sine function of (x/2). The second curve is y = 0, which is just the x-axis, and the third curve is x = 2, which is a vertical line passing through x = 2.

The graph of y = sin^-1(x/2) is a curve that starts at (0, 0), increases until reaching its maximum at (2, π/2), and then decreases back to 0 as x approaches 4.

To find the region bounded by these curves, we need to find the values of y where the curves intersect. In this case, the curves intersect at two points: (0, 0) and (2, π/2).

To compute the area between the curves using integration, we need to integrate the difference between the upper and lower curves. In this case, the upper curve is y = sin^-1(x/2), and the lower curve is y = 0.

Let's denote the region between y = sin^-1(x/2) and y = 0 as R. To find the area of R, we integrate the difference between the two curves over the interval where they intersect.

The definite integral representing the area of R is:

∫[0, π/2](sin^-1(x/2) - 0) dy

We integrate with respect to y because we are given the curves in terms of y, and we need to determine the area between them.

Now, let's find the antiderivative of sin^-1(x/2). To do this, we use integration techniques like substitution or integration by parts. Applying the substitution u = x/2, we have:

∫sin^-1(x/2) dy = 2∫sin^-1(u) du

Let's denote the antiderivative of sin^-1(u) as F(u). Then, the antiderivative of sin^-1(x/2) can be represented as:

2F(u) + C

where C is the constant of integration.

Finally, substituting back u = x/2, we get:

2F(x/2) + C

So the definite integral representing the area between the curves is:

∫[0, π/2](2F(x/2) + C) dy

I hope this helps! Let me know if you have any further questions.

You have things messed up. The curve does not go past x=1, so it cannot be bounded on the right by the line x=2.

Either set a y boundary which intersects the curve above the x-axis, or shift the x boundary.