Hey guys! Today, we're diving deep into the world of calculus to tackle a fascinating integral problem. We're going to evaluate the definite integral of ∫3^5 [2(x+1)]/[2x^2-3x+1] dx. This might look intimidating at first, but don't worry, we'll break it down step-by-step, making sure everyone can follow along. So, grab your pencils, and let's get started!
Understanding the Integral
At its core, this integral asks us to find the area under the curve of the function f(x) = [2(x+1)]/[2x^2-3x+1] between the limits x = 3 and x = 5. To do this effectively, we'll need to employ a combination of algebraic manipulation and integral calculus techniques. The key to solving this integral lies in recognizing the structure of the integrand and choosing the appropriate method. Before we jump into the solution, let's talk about why integrals like these are so important. Integrals aren't just abstract math problems; they have real-world applications in physics, engineering, economics, and many other fields. For example, they can be used to calculate the displacement of an object, the average value of a function, or the probability of an event. So, mastering integral calculus is a valuable skill that can open doors to many exciting opportunities. Now, let's circle back to our specific integral. What makes it interesting? Well, the denominator is a quadratic expression, and the numerator is a linear expression. This suggests that we might be able to use techniques like partial fraction decomposition or a clever u-substitution to simplify the integral. We'll explore both of these possibilities as we move through the solution. Remember, the goal isn't just to get the right answer, but also to understand the underlying concepts and techniques. So, let's approach this problem with curiosity and a willingness to learn. Are you ready to dive in? Let's do it!
Step 1: Factoring the Denominator
The first crucial step in evaluating this integral is to simplify the denominator of our integrand. We have 2x^2 - 3x + 1. To make things easier, let's factor this quadratic expression. Factoring involves finding two binomials that multiply together to give us the original quadratic. We're looking for two numbers that multiply to 2 * 1 = 2 and add up to -3. These numbers are -2 and -1. So, we can rewrite the quadratic as 2x^2 - 2x - x + 1. Now, we can factor by grouping: 2x(x - 1) - 1(x - 1). This gives us (2x - 1)(x - 1). Ah, that's much cleaner! Now our integral looks like this: ∫3^5 [2(x+1)]/[(2x - 1)(x - 1)] dx. Why is factoring so important? Well, it allows us to break down the complex fraction into simpler fractions using a technique called partial fraction decomposition. This technique is a powerful tool for integrating rational functions, which are functions that can be expressed as a ratio of two polynomials. By factoring the denominator, we've essentially laid the groundwork for using partial fraction decomposition. Think of it like breaking a big, complicated task into smaller, more manageable tasks. Factoring is the first step in simplifying our integral and making it easier to solve. Now that we've factored the denominator, we're ready to move on to the next step: partial fraction decomposition. But before we do that, let's take a moment to appreciate the power of factoring. It's a fundamental algebraic technique that pops up in all sorts of math problems, from solving equations to simplifying expressions to, yes, evaluating integrals! So, mastering factoring is a key skill for any aspiring mathematician or scientist. Okay, let's keep the momentum going and tackle the next step!
Step 2: Partial Fraction Decomposition
With the denominator nicely factored, we can now apply partial fraction decomposition. This technique allows us to rewrite the integrand as a sum of simpler fractions, each with a linear denominator. This will make the integration process much easier. Our goal is to express [2(x+1)]/[(2x - 1)(x - 1)] in the form A/(2x - 1) + B/(x - 1), where A and B are constants that we need to determine. To find A and B, we multiply both sides of the equation by the common denominator (2x - 1)(x - 1). This gives us 2(x + 1) = A(x - 1) + B(2x - 1). Now, we can use a couple of clever tricks to solve for A and B. One way is to substitute specific values of x that will eliminate one of the variables. For example, if we let x = 1, the term with A becomes zero, and we get 2(1 + 1) = B(2(1) - 1), which simplifies to 4 = B. So, we've found that B = 4. Next, let's let x = 1/2. This will make the term with B zero, and we get 2(1/2 + 1) = A(1/2 - 1), which simplifies to 3 = A(-1/2). Solving for A, we find that A = -6. Great! We've found both A and B. So, we can rewrite our integrand as -6/(2x - 1) + 4/(x - 1). See how much simpler this looks? Now we have two separate fractions, each of which is much easier to integrate than the original expression. This is the power of partial fraction decomposition! It allows us to break down complex rational functions into manageable pieces. Before we move on to the next step, let's just take a moment to appreciate the elegance of this technique. It's a classic example of how a little bit of algebraic manipulation can make a big difference in solving a calculus problem. Okay, with our integrand nicely decomposed, we're ready to integrate. Let's go!
Step 3: Integrating the Decomposed Fractions
Now comes the fun part – integrating the simplified fractions! We've successfully decomposed our original integrand into -6/(2x - 1) + 4/(x - 1). So, we need to evaluate the integral of each term separately: ∫3^5 [-6/(2x - 1) + 4/(x - 1)] dx = -6∫3^5 [1/(2x - 1)] dx + 4∫3^5 [1/(x - 1)] dx. Let's tackle the first integral: -6∫3^5 [1/(2x - 1)] dx. To solve this, we can use a simple u-substitution. Let u = 2x - 1. Then, du = 2 dx, so dx = (1/2) du. When x = 3, u = 2(3) - 1 = 5, and when x = 5, u = 2(5) - 1 = 9. So, our integral becomes -6∫5^9 [1/u] (1/2) du = -3∫5^9 [1/u] du. The integral of 1/u is ln|u|, so we have -3[ln|u|]5^9 = -3(ln(9) - ln(5)). Now, let's move on to the second integral: 4∫3^5 [1/(x - 1)] dx. Again, we can use a u-substitution. Let v = x - 1. Then, dv = dx. When x = 3, v = 3 - 1 = 2, and when x = 5, v = 5 - 1 = 4. So, our integral becomes 4∫2^4 [1/v] dv. The integral of 1/v is ln|v|, so we have 4[ln|v|]2^4 = 4(ln(4) - ln(2)). We've successfully integrated both terms! Now, we just need to combine the results and simplify. Remember, the key to integrating these types of fractions is recognizing the form and using appropriate substitution techniques. U-substitution is a powerful tool that allows us to transform complex integrals into simpler ones. By carefully choosing our substitutions, we can often make the integration process much more manageable. Okay, let's bring it all together in the next step!
Step 4: Evaluating and Simplifying
Alright, we're in the home stretch! We've integrated each part, and now it's time to evaluate the definite integral and simplify the result. We found that -6∫3^5 [1/(2x - 1)] dx = -3(ln(9) - ln(5)) and 4∫3^5 [1/(x - 1)] dx = 4(ln(4) - ln(2)). So, our original integral ∫3^5 [2(x+1)]/[(2x^2-3x+1)] dx is equal to -3(ln(9) - ln(5)) + 4(ln(4) - ln(2)). Now, let's use some logarithm properties to simplify this expression. Recall that ln(a) - ln(b) = ln(a/b). Applying this to our result, we get -3ln(9/5) + 4ln(4/2) = -3ln(9/5) + 4ln(2). We can also use the property aln(b) = ln(b^a) to further simplify: ln((9/5)^-3) + ln(2^4) = ln((5/9)^3) + ln(16). Now, using the property ln(a) + ln(b) = ln(ab), we get ln(((5/9)^3) * 16) = ln((125/729) * 16) = ln(2000/729). So, the final value of our integral is ln(2000/729). Woohoo! We made it! We've successfully evaluated the definite integral. This problem demonstrates the power of combining algebraic techniques like factoring and partial fraction decomposition with calculus techniques like u-substitution. By breaking down the problem into smaller, more manageable steps, we were able to arrive at the solution. Remember, practice makes perfect. The more you work through problems like this, the more comfortable you'll become with the techniques and the more confident you'll feel in your ability to tackle even the most challenging integrals. So, keep practicing, keep exploring, and keep having fun with math! You've got this!
Conclusion
So, guys, we've successfully navigated the world of integral calculus and evaluated the definite integral ∫3^5 [2(x+1)]/[2x^2-3x+1] dx. We started by factoring the denominator, then used partial fraction decomposition to simplify the integrand. We employed u-substitution to integrate the resulting fractions and finally, used logarithm properties to simplify the answer. The final result? ln(2000/729). This journey highlights the importance of a solid foundation in algebra and calculus techniques. Each step built upon the previous one, demonstrating the interconnectedness of mathematical concepts. Remember, evaluating integrals isn't just about finding the right answer; it's about developing problem-solving skills and gaining a deeper understanding of mathematical principles. By breaking down complex problems into smaller, manageable steps, we can conquer any mathematical challenge. Keep practicing, stay curious, and never stop exploring the fascinating world of mathematics!