2. Let f(x) = ksin(kx), where k is a positive constant. Find the area of the region bounded by one arch of the graph of f and the x-axis

let's look at the first arch starting at the origin.

the period of ksin(kx) is 2π/k
so the arch goes from 0 to π/k

Area = ∫(ksin(kx) dx from 0 to π/k
=[ - cos(kx) ] form 0 to π/k
= -cos(π) - (-cos0))
= -(-1) + 1
= 2

To find the area of the region bounded by one arch of the graph of f(x) = ksin(kx) and the x-axis, we will use definite integration.

Here are the steps to find the area:

Step 1: Determine the boundaries of integration. To find the area of one arch, we need to find the values of x where the function f(x) intersects or touches the x-axis. Recall that sin(x) = 0 when x = nπ, where n is an integer. Since we have f(x) = ksin(kx), we get kx = nπ, and therefore, x = nπ/k. Let's denote these values as x1 and x2.

Step 2: Set up the definite integral. The area between the curve and the x-axis can be found by integrating the function f(x) over the interval [x1, x2].

∫[x1, x2] f(x) dx = ∫[x1, x2] ksin(kx) dx

Step 3: Evaluate the integral. To integrate ksin(kx), we can use the substitution u = kx, du = k dx.

Now, the integral becomes:

∫[x1, x2] ksin(kx) dx = ∫[x1, x2] sin(u) du

Using the antiderivative of sin(u), which is -cos(u), the integral becomes:

= -cos(u) ∣ [x1, x2]
= -cos(kx) ∣ [x1, x2]

Step 4: Compute the area. Finally, we can calculate the area by substituting the values x1 and x2 into the integrated function:

Area = -cos(kx) ∣ [x1, x2] = -[cos(kx2) - cos(kx1)]

This gives us the area of the region bounded by one arch of f(x) and the x-axis.