Mastering Definite Integrals & Marginal Production
Hey guys! Today, we're diving deep into the fascinating world of integral calculus, specifically focusing on how to solve definite integrals. Definite integrals, unlike their indefinite counterparts, give us a specific numerical value, representing the area under a curve between two defined limits. We'll break down the process step-by-step, making sure you not only understand the how but also the why behind each calculation. So, grab your calculators, and let's get started!
1. Evaluating the Definite Integral of (15x² + 12x - 5) from 1 to 2
Let's kick things off with our first definite integral: ∫[1,2] (15x² + 12x - 5) dx. This might look intimidating at first, but don't worry; we'll tackle it together. The key here is the power rule for integration, which states that ∫xⁿ dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. However, since we're dealing with a definite integral, we won't need to worry about the constant of integration because it will cancel out during the evaluation process. Our main keywords for this section are definite integrals, power rule, and integral calculus.
First, let's find the indefinite integral of the given expression. We'll apply the power rule to each term individually:
- ∫15x² dx = 15 * (x^(2+1))/(2+1) = 15 * (x³/3) = 5x³
- ∫12x dx = 12 * (x^(1+1))/(1+1) = 12 * (x²/2) = 6x²
- ∫-5 dx = -5x
Now, we combine these results to get the indefinite integral:
∫(15x² + 12x - 5) dx = 5x³ + 6x² - 5x
Next, we need to evaluate this indefinite integral at our limits of integration, which are 2 and 1. This means we'll plug in 2 and 1 into the expression and subtract the result at the lower limit (1) from the result at the upper limit (2). This process is the cornerstone of evaluating definite integrals, and understanding it is crucial for mastering integral calculus. Think of it as finding the difference in the antiderivative at two specific points, which gives us the net change, or the area under the curve, between those points.
So, let's do the calculations:
- At x = 2: 5(2)³ + 6(2)² - 5(2) = 5(8) + 6(4) - 10 = 40 + 24 - 10 = 54
- At x = 1: 5(1)³ + 6(1)² - 5(1) = 5 + 6 - 5 = 6
Finally, we subtract the value at the lower limit from the value at the upper limit:
54 - 6 = 48
Therefore, ∫[1,2] (15x² + 12x - 5) dx = 48. Woohoo! We've successfully solved our first definite integral. Remember, the power rule is your friend, and meticulous calculations are key to avoiding errors. The beauty of integral calculus lies in its ability to transform complex problems into manageable steps, and this example perfectly illustrates that principle.
2. Tackling the Integral of (x² - 2/√[3]x) from 1 to 3
Okay, let's move on to our next challenge: ∫[1,3] (x² - 2/√[3]x) dx. This one introduces a fractional exponent, which might seem tricky, but fear not! We'll use the same power rule we used before, just with a slightly different twist. The core concept remains the same: find the antiderivative and then evaluate it at the limits of integration. Let's break it down. Our main keywords for this part are fractional exponents, antiderivative, and the ever-present power rule. We're essentially building on our understanding of integral calculus, adding more tools to our toolbox.
First, we need to rewrite the second term, 2/√[3]x, using exponents. Remember that √[n]x is the same as x^(1/n). So, √[3]x is x^(1/3), and 2/√[3]x is 2x^(-1/3). This manipulation is crucial because it allows us to apply the power rule. Handling fractional exponents is a fundamental skill in calculus, and this example provides a great opportunity to practice it.
Now our integral looks like this: ∫[1,3] (x² - 2x^(-1/3)) dx.
Let's find the indefinite integral of each term:
- ∫x² dx = (x^(2+1))/(2+1) = x³/3
- ∫-2x^(-1/3) dx = -2 * (x^(-1/3 + 1))/(-1/3 + 1) = -2 * (x^(2/3))/(2/3) = -2 * (3/2) * x^(2/3) = -3x^(2/3)
Combining these, we get the indefinite integral:
∫(x² - 2x^(-1/3)) dx = x³/3 - 3x^(2/3)
Now comes the exciting part: evaluating the antiderivative at the limits of integration, 3 and 1. Remember, we're finding the difference between the values at these two points, which represents the net change or the area under the curve. This step is the heart and soul of definite integration.
- At x = 3: (3)³/3 - 3(3)^(2/3) = 27/3 - 3 * 3^(2/3) = 9 - 3 * 3^(2/3)
- At x = 1: (1)³/3 - 3(1)^(2/3) = 1/3 - 3
Now, subtract the value at the lower limit (1) from the value at the upper limit (3):
(9 - 3 * 3^(2/3)) - (1/3 - 3) = 9 - 3 * 3^(2/3) - 1/3 + 3 = 12 - 1/3 - 3 * 3^(2/3) = 35/3 - 3 * 3^(2/3)
Therefore, ∫[1,3] (x² - 2/√[3]x) dx = 35/3 - 3 * 3^(2/3). We did it! This problem highlights the importance of understanding exponent rules and how they play a crucial role in integral calculus. By mastering these techniques, you'll be well-equipped to tackle a wide range of definite integrals.
3. Determining Total Production from Marginal Production
Let's switch gears a bit and explore a real-world application of integration. We're given the marginal production of labor (MP_L) as MP_L = 3/L² - 0.01L, and we want to find the total production. This is a classic example of how integral calculus can be used in economics and other fields. The key concept here is that the integral of a marginal function gives us the total function. Our main keywords for this section are marginal production, total production, and the powerful connection between marginal and total using integration.
Marginal production represents the additional output produced by adding one more unit of labor. To find the total production, we need to integrate the marginal production function with respect to labor (L). This process essentially sums up all the small increments of production to give us the overall production level. It's like piecing together a puzzle, where each piece represents a small contribution to the total output.
So, we need to find ∫(3/L² - 0.01L) dL. Let's rewrite 3/L² as 3L^(-2) to make it easier to apply the power rule.
Now our integral looks like this: ∫(3L^(-2) - 0.01L) dL.
Let's find the indefinite integral of each term:
- ∫3L^(-2) dL = 3 * (L^(-2+1))/(-2+1) = 3 * (L^(-1))/(-1) = -3L^(-1) = -3/L
- ∫-0.01L dL = -0.01 * (L^(1+1))/(1+1) = -0.01 * (L²/2) = -0.005L²
Combining these, we get the indefinite integral:
∫(3/L² - 0.01L) dL = -3/L - 0.005L² + C
Notice the + C, the constant of integration. In this context, the constant represents the fixed costs or the production level when labor input is zero. To find the specific total production function, we would need some additional information, such as the initial production level for a given level of labor. Without that information, we can only express the total production function in terms of this constant. The constant of integration is a crucial concept in integral calculus, and understanding its significance in different applications is vital.
Therefore, the total production function is Total Production = -3/L - 0.005L² + C. This result demonstrates the power of integral calculus in bridging the gap between marginal and total concepts. By integrating the marginal production, we've successfully derived the total production function, providing valuable insights into the relationship between labor input and overall output. The importance of understanding the constant of integration is also highlighted in the context of this practical application.
Conclusion: The Power of Integral Calculus
We've covered a lot of ground today, guys! From evaluating definite integrals using the power rule to applying integration in a real-world economic scenario, we've seen the versatility and power of integral calculus. Remember, practice makes perfect, so keep solving those integrals, and you'll be a pro in no time! We've explored how to handle fractional exponents, the significance of the antiderivative, and the crucial role of the constant of integration. These concepts are the building blocks of more advanced calculus, and a solid understanding of them will set you up for success in your future mathematical endeavors. So, keep exploring, keep learning, and keep integrating!
