Find the area of the region bounded by the curves y = sin^-1(x/4) , y = 0, and x = 4 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.
Note that x=4siny and you want the area "above" the curve
a = ∫[0,pi/2] 4-4siny dy = 4y+4cosy [0,pi/2] = 2pi-4
To find the area of the region bounded by the given curves, we need to integrate with respect to y. Let's break down the problem into two separate parts: the curve y = sin^(-1)(x/4) and the line y = 0.
First, let's find the y-values of the points where the curve and the line intersect. At the intersection points, y = sin^(-1)(x/4) = 0. This implies that x/4 = sin(0) = 0, which gives us x = 0. So the left intersection point is (0, 0).
Next, we need to find the right intersection point. Since y = sin^(-1)(x/4), for the right intersection point, y = 0, which means sin^(-1)(x/4) = 0. Taking the inverse sine of both sides gives us x/4 = sin(0) = 0, and thus x = 0 again. So the right intersection point is also (0, 0).
Therefore, the region bounded by the curves is a triangle with base x = 0 and height y = sin^(-1)(x/4). To find the area of this region, we integrate the function y = sin^(-1)(x/4) from y = 0 to y = sin^(-1)(0). The upper limit y = sin^(-1)(0) is easy to evaluate because sin^(-1)(0) = 0.
Now, let's integrate with respect to y:
∫[0 to sin^(-1)(0)] sin^(-1)(x/4) dy
Since we are integrating with respect to y, we need to express x in terms of y. Rewriting the original equation y = sin^(-1)(x/4), we have x = 4sin(y).
The integral becomes:
∫[0 to sin^(-1)(0)] sin^(-1)(4sin(y)/4) dy
Now, we can simplify this expression by noticing that sin^(-1)(4sin(y)/4) is just y, as the arcsine and sine functions cancel each other out.
So, the integral can be simplified to:
∫[0 to sin^(-1)(0)] y dy
To integrate y with respect to y, we treat y as a simple power function where the exponent is 1.
The antiderivative of y with respect to y is (1/2)y^2.
So, the definite integral becomes:
(1/2)∫[0 to sin^(-1)(0)] y^2 dy
Evaluating this definite integral from 0 to sin^(-1)(0) gives:
(1/2)[(1/3)y^3] evaluated from 0 to sin^(-1)(0)
Now, we substitute y = sin^(-1)(0) = 0 into the formula:
(1/2)[(1/3)(sin^(-1)(0))^3] - (1/2)[(1/3)(0)^3]
Simplifying the expression, we find that (1/3)(sin^(-1)(0))^3 = 0.
Thus, the area of the region bounded by the curves y = sin^(-1)(x/4), y = 0, and x = 4, obtained by integrating with respect to y, is 0.
Therefore, the area of the region is zero.