Sketch the region on paper. If it is a finite region, find its area. Round your answer to three decimal places. (If the area is not finite, enter NONE.)
S = {(x,y) | x ≥ −4, 0 ≤ y ≤ e-x/2}
∫[-4,∞] e^-(x/2) dx
= -2 e^(-x/2) [4,∞]
= -2 (e^(-∞) - e^(-2))
= -2 (0 - 1/e^2)
= 2/e^2
Well, let me give you a visual description of the region S. Imagine a coordinate plane, and draw a line passing through the point (-4, 0) parallel to the y-axis.
The region S is everything above this line, bounded by the x-axis and the graph of y = e^(-x/2) (starting at x = -4). So, it looks like a sort of curvy triangle or a hill.
Now, to find the area of this region, we need to integrate the function e^(-x/2) over the interval [-4, ∞]. However, since the upper limit is infinity, the area is unbounded and thus not finite.
Therefore, the area of region S is NONE. It's like trying to measure the size of a clown's smile - it just goes on forever!
To sketch the region S on paper, we can start by drawing the coordinate axes x and y.
Since x ≥ -4, we can draw a vertical line passing through x = -4.
To determine the region for y, note that y lies between 0 and e^(-x/2).
For x = -4, the upper bound of y is e^(-(-4)/2) = e^2.718 ≈ 14.778.
So, the region S can be represented as the area between the line x = -4 and the curve y = e^(-x/2), with y values ranging from 0 to approximately 14.778.
To find the area of this finite region, we need to integrate the given function from x = -4 to x = ∞ and then calculate the definite integral.
The area A can be calculated as follows:
A = ∫[0 to ∞] e^(-x/2) dx
This integral can be evaluated using integration techniques.
However, since it extends to ∞, the area is not finite. Therefore, the answer is NONE.
To sketch the region S on paper, we need to plot the points that satisfy the given conditions.
The condition x ≥ -4 tells us that the region starts from the vertical line x = -4 and extends towards the right. So, we can draw a vertical line passing through x = -4.
The condition 0 ≤ y ≤ e^(-x/2) tells us that the y-coordinate of any point in the region S must be greater than or equal to 0 and less than or equal to e^(-x/2), where e is the base of the natural logarithm.
To plot the region, we can start by choosing some specific x-values, calculating the corresponding y-values using the equation y = e^(-x/2), and then plotting the points (x, y). We can also choose additional points to ensure that we capture the shape of the region accurately.
Let's choose a few x-values and calculate the corresponding y-values:
For x = -4: y = e^(-(-4)/2) = e^2 ≈ 7.389
For x = -2: y = e^(-(-2)/2) = e^1 ≈ 2.718
For x = 0: y = e^(0/2) = e^0 = 1
For x = 2: y = e^(-2/2) = e^(-1) ≈ 0.368
For x = 4: y = e^(-4/2) = e^(-2) ≈ 0.135
Plotting these points on a graph, we can see how the region S looks like. It will start from the vertical line x = -4 and extend towards the right, with the y-values being bounded by the curve y = e^(-x/2).
To find the area of the finite region S, we need to find the integral of the function y = e^(-x/2) within the range of x ≥ -4. Integrating this function will give us the area under the curve between the vertical line x = -4 and the shape defined by the curve y = e^(-x/2).
To evaluate this integral, we can use calculus techniques or numerical methods such as numerical integration. However, since the question asks for rounding the answer to three decimal places, it is likely that numerical methods are expected.
Using a numerical integration method such as the trapezoidal rule or Simpson's rule, we can approximate the area under the curve.