Find the area of the region bounded by the curves y = sin x, y = csc^2x, x = pi/4, and x = (3pi)/4.
A quick peek at our problem:
https://www.wolframalpha.com/input/?i=plot+y+%3D+sin+x%2C+y+%3D+csc%5E2x
looks like they intersect at x = π/2 and there is symmetry
so we can take
area = 2∫ (csc^2 x - sinx) dx from π/4 to π/2
recalling that the derivative of cotx = -sec^2 x
= 2[ (- cotx + cosx) ] from π/4 to π/2
= 2 ( (-cot π/2 + cos π/2 - (-cot π/4 + cos π/4) )
= 2( 0 + 0 - (-1 + √2/2) )
= 2(1 - √2/2)
= 2 - √2
Thank you!
Sure, let's find the area of that region together! Get ready for some mathematical circus fun.
First, let's find the points of intersection between y = sin(x) and y = csc^2(x). We can do this by setting the two equations equal to each other:
sin(x) = csc^2(x)
Now, we can express csc^2(x) as 1/sin^2(x) and solve for sin(x):
sin(x) = 1/sin^2(x)
Taking the reciprocal of both sides gives us:
1/sin(x) = sin^2(x)
Simplifying further by using the identity sin^2(x) + cos^2(x) = 1, we get:
cos^2(x) = 1
Since the cosine squared of an angle is always less than or equal to 1, the equation cos^2(x) = 1 is satisfied only when cos(x) = ±1. This means that the points of intersection between the two curves occur when sin(x) = ±1.
Now, let's find the x-values of these points. Taking the inverse sine of ±1 gives us:
x = pi/2 for sin(x) = 1
x = (3pi)/2 for sin(x) = -1
Since we are only interested in the region between x = pi/4 and x = (3pi)/4, the points we're looking for are (pi/2, 1) and ((3pi)/2, -1).
Next, we need to find the y-values of these points. We'll substitute the x-values we just found into the equations to get:
y = sin(pi/2) = 1
y = sin((3pi)/2) = -1
Now we can calculate the area of the region by integrating y = sin(x) - y = csc^2(x) between x = pi/4 and x = (3pi)/4:
Area = ∫ (sin(x) - csc^2(x)) dx
However, since we're dealing with curves that can't be expressed by elementary functions, finding the definite integral becomes a bit of a circus act. So let's call in the integral acrobat to help us with this one!
*Acrobat bot flips into the scene*
Acrobat bot: "Hey there! It looks like you need some help with this integral. Sadly, I can't do it due to its complexity, but if you have any other math problems or need more mathematical fun, I'm here to assist you!"
Clown bot: "Thanks for your honesty, Acrobat bot! It seems like our integral has proved to be a bit of a daredevil. In this case, I recommend using numerical methods or graphing software to approximate the area. Remember, even for a clown bot, there are limits to what we can juggle. Keep the math circus going!"
Acrobat bot: "Absolutely, Clown bot! It's all about finding the right tools to keep the show rolling. So, if you have any other math inquiries or need more amusement, just let us know!"
And so, the quest to find the exact area of the region becomes an adventure for another day. Keep exploring, and remember that math can always surprise and amaze us with its complexity!
To find the area of the region bounded by the given curves, we need to compute the definite integral between the two given values of x.
Step 1: Determine the intersection points.
To find the bounds of integration, we need to find where the curves intersect.
The first curve is y = sin(x), and the second curve is y = csc^2(x).
Since csc^2(x) = 1/sin^2(x), we have y = 1/sin^2(x).
Setting sin(x) equal to 1/sin^2(x), we get:
sin(x) = 1/sin^2(x)
Multiplying both sides by sin^2(x), we get:
sin^3(x) = 1
Taking the cube root of both sides, we get:
sin(x) = 1
The solution to this equation is x = pi/2 (since sin(pi/2) = 1). So, the curves intersect at x = pi/2.
Step 2: Set up the definite integral.
Since we are calculating the area of the region bounded by the curves, we need to integrate the difference between the two functions over the given interval.
The area bounded by the curves can be expressed as the integral from x = pi/4 to x = (3pi)/4 of the difference between the two functions: A = ∫[pi/4, 3pi/4] (sin(x) - 1/sin^2(x)) dx.
Step 3: Evaluate the integral.
To calculate the integral, we can split it into two parts:
∫[pi/4, pi/2] (sin(x) - 1/sin^2(x)) dx + ∫[pi/2, 3pi/4] (sin(x) - 1/sin^2(x)) dx
For the first integral, we can rewrite it as:
∫[pi/4, pi/2] (sin(x) - csc^2(x)) dx
Calculating the integral for this part:
∫[pi/4, pi/2] (sin(x) - csc^2(x)) dx = [-cos(x) + cot(x)] from x = pi/4 to x = pi/2
For the second integral, we can rewrite it as:
∫[pi/2, 3pi/4] (sin(x) - csc^2(x)) dx
Calculating the integral for this part:
∫[pi/2, 3pi/4] (sin(x) - csc^2(x)) dx = [cos(x) + cot(x)] from x = pi/2 to x = 3pi/4
Step 4: Compute the final result.
Now, we can evaluate the individual parts and subtract them:
[-cos(x) + cot(x)] from pi/4 to pi/2 + [cos(x) + cot(x)] from pi/2 to 3pi/4
Calculating the first part at the upper bound (pi/2):
[-cos(pi/2) + cot(pi/2)] - [-cos(pi/4) + cot(pi/4)]
Cos(pi/2) is equal to 0, cot(pi/2) is undefined, cos(pi/4) is 1/sqrt(2), and cot(pi/4) is 1. Thus, the first part evaluates to:
[0 + undefined] - [-1/sqrt(2) + 1]
Next, we calculate the second part at the upper bound (3pi/4):
[cos(3pi/4) + cot(3pi/4)] - [cos(pi/2) + cot(pi/2)]
Cos(3pi/4) is -1/sqrt(2), cos(pi/2) is 0, and cot(3pi/4) is -1. Thus, the second part evaluates to:
[-1/sqrt(2) + (-1)] - [0 + undefined]
However, note that cot(pi/2) and cot(3pi/4) are both undefined. Therefore, the second part evaluates to undefined as well.
Since one part of the integral evaluates to undefined, we cannot find the area bounded by the curves using this method. It is possible that the curves do not completely enclose a region, or there may be some other error in the problem statement.