Find the area of the region bounded by the curves y equals the inverse sine of x divided by 4, y = 0, and x = 4 obtained by integrating with respect to y. Your work must include the definite integral and the antiderivative.
To find the area of the region bounded by the curves, we need to set up the definite integral by integrating with respect to y.
First, let's graph the curves y = arcsin(x)/4, y = 0, and x = 4 to visualize the region.
The curve y = arcsin(x)/4 is an arcsine function scaled vertically by a factor of 1/4.
Since the arcsine function is only defined for -1 ≤ x ≤ 1, we will find the intersection points between y = 0 and y = arcsin(x)/4 within that range.
To find the intersection points, we set y = 0 and solve for x:
0 = arcsin(x)/4
Since arcsin(x) is equal to zero when x = 0, we have:
0 = 0/4
Since this is true for any value of x, the line y = 0 intersects the curve y = arcsin(x)/4 at x = 0.
Now let's find the other intersection point by setting x = 4 and solving for y:
y = arcsin(4)/4
Using a calculator, we find that arcsin(4) is undefined since the arcsine function has a range of -π/2 ≤ arcsin(x) ≤ π/2.
Therefore, there is no intersection point between x = 4 and y = arcsin(x)/4.
Now that we have identified the range of integration, from x = 0 to x = 1, we can express the region bounded by the curves as:
A = ∫[0,1] (upper curve - lower curve) dy
To determine the upper and lower curves, we need to express the equations in terms of y.
For the upper curve, y = arcsin(x)/4, we solve for x:
4y = arcsin(x)
x = sin(4y)
For the lower curve, y = 0, x remains constant.
Therefore, the integral becomes:
A = ∫[0,1] (sin(4y) - 0) dy
A = ∫[0,1] sin(4y) dy
Now let's find the antiderivative of sin(4y):
The antiderivative of sin(4y) is -(1/4)cos(4y).
So, the integral becomes:
A = [-cos(4y)/4] evaluated from 0 to 1
Plugging in the upper and lower limits:
A = [-cos(4) - (-cos(0))]/4
A = [-cos(4) + 1]/4
Therefore, the area of the region bounded by the curves y = arcsin(x)/4, y = 0, and x = 4 obtained by integrating with respect to y is (-cos(4) + 1)/4.
To find the area of the region bounded by the curves, we will integrate with respect to y. We need to find the x-values that correspond to the given y-values and use those as the limits of integration.
First, let's solve the equation y = arcsin(x)/4 for x.
Multiply both sides of the equation by 4:
4y = arcsin(x)
Now, take the sine of both sides to find x:
x = sin(4y)
The limits of integration will be found by determining the y-values at which the curves intersect. In this case, y = 0 is the lower boundary, and x = 4 will be the upper boundary.
Now, we can proceed to the definite integral:
The area (A) can be calculated by integrating the function (x) with respect to (y) from y = 0 to y = (arcsin(4)/4):
A = ∫[from 0 to (arcsin(4)/4)] x dy
Since x = sin(4y), we can rewrite the integral as:
A = ∫[from 0 to (arcsin(4)/4)] sin(4y) dy
To calculate the definite integral, we need to find the antiderivative of sin(4y).
The antiderivative of sin(4y) is -(1/4)cos(4y).
Therefore, the area can be calculated as:
A = [-1/4 * cos(4y)] evaluated from 0 to (arcsin(4)/4)
Substituting the values gives:
A = [-1/4 * cos(4(arcsin(4)/4))] - [-1/4 * cos(4(0))]
Simplifying further:
A = [-1/4 * cos(arcsin(4))] - [-1/4 * cos(0)]
Since cos(arcsin(4)) = cos(0), the equation simplifies to:
A = [-1/4 * cos(0)] - [-1/4 * cos(0)]
Finally, we solve for cos(0) which equals 1:
A = (-1/4 * 1) - (-1/4 * 1)
A = -1/4 + 1/4
A = 0
Therefore, the area of the region bounded by the given curves is 0.
why all the words?
y = arcsin(x/4)
using thin horizontal strips, the area is
a = ∫[0,π/2] (4-x) dy
But x = 4siny, so
a = ∫[0,π/2] (4-4siny) dy
= 2(π-2)
You can check your work using vertical strips:
a = ∫[0,4] y dx
= ∫[0,4] arcsin(x/4) dx
= 2(π-2)