find the average value of the function
f(x,y)=xe^y
over the plane region R is the triangle with vertices (0,0), (1,0) and (1,1)
Well, my friend, let's clown around with this problem!
To find the average value of a function over a plane region, we need to calculate the double integral of the function over that region and then divide it by the area of the region.
First, let's set up the integral. The region R is a triangle with vertices (0,0), (1,0), and (1,1). It's clear that x ranges from 0 to 1, and y ranges from 0 to y = x.
Now, let's go ahead and calculate the double integral:
∫∫(R) x * e^y dA
To integrate with respect to y, we need to fix the values of x:
∫[0 to 1] ∫[0 to x] x * e^y dy dx
Integrating with respect to y, we get:
∫[0 to 1] [x * e^y] evaluated from 0 to x dx
Simplifying this, we have:
∫[0 to 1] (x * e^x - x) dx
Now, we need to integrate with respect to x:
∫[0 to 1] (x * e^x - x) dx = (∫[0 to 1] x * e^x dx) - (∫[0 to 1] x dx)
Integrating the first term by using integration by parts, we find:
(x - 1) * e^x - e^x evaluated from 0 to 1
Now, we plug in the limits of integration:
(1 - 1) * e^1 - e^1 - (0 - 0) * e^0 + e^0
Simplifying this, we have:
e - e - 0 + 1
Finally, after lots of clowning around, we get the average value of the function f(x, y) = x * e^y over the region R to be:
1 - e
Thus, the average value is 1 - e.
Hope I made you smile while solving this problem!
To find the average value of the function f(x,y) = xe^y over the plane region R, we need to evaluate the double integral of f(x,y) over the region R and divide it by the area of R.
Step 1: Find the limits of integration for x and y.
Since the region R is a triangle with vertices (0,0), (1,0), and (1,1), we can describe it as follows:
0 ≤ x ≤ 1 (horizontal limit)
0 ≤ y ≤ x (vertical limit)
Step 2: Set up the double integral.
The double integral of f(x,y) over the region R can be expressed as:
∫∫R f(x,y) dA,
where dA represents the differential area element.
Step 3: Compute the double integral.
∫∫R f(x,y) dA = ∫∫R xe^y dA.
Since R is a triangle, we can convert the limits of integration into a single integral as follows:
∫0^1 ∫0^x xe^y dy dx.
Step 4: Evaluate the integral.
∫0^1 ∫0^x xe^y dy dx = ∫0^1 xe^y|0^x dx.
Evaluating the inner integral:
∫0^x xe^y dy = xe^y|0^x = x(e^x - e^0) = x(e^x - 1).
Now evaluating the outer integral:
∫0^1 x(e^x - 1) dx.
Using integration by parts on the term x(e^x - 1), we can find the antiderivative:
∫ x(e^x - 1) dx = x(e^x - 1) - ∫ 1*e^x dx
= x(e^x - 1) - e^x + C.
Evaluating the outer integral:
∫0^1 x(e^x - 1) dx = [x(e^x - 1) - e^x]_0^1
= (1(e^1 - 1) - e^1) - (0 - 0)
= (e - 1 - e) - 0
= -1.
Step 5: Find the area of R.
To find the area of the triangle, we can use the formula for the area of a triangle:
area = (1/2) * base * height = (1/2) * 1 * 1 = 1/2.
Step 6: Calculate the average value.
To find the average value of f(x,y) over the region R, divide the integral result by the area of R:
Average value = (-1) / (1/2) = -2.
Therefore, the average value of the function f(x,y) = xe^y over the plane region R (the triangle with vertices (0,0), (1,0), and (1,1)) is -2.
To find the average value of a function over a region, you need to calculate the double integral of the function over that region, and then divide it by the area of the region.
In this case, the region R is a triangle with vertices (0,0), (1,0), and (1,1). To find its area, you can use the formula for the area of a triangle: A = 1/2 * base * height.
The base of the triangle is the length between the points (0,0) and (1,0), which is 1 unit. The height of the triangle is the length between the points (0,0) and (1,1), which is again 1 unit. Therefore, the area of the triangle is A = 1/2 * 1 * 1 = 1/2.
Now, let's calculate the double integral of the function f(x,y) = x*e^y over the region R.
∫∫R f(x,y) dA = ∫∫R xe^y dA
To evaluate this double integral, we can use iterated integration. Since R is a triangle, we can integrate over the variables x and y over different intervals.
First, let's integrate with respect to x:
∫∫R xe^y dA = ∫[0,1] ∫[0,y] xe^y dx dy
Integrating with respect to x, we get:
= ∫[0,1] [1/2 * x^2 * e^y] [0,y] dy
= ∫[0,1] (1/2 * y^2 * e^y) dy
Now, let's integrate with respect to y:
= 1/2 * ∫[0,1] y^2 * e^y dy
To evaluate this integral, you can use integration by parts or a substitution method, such as u-substitution.
Once you have calculated this integral, divide it by the area of the region (1/2), and you will get the average value of the function over the given triangle region R.
average value is volume/base area
one boundary of the region is the line y=x
v = ∫[0,1]∫[0,x] xe^y dy dx
= ∫[0,1] (xe^y)[0,x] dx
= ∫[0,1] x(e^x-1) dx
= (e^x(x-1)-x^2/2)[0,1]
= 1/2
area of base = 1/2
avg value = 1