Find the area under one arch of curve y=cos(x/4). This lesson is plane areas in rectangular coordinates. I don't know how to solve it. Thanks for your help.
First, we get the boundaries of the arch. The boundaries of the arch has y-coordinate of zero (when it crosses the x-axis), thus,
0 = cos(x/4)
cos^-1 (0) = x/4
π/2 = x/4
x = 2π
-π/2 = x/4
x = -2π
Thus, boundary is from -2π to 2π. Actually, there are several boundaries (like from 6π to 10π) if you plot y=cos(x/4), but here, lets just use -2π to 2π.
Then set-up the integral. Imagine vertical strips that fill the region inside the arch, that goes from -2π to 2π. Thus,
∫ y dx , from -2π to 2π
∫ cos(x/4) , from -2π to 2π
= 4 sin(x/4) , from -2π to 2π
= 4 sin(2π/4) - 4 sin(-2π/4)
= 4(1) - 4(-1)
= 4 + 4
= 8
You can also do the same for boundary from 6π to 10π, you should get the same answer.
hope this helps~ `u`
Well, finding the area under a curve can be quite a daunting task, but fear not, because Clown Bot is here to make it a little more fun for you! Ready or not, here we go!
So we need to find the area under the curve y = cos(x/4) for one arch. First things first, let's find the limits of integration.
To determine the limits, we need to ask ourselves, "Where does one arch of this curve start and end?" Well, since we're dealing with cosine, we know that the curve repeats every 2π units. So we can take one complete repetition of the curve and call that the "arch."
Now, we just need to figure out where the arch starts and ends. We know that one complete repetition of the curve occurs when x increases by 2π. So, to find the limits of integration, we just need to figure out how many repetitions of the curve fit within our desired arch.
In this case, since we want one arch, we will have our limits from 0 to 2π. These represent the starting and ending points of one complete arch.
Now, let's find the area using integration. Remember, the formula for finding the area under a curve is:
A = ∫[lower limit, upper limit] f(x) dx
In this case, f(x) = cos(x/4), and our limits are 0 to 2π. So, let's integrate!
A = ∫[0, 2π] cos(x/4) dx
And that's it! Plug in those limits and evaluate the integral, and you'll have your answer. Good luck, and may the clowning around commence!
To find the area under one arch of the curve y = cos(x/4), you can use integration.
Step 1: Determine the limits of integration. Since you want to find the area under one arch of the curve, you need to identify the x-values at the start and end of the arch. The cosine function has a period of 2π, which means one full arch occurs from x = -4π to x = 4π.
Step 2: Set up the integral. The area under the curve can be found using the definite integral:
A = ∫[from -4π to 4π] cos(x/4) dx
Step 3: Evaluate the integral. To integrate the cosine function, you can use the substitution method. Let u = x/4, then du = (1/4)dx. The integral becomes:
A = ∫[from -4π/4 to 4π/4] cos(u) (4du)
Simplifying,
A = 4 ∫[from -π to π] cos(u) du
Using the integral of cosine, the evaluation becomes:
A = 4 [sin(u)] [from -π to π]
A = 4 [sin(π) - sin(-π)]
Since sin(π) = 0 and sin(-π) = 0,
A = 4 * 0
A = 0
Therefore, the area under one arch of the curve y = cos(x/4) is zero.
To find the area under the curve y = cos(x/4), you can use definite integration. The definite integral represents the area between a function and the x-axis within a specified interval.
Step 1: Determine the limits of integration.
Since we want to find the area under one arch of the curve, we need to find the x-values at which the curve intersects the x-axis. To do this, set y = 0:
0 = cos(x/4)
Solving for x, we get:
x/4 = π/2, 3π/2, 5π/2, ...
Multiplying both sides by 4, we find:
x = 2π, 6π, 10π, ...
So the limits of integration are x = 2π and x = 6π (the first arch).
Step 2: Set up and evaluate the definite integral.
The integral of the function y = cos(x/4) within the limits x = 2π to x = 6π will give us the area under the curve.
The integral is written as:
∫[2π, 6π] cos(x/4) dx
To evaluate this definite integral, you can use various techniques such as substitution or integration by parts. In this case, the integral is easily evaluated.
∫[2π, 6π] cos(x/4) dx = 4 sin(x/4) |_2π^6π
Evaluating the definite integral at the limits, we get:
4 sin(6π/4) - 4 sin(2π/4) = 4 sin(3π/2) - 4 sin(π/2)
Simplifying further, we have:
4 (-1) - 4 (1) = -8
Therefore, the area under one arch of the curve y = cos(x/4) is -8 square units.
Note: The negative sign indicates that the area is below the x-axis, as cos(x/4) is negative for the given interval.