Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer. (Round your answer to four significant digits.) HINT [See Quick Example page 1028.]

Enclosed by y = 2x, y = x + 2, x = −2, and x = 2

To find the area of the region enclosed by the given curves, we need to find the points of intersection between them and integrate the difference in their equations over that range.

First, let's graph the curves y = 2x, y = x + 2, x = -2, and x = 2 to visualize the region:

GRAPH

From the graph, we can see that the region in question is the triangular region bounded by the x-axis, the line y = 2x, and the line y = x + 2.

To find the points of intersection between these curves, we can set their equations equal to each other:

2x = x + 2

Subtracting x from both sides gives:

x = 2

So, the lines y = 2x and y = x + 2 intersect at x = 2.

Now, we can integrate the difference between the equations of the curves over the range from x = -2 to x = 2 to find the area:

A = ∫[from -2 to 2] (2x - (x + 2)) dx

A = ∫[from -2 to 2] (x - 2) dx

Using the fundamental theorem of calculus, we can evaluate the integral:

A = [(x^2 / 2) - 2x] [from -2 to 2]

A = [(2^2 / 2) - 2(2)] - [((-2)^2 / 2) - 2(-2)]

A = (4/2 - 4) - (4/2 + 4)

A = 2 - 4 - 2 + 4

A = 0

Therefore, the area of the region enclosed by the curves y = 2x, y = x + 2, x = -2, and x = 2 is 0 square units.

To find the area of the indicated region, we need to calculate the area between the curves y = 2x and y = x + 2, bounded by the lines x = -2 and x = 2.

First, let's graph the two curves and see how they intersect or overlap each other. This will help us visualize the region and understand what we need to do next.

To graph the curves, plot several points for each equation and then connect them to get the overall shape.

For y = 2x:
When x = -2, y = 2(-2) = -4
When x = 0, y = 2(0) = 0
When x = 2, y = 2(2) = 4

For y = x + 2:
When x = -2, y = (-2) + 2 = 0
When x = 0, y = 0 + 2 = 2
When x = 2, y = 2 + 2 = 4

Plotting these points and connecting them, we have a straight line for y = 2x slanting upwards from (-2,-4) to (2, 4), and a straight line for y = x + 2 slanting upwards from (-2,0) to (2,4).

Now, we need to determine the points of intersection between these two curves. To do this, we set them equal to each other and solve for x:

2x = x + 2
x = 2

So, at x = 2, the two curves intersect.

Next, we calculate the area by integrating the difference of the two functions over the given interval.

The area is given by the definite integral of (top function - bottom function) with respect to x and evaluated from the leftmost bound to the rightmost bound.

To find the area, we integrate (2x - (x + 2)) with respect to x from x = -2 to x = 2:

∫(2x - (x + 2))dx from -2 to 2

Simplifying the integral, we have:

∫(x - 2) dx from -2 to 2

Integrating, we get:

∫x dx - ∫2 dx from -2 to 2

Integrating x gives us (1/2)x^2, and integrating 2 gives us 2x. Evaluating the integral from -2 to 2:

[1/2 * x^2] from -2 to 2 - [2 * x] from -2 to 2

Substituting the limits of integration:

(1/2 * 2^2) - (1/2 * (-2)^2) - (2 * 2) + (2 * (-2))

Simplifying:

2 - 2 - 4 + (-4) = -8

Since the area cannot be negative, we take the absolute value of the result, which gives us:

| -8 | = 8

Therefore, the area of the region enclosed by the curves y = 2x, y = x + 2, x = -2, and x = 2 is 8 square units.