Evaluating Double Integral Of F(x, Y) = 2 + Xy Over A Region
Hey guys! Today, we're diving deep into a fascinating problem in multivariable calculus. We're going to explore the integral of the function f(x, y) = 2 + xy over a specific region in the XY-plane. This region is bounded by some interesting inequalities, and figuring out the integral will require a solid understanding of double integrals and how to set them up correctly. So, buckle up and let's get started!
Defining the Battlefield: The Region of Integration
First, let's understand the battlefield we're working on, which is the region in the XY-plane. We're given the following boundaries:
- 1 ≤ x ≤ 2
- 1 ≤ y ≤ x
These inequalities define a region that's bounded by vertical lines at x = 1 and x = 2, and by the horizontal line y = 1 and the line y = x. To really grasp this, it's super helpful to sketch this region. Go ahead, grab a piece of paper or your favorite graphing tool and draw these lines. You'll see that the region is a trapezoid-like shape nestled in the first quadrant.
Understanding this region is crucial because it dictates the limits of integration in our double integral. The limits tell us the range of x and y values we need to consider when summing up the function f(x, y) over this area. If we get these limits wrong, the entire calculation goes sideways, and we definitely don't want that!
Think of it this way: we're trying to find the volume under the surface defined by f(x, y) and above this trapezoidal region. To do this, we'll slice the region into tiny rectangles and calculate the contribution of f(x, y) on each rectangle. The double integral is just a way of adding up all these tiny contributions in a precise and systematic way. To do this, the main keywords are double integrals, limits of integration, and bounded region.
Setting Up the Double Integral: Our Weapon of Choice
Now that we know the region, let's set up the double integral, our primary weapon in this calculus quest. This is where things can get a little tricky, but don't worry, we'll break it down step by step. Since we have y bounded between 1 and x, and x bounded between 1 and 2, we can write the double integral as:
∫(from 1 to 2) ∫(from 1 to x) (2 + xy) dy dx
Notice the order of integration: we're integrating with respect to y first (inner integral) and then with respect to x (outer integral). This order is dictated by the limits we have. Since the limits for y are functions of x (1 to x), we must integrate with respect to y first. If we tried to integrate with respect to x first, we'd run into trouble because our limits for x would need to be expressed in terms of y, which is possible but a bit more complicated in this case. So, it's better to stick with the order that the problem naturally suggests.
Remember, the inner integral treats x as a constant while we integrate with respect to y. This is a key concept in multivariable calculus. We're essentially finding the area under the curve of (2 + xy) as y varies from 1 to x, for a fixed value of x. Then, the outer integral sums up these areas over the range of x values from 1 to 2, giving us the total volume. To achieve this, the most important keywords are order of integration, double integral setup, and limits of integration.
The Inner Integral: Conquering the First Challenge
Let's tackle the inner integral first. This is where we integrate (2 + xy) with respect to y, treating x as a constant. So, we have:
∫(from 1 to x) (2 + xy) dy = [2y + (1/2)xy^2](from 1 to x)
We've found the antiderivative of (2 + xy) with respect to y, which is 2y + (1/2)xy^2. Now, we need to evaluate this at the limits of integration, y = x and y = 1. This means we'll substitute these values into the expression and subtract:
[2x + (1/2)x(x^2)] - [2(1) + (1/2)x(1)^2] = 2x + (1/2)x^3 - 2 - (1/2)x
Simplifying this, we get:
(3/2)x + (1/2)x^3 - 2
This expression represents the result of the inner integral. It's a function of x, which makes sense because we integrated out the y variable. Now, this expression becomes the integrand for our outer integral. Think of it as the area we calculated for a particular slice at a fixed x value. The keywords here are inner integral, antiderivative, and evaluating limits.
The Outer Integral: The Final Showdown
Now, for the outer integral, we need to integrate the result from the inner integral with respect to x, from 1 to 2. So, we have:
∫(from 1 to 2) [(3/2)x + (1/2)x^3 - 2] dx
Let's find the antiderivative of this expression with respect to x:
[(3/4)x^2 + (1/8)x^4 - 2x](from 1 to 2)
Now, we evaluate this at the limits of integration, x = 2 and x = 1:
[(3/4)(2)^2 + (1/8)(2)^4 - 2(2)] - [(3/4)(1)^2 + (1/8)(1)^4 - 2(1)]
Simplifying this, we get:
[3 + 2 - 4] - [3/4 + 1/8 - 2] = 1 - [-9/8] = 1 + 9/8 = 17/8
So, the final result of the double integral is 17/8. This represents the volume under the surface f(x, y) = 2 + xy and above the region defined by our inequalities. Congratulations, guys! We've conquered this integral beast! Key terms here include outer integral, antiderivative, and final result.
Analyzing the Statements: Truth or False?
Now, let's circle back to the statements mentioned in the original problem. These statements are like little challenges, and we need to use our knowledge to determine if they're true or false.
The statements are:
I - The area defined in the XY plane is 1/2
II - The integral of the function f(x, y) in the region of the plane
Let's break them down:
Statement I: The Area of the Region
To find the area of the region, we can use a double integral where the integrand is simply 1. This is because integrating 1 over a region gives you the area of that region. So, we need to calculate:
∫(from 1 to 2) ∫(from 1 to x) 1 dy dx
This is similar to our previous integral, but much simpler. The inner integral is:
∫(from 1 to x) 1 dy = [y](from 1 to x) = x - 1
Now, the outer integral is:
∫(from 1 to 2) (x - 1) dx = [(1/2)x^2 - x](from 1 to 2)
Evaluating this at the limits, we get:
[(1/2)(2)^2 - 2] - [(1/2)(1)^2 - 1] = [2 - 2] - [1/2 - 1] = 0 - [-1/2] = 1/2
So, Statement I is TRUE! The area of the region is indeed 1/2. Keywords: area calculation, double integral, and statement analysis.
Statement II: The Integral of the Function
We already calculated the integral of the function f(x, y) over the region in the previous sections. We found that the integral is 17/8. Statement II would need to provide a specific value or a comparison to be evaluated as true or false. Since it's incomplete, we can't definitively say if it's true or false without more information. However, we know the correct value of the integral is 17/8. The important key terms are integral calculation, previous result, and statement completion.
Conclusion: Victory Lap!
And there you have it, guys! We've successfully navigated the world of double integrals, calculated the volume under a surface, and analyzed the given statements. This problem highlights the power of multivariable calculus in solving real-world problems and understanding complex functions. Remember, the key is to break down the problem into smaller, manageable steps, and always visualize the region of integration. Keep practicing, and you'll become a calculus master in no time! Most important keywords are problem-solving, multivariable calculus, and understanding concepts. The key takeaway is how to apply these concepts effectively.
