Sketch the region in the first quadrant enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

y=10cosx, y=10sin2x,x=0

first solve the two to find where they cross

10sin2x = 10cosx
sin2x - cosx = 0
2sinxcox - cosx = 0
cosx(2sinx - 1) = 0
cosx = 0 or sinx = 1/2
x = π/2 or x = π/4 in the first quadrant

so I would do it in two integrals
1. integral [ 10sin2x] from 0 to π/4
2. integral [10cosx] from π/4 to π/2

Area = -20cos2x | from 0 to π/4 + 10sinx | from π/4 to π/2
= (-20cos2(π/4) + 20cos 0) + (10sinπ/2 - (10sinπ/4))
= 0 + 20 + 0 - 10/√2
= 20 - 10/√2
= (20√2 - 10)/√2 = appr. 12.929

To sketch the region in the first quadrant enclosed by the given curves, we first need to determine the points where these curves intersect.

Start by setting the two functions equal to each other:
10cos(x) = 10sin(2x)

Divide both sides by 10:
cos(x) = sin(2x)

Now, using the double-angle identity for sine, we can rewrite sin(2x) as 2sin(x)cos(x):

cos(x) = 2sin(x)cos(x)

Simplify by canceling out common factors:
1 = 2sin(x)

Divide both sides by 2:
sin(x) = 1/2

To find the solutions to this equation, we need to determine in which interval the sine function takes the value of 1/2. The sine function is positive in the first and second quadrants, so we look for the first positive value in the first quadrant.

In the first quadrant, x = π/6 is the angle that satisfies sin(x) = 1/2.

Now, let's sketch the region enclosed by the curves:

First, let's plot the curves separately:

Curve 1: y = 10cos(x)
This is a cosine curve that starts at y = 10 (when x = 0), decreases to its minimum value at x = π/2, and then increases again.

Curve 2: y = 10sin(2x)
This is a sine curve with double the frequency of the standard sine curve. It starts at y = 0 (when x = 0), reaches its maximum at x = π/6, and then decreases symmetrically.

Now, we draw the line x = 0, which is a vertical line passing through the y-axis.

The region enclosed by the curves is the area between the curves y = 10cos(x) and y = 10sin(2x), bounded by the y-axis on the left and the line x = 0 on the right.

The next step is to decide whether to integrate with respect to x or y. In this case, it is more convenient to integrate with respect to x since the given equations are given in terms of x.

To find the area of the region, we need to integrate the difference between the two curves from x = 0 to x = π/6:

Area = ∫[0, π/6] (10cos(x) - 10sin(2x)) dx

By evaluating this integral, we can find the exact area of the region between the curves in the first quadrant.