find the average value of the function f(x,y)=e^(-x^2) over the plane region R which is the triangle with vertices (0,0), (1,0) and (1,1)
Please check the function for typo, since
f(x,y)=e^(-x²) is independent of y.
Assuming no typo,
the region R is bounded between x=0 and x=1, and y=0 and y=x.
So the integegration
dy from 0 to x
dx from 0 to 1.
I=∫∫ye^(-x²)dy dx
=∫xe^-x² dx
=1/2-e^(-1)/2
The area of R is 1/2, so
Average value
= I/(1/2)
=(e-1)/e
To find the average value of the function f(x, y) = e^(-x^2) over the given plane region R, we need to calculate the double integral of the function over the region R and then divide it by the area of R.
Step 1: Find the limits of integration for x and y.
Since the triangle has vertices (0, 0), (1, 0), and (1, 1), the limits for x will be from 0 to 1, and for y, it will be from 0 to y = x.
Step 2: Set up the double integral.
The average value of the function can be calculated as:
Average value = (1/Area of R) * ∬(R) [f(x, y) dA]
Here, dA represents the differential element of area.
We can rewrite the expression as:
Average value = (1/Area of R) * ∫∫(R) [e^(-x^2) dx dy]
Step 3: Evaluate the double integral.
∫∫(R) [e^(-x^2) dx dy] = ∫(0 to 1) ∫(0 to x) [e^(-x^2) dy dx]
= ∫(0 to 1) [e^(-x^2) * (x - 0)] dx
= ∫(0 to 1) [x * e^(-x^2)] dx
Step 4: Calculate the area of the region.
The area of the triangle can be calculated as half of the base times the height.
Area of R = (1/2) * (1 - 0) * (1 - 0) = 1/2
Step 5: Plug in the values to calculate the average value.
Average value = (1/Area of R) * ∫(0 to 1) [x * e^(-x^2)] dx
= (2/1) * ∫(0 to 1) [x * e^(-x^2)] dx
= 2 * [-1/2 * e^(-x^2)] from 0 to 1
= 2 * [-1/2 * e^(-1^2) - (-1/2 * e^(-0^2))]
= 2 * [-1/2 * e^(-1) - (-1/2 * e^0)]
= 2 * [-1/2 * (1/e) - (-1/2 * 1)]
= 2 * [-1/2e + 1/2]
= -1/e + 1
Therefore, the average value of the function f(x, y) = e^(-x^2) over the triangle with vertices (0, 0), (1, 0), and (1, 1) is -1/e + 1.
To find the average value of the function f(x, y) = e^(-x^2) over the plane region R, you need to evaluate the double integral of the function over the region R and then divide it by the area of R.
Here's how you can do it step by step:
Step 1: Determine the boundaries of the integration for x and y. We are given that the region R is a triangle with vertices (0,0), (1,0), and (1,1). Hence, for x, the boundaries are from 0 to 1 and for y, the boundaries are from 0 to x (since y is bounded by the line y = x within the triangle).
Step 2: Set up the integral. We need to integrate the function f(x, y) = e^(-x^2) over the region R. So the integral becomes:
I = ∫∫R e^(-x^2) dy dx
Where R represents the triangle region.
Step 3: Evaluate the integral. We can integrate f(x, y) = e^(-x^2) with respect to y, treating x as a constant, and then integrate with respect to x. Let's start by integrating with respect to y:
∫ e^(-x^2) dy = e^(-x^2) * y + C
Evaluating this integral from y = 0 to y = x yields (e^(-x^2) * x) - (e^(-x^2) * 0) = x * e^(-x^2)
Step 4: Integrate with respect to x. Integrate x * e^(-x^2) with respect to x from x = 0 to x = 1:
∫[0,1] x * e^(-x^2) dx
To integrate this, you can use a substitution. Let u = -x^2 and du = -2x dx:
∫ e^u (-du/2) = -(1/2) e^u + C
Plugging back in for u and applying the limits:
-(1/2) e^(-x^2) | [0,1] = -(1/2) e^(-1) + (1/2) e^0 = -(1/2) e^(-1) + (1/2)
Step 5: Calculate the area of the region R. The region R is a triangle with a base of length 1 and a height of 1, so the area is (1/2) * 1 * 1 = 1/2.
Step 6: Calculate the average value. Finally, divide the value obtained in Step 4 by the area obtained in Step 5:
Average value = [-(1/2) e^(-1) + (1/2)] / (1/2) = -e^(-1) + 1
Therefore, the average value of the function f(x, y) = e^(-x^2) over the triangle region R is -e^(-1) + 1.