Hey there, math enthusiasts! Ever stumbled upon a double integral that looks like it belongs in a sci-fi movie rather than a textbook? Fear not! Today, we're diving headfirst into one such integral, breaking it down step-by-step so you can conquer these mathematical beasts with confidence. We're going to be tackling the integral ∫₂³ ∫₂⁸ xye^(x+y) dy dx, a classic example that combines polynomial and exponential functions, making it a thrilling challenge. So, grab your calculators, sharpen your pencils, and let's get started!
Understanding Double Integrals: The Building Blocks
Before we jump into the nitty-gritty, let's take a moment to understand what double integrals are all about. Think of them as the superheroes of multivariable calculus, allowing us to calculate volumes under surfaces, areas of complex regions, and even average values of functions over two-dimensional spaces. Essentially, a double integral is an integral of an integral – we first integrate with respect to one variable, treating the other as a constant, and then integrate the result with respect to the second variable. The order of integration matters, and in our case, we'll be integrating with respect to y first and then x. This means we'll initially treat x as a constant while we integrate xye^(x+y) with respect to y, and then we'll integrate the resulting expression with respect to x. This process might sound intimidating, but trust me, it's like learning to ride a bike – a bit wobbly at first, but smooth sailing once you get the hang of it. The key to mastering double integrals lies in understanding the limits of integration and the order in which to perform the integration. These limits define the region over which we are performing the integration, and they play a crucial role in setting up the integral correctly. For instance, in our problem, the limits 2 and 8 for the inner integral (with respect to y) indicate that we are integrating over the interval [2, 8] along the y-axis, while the limits 2 and 3 for the outer integral (with respect to x) indicate integration over the interval [2, 3] along the x-axis. Visualizing this region in the xy-plane can often provide valuable insights into the problem. Furthermore, understanding the properties of integrals, such as linearity and the ability to split integrals over sums, is essential for simplifying the integration process. Don't worry if you don't grasp everything immediately; we'll reinforce these concepts as we work through the problem step-by-step.
Setting the Stage: Our Integral Challenge
Now, let's revisit our star player: ∫₂³ ∫₂⁸ xye^(x+y) dy dx. This integral looks a bit daunting, doesn't it? But don't worry, we're going to break it down into manageable chunks. The presence of both x and y in the integrand, along with the exponential function e^(x+y), suggests that we'll need to employ some clever integration techniques. Specifically, we'll be using integration by parts, a method that's particularly useful when dealing with products of functions. Before we dive into the integration, let's make a strategic move: we can rewrite e^(x+y) as e^x * e^y. This seemingly simple step will make our lives much easier, as it allows us to separate the variables and treat e^x as a constant when integrating with respect to y. This separation is a crucial trick in solving many double integrals, and it's something you should keep in your toolbox. Think of it as a mathematical ninja move – a subtle change that leads to a significant simplification. By separating the variables, we transform the integral into a form that's more amenable to integration by parts. We'll then focus on the inner integral first, treating x as a constant, and apply integration by parts to the product ye^y. This will involve choosing appropriate u and dv terms and carefully applying the integration by parts formula. Remember, the goal is to reduce the complexity of the integrand at each step, making the overall integration process more manageable. So, with our strategic move in place, we're ready to embark on the integration journey!
Integration by Parts: Our Trusty Tool
The key to cracking this integral lies in a technique called integration by parts. Remember the formula? It's ∫ u dv = uv - ∫ v du. This formula is our secret weapon for handling integrals involving products of functions. In our case, we'll be applying it to the inner integral, ∫₂⁸ xye^(x+y) dy. Since we've already rewritten e^(x+y) as e^x * e^y, our integral now looks like ∫₂⁸ x e^x * ye^y dy. Now, we'll focus on the ye^y part. Let's choose u = y and dv = e^y dy. This means du = dy and v = e^y. Plugging these into the integration by parts formula, we get:
∫ ye^y dy = ye^y - ∫ e^y dy = ye^y - e^y + C
Notice how we strategically chose u and dv. We picked u = y because its derivative, du = dy, is simpler than y. This is the general strategy when using integration by parts – choose u such that its derivative simplifies the integral. Now, let's apply this result back to our inner integral. We have:
∫₂⁸ x e^x * ye^y dy = x e^x [ ye^y - e^y ]₂⁸
We've successfully integrated with respect to y! Now, it's just a matter of plugging in the limits of integration and simplifying. Remember, we're treating x as a constant during this inner integration, so x e^x simply acts as a coefficient. By carefully applying the integration by parts formula and choosing our u and dv terms wisely, we've made significant progress in solving this integral. This step demonstrates the power of strategic thinking in calculus – by selecting the right technique and applying it methodically, we can transform a complex problem into a manageable one. Now, let's move on to the next step and evaluate the result at the limits of integration.
Evaluating the Inner Integral: A Step Closer to Victory
Now, let's plug in those limits of integration (8 and 2) into our result from the previous step. We had:
x e^x [ ye^y - e^y ]₂⁸
Plugging in y = 8, we get:
x e^x (8e⁸ - e⁸)
And plugging in y = 2, we get:
x e^x (2e² - e²)
Subtracting the second expression from the first, we have:
x e^x (8e⁸ - e⁸ - 2e² + e²) = x e^x (7e⁸ - e²)
Wow, that looks much simpler! We've successfully evaluated the inner integral, and now we're left with a much more manageable expression to integrate with respect to x. This step highlights the importance of careful evaluation of limits, as it often leads to significant simplification. The process of plugging in the limits and subtracting can be prone to errors if not done meticulously, so it's crucial to double-check your calculations. The result we obtained, x e^x (7e⁸ - e²), is now a function of x only, which is exactly what we need to proceed with the outer integral. Notice how the constant term (7e⁸ - e²) arose from the evaluation of the limits, and it will simply act as a coefficient in the outer integral. This constant term might seem large, but don't be intimidated by it; it's just a number, and we'll handle it just like any other constant. With the inner integral evaluated, we've cleared a major hurdle, and we're now well-positioned to tackle the final step – integrating with respect to x.
The Outer Integral: The Final Showdown
We're in the home stretch now! We need to integrate the result from the previous step with respect to x: ∫₂³ x e^x (7e⁸ - e²) dx. Notice that (7e⁸ - e²) is just a constant, so we can pull it out of the integral:
(7e⁸ - e²) ∫₂³ x e^x dx
Now, we're faced with another integration by parts! This time, we're integrating x e^x with respect to x. Let's choose u = x and dv = e^x dx. This gives us du = dx and v = e^x. Applying integration by parts:
∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C
Now, we need to evaluate this from 2 to 3:
[x e^x - e^x]₂³ = (3e³ - e³) - (2e² - e²) = 2e³ - e²
Finally, we multiply this by the constant we pulled out earlier:
(7e⁸ - e²) (2e³ - e²)
And there you have it! We've successfully navigated the treacherous waters of double integration. This final step demonstrates the power of perseverance and attention to detail. After all the complex calculations and strategic maneuvers, we arrive at a relatively simple expression that represents the value of the double integral. The key to success in these problems lies in breaking them down into smaller, manageable steps and applying the appropriate techniques at each stage. By carefully tracking our progress and avoiding errors along the way, we can conquer even the most intimidating integrals. So, congratulations on making it to the end! You've earned your stripes as a double integral solver.
Final Answer and Reflections
The final answer to the integral ∫₂³ ∫₂⁸ xye^(x+y) dy dx is (7e⁸ - e²) (2e³ - e²). That's a pretty hefty number, but it's the precise value of the integral. We've journeyed through the lands of double integrals, battled with integration by parts, and emerged victorious. Remember, the key to solving these problems is to break them down into smaller steps, use the right techniques, and be patient. Double integrals can seem scary at first, but with practice, they become just another tool in your mathematical arsenal. So, keep practicing, keep exploring, and keep conquering those integrals!
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Conquering Double Integrals A Step-by-Step Solution to ∫₂³ ∫₂⁸ xyeˣ⁺ʸ dy dx