To find the area of the region bounded by the graphs of sin(x) and cos(x) between x = pi/4 and x = 5*pi/4 using horizontal strips (dxdy order of integration), we can follow these steps:
Step 1: Visualize the region
Draw a graph of sin(x) and cos(x) on the same coordinate plane, and shade the region between them. Make sure to mark the bounds x = pi/4 and x = 5*pi/4 on the x-axis.
Step 2: Determine the limits of integration
Since we are using horizontal strips, the limits of integration will be the y-values between the intersections of the two graphs. To find these intersections, set sin(x) equal to cos(x) and solve for x:
sin(x) = cos(x)
Divide both sides by cos(x):
tan(x) = 1
Take the inverse tangent of both sides:
x = pi/4
So the intersection is at x = pi/4.
Similarly, we want to find the other intersection point. Since sin(x) and cos(x) have a period of 2pi, we can add a full period (2pi) to x = pi/4 to find the second intersection:
x = pi/4 + 2pi = 9*pi/4
Therefore, the limits of integration for x will be pi/4 and 9*pi/4.
Step 3: Set up the integral
To set up the double integral in the dx dy order, we need to rewrite the region in terms of x and y. The region is bounded by sin(x) and cos(x), so the lower and upper bounds for y should be sin(x) and cos(x), respectively. The outer integral will be with respect to y, so our integral becomes:
∫[from sin(x) to cos(x)] ∫[from x = pi/4 to x = 9*pi/4] 1 dxdy
Step 4: Calculate the integral
Integrate with respect to x first:
∫[from sin(x) to cos(x)] 1 dx = [x] [from sin(x) to cos(x)] = cos(x) - sin(x)
Now substitute the limits of integration:
∫[from pi/4 to 9*pi/4] (cos(x) - sin(x)) dx
Integrating this expression gives us:
[sin(x) + cos(x)] [from pi/4 to 9*pi/4] = [sin(9*pi/4) + cos(9*pi/4)] - [sin(pi/4) + cos(pi/4)]
Simplifying further:
[1/sqrt(2) - 1/sqrt(2)] - [1/sqrt(2) + 1/sqrt(2)] = 0
Therefore, the area of the region bounded by sin(x) and cos(x) between x = pi/4 and x = 5*pi/4 using horizontal strips is 0.
Contrary to what was suggested by the textbook, this task was not formidable; it is just that the area between these two functions happens to be zero in this particular range.