Evaluating Definite Integral: ∫(5x³ - 5x⁻¹/²) Dx From 3 To 4
Understanding how to solve definite integrals is a fundamental skill in calculus. This article will guide you through the step-by-step process of evaluating the definite integral of the function 5x³ - 5x⁻¹/² from the lower limit of 3 to the upper limit of 4. We'll break down each step, making it easy to follow and understand, even if you're just beginning your journey in calculus. Mastering definite integrals like this one is crucial for numerous applications in physics, engineering, and other scientific fields. So, let's dive in and explore the solution together! Understanding the process not only helps in solving this particular problem but also lays a strong foundation for tackling more complex integrals in the future. Remember, practice makes perfect, and by working through examples like this, you'll build confidence and proficiency in integral calculus.
1. Understanding the Integral
Before we jump into the calculations, let's take a moment to understand what we're working with. The definite integral ∫₃⁴ (5x³ - 5x⁻¹/²) dx represents the net signed area between the curve of the function f(x) = 5x³ - 5x⁻¹/² and the x-axis, from x = 3 to x = 4. In simpler terms, we're finding the area bounded by this curve, the x-axis, and the vertical lines x = 3 and x = 4. The integral sign ∫ is a stylized S, representing summation, and dx indicates that we're integrating with respect to the variable x. The numbers 3 and 4 are the limits of integration, defining the interval over which we're calculating the area. The function inside the integral, 5x³ - 5x⁻¹/², is the integrand. To evaluate this integral, we'll use the power rule for integration and the fundamental theorem of calculus. These are essential tools in integral calculus, and understanding them will greatly simplify the process.
The power rule states that the integral of xⁿ is (x^(n+1))/(n+1), provided that n is not equal to -1. We'll apply this rule to both terms in our integrand. The fundamental theorem of calculus tells us that to evaluate a definite integral, we first find the antiderivative of the integrand and then evaluate it at the upper and lower limits of integration. The difference between these values gives us the value of the definite integral. This theorem connects the concepts of differentiation and integration, showing them to be inverse operations of each other. By mastering these concepts, you'll be well-equipped to handle a wide range of integration problems. Remember, the key is to break down the problem into manageable steps and apply the appropriate rules and theorems.
2. Finding the Antiderivative
The next step in evaluating the definite integral is to find the antiderivative of the function 5x³ - 5x⁻¹/². An antiderivative is a function whose derivative is equal to the original function. In other words, we're looking for a function F(x) such that F'(x) = 5x³ - 5x⁻¹/². To find the antiderivative, we'll apply the power rule for integration to each term separately. The power rule states that ∫xⁿ dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. For the first term, 5x³, we have n = 3. Applying the power rule, we get ∫5x³ dx = 5 * (x^(3+1))/(3+1) + C₁ = 5 * (x⁴)/4 + C₁ = (5/4)x⁴ + C₁. For the second term, -5x⁻¹/², we have n = -1/2. Applying the power rule, we get ∫-5x⁻¹/² dx = -5 * (x^(-1/2 + 1))/(-1/2 + 1) + C₂ = -5 * (x¹/²)/(1/2) + C₂ = -10x¹/² + C₂. Combining these results, the antiderivative of 5x³ - 5x⁻¹/² is (5/4)x⁴ - 10x¹/² + C, where C = C₁ + C₂ is the overall constant of integration. However, when evaluating definite integrals, the constant of integration cancels out, so we can ignore it for now. Therefore, our antiderivative is F(x) = (5/4)x⁴ - 10x¹/². Understanding how to find antiderivatives is crucial for evaluating both definite and indefinite integrals. It involves reversing the process of differentiation and applying the appropriate rules, such as the power rule, sum rule, and constant multiple rule. With practice and familiarity, you'll become more adept at finding antiderivatives and solving integration problems.
3. Applying the Fundamental Theorem of Calculus
Now that we've found the antiderivative, the next step is to apply the fundamental theorem of calculus. This theorem provides a direct method for evaluating definite integrals. It states that if F(x) is an antiderivative of f(x), then the definite integral of f(x) from a to b is given by ∫ₐᵇ f(x) dx = F(b) - F(a). In our case, f(x) = 5x³ - 5x⁻¹/², F(x) = (5/4)x⁴ - 10x¹/², a = 3, and b = 4. So, we need to evaluate F(4) and F(3) and then subtract the latter from the former. First, let's find F(4): F(4) = (5/4)(4)⁴ - 10(4)¹/² = (5/4)(256) - 10(2) = 320 - 20 = 300. Next, let's find F(3): F(3) = (5/4)(3)⁴ - 10(3)¹/² = (5/4)(81) - 10√3 = (405/4) - 10√3. Now, we subtract F(3) from F(4): ∫₃⁴ (5x³ - 5x⁻¹/²) dx = F(4) - F(3) = 300 - [(405/4) - 10√3] = 300 - (405/4) + 10√3. To simplify this expression, we can convert 300 to a fraction with a denominator of 4: 300 = 1200/4. So, the expression becomes: (1200/4) - (405/4) + 10√3 = (1200 - 405)/4 + 10√3 = (795/4) + 10√3. Therefore, the value of the definite integral is (795/4) + 10√3. The fundamental theorem of calculus is a cornerstone of calculus, linking differentiation and integration in a profound way. It allows us to evaluate definite integrals by simply finding an antiderivative and evaluating it at the limits of integration. This theorem simplifies the process of finding areas under curves and has numerous applications in various fields.
4. Simplifying the Result
We've arrived at the solution (795/4) + 10√3, but let's take a moment to see if we can simplify it further or express it in a more practical form. The expression consists of a fraction and a term involving a square root, which are both in their simplest forms individually. However, we can approximate the value to get a decimal representation, which might be more intuitive for some applications. To do this, we can use a calculator to find the approximate value of 10√3. The square root of 3 is approximately 1.732, so 10√3 is approximately 10 * 1.732 = 17.32. Now, we can add this to the fraction 795/4. First, let's convert the fraction to a decimal: 795/4 = 198.75. Adding the two values, we get 198.75 + 17.32 = 216.07. So, the approximate value of the definite integral is 216.07. This decimal approximation gives us a better sense of the magnitude of the area represented by the integral. While the exact answer (795/4) + 10√3 is precise, the decimal approximation can be more useful in real-world applications where numerical values are needed. In this case, both the exact and approximate values are valid ways to express the solution. The choice of which to use often depends on the context and the level of precision required. Understanding how to simplify and approximate results is an important skill in calculus and mathematics in general. It allows you to express solutions in the most appropriate form for the given situation. Remember, mathematics is not just about finding the right answer, but also about understanding its meaning and implications.
5. Conclusion
In this article, we've walked through the process of evaluating the definite integral ∫₃⁴ (5x³ - 5x⁻¹/²) dx. We started by understanding the integral and what it represents geometrically. Then, we found the antiderivative of the integrand using the power rule for integration. Next, we applied the fundamental theorem of calculus to evaluate the integral at the limits of integration. Finally, we simplified the result and obtained both the exact answer, (795/4) + 10√3, and a decimal approximation, 216.07. Evaluating definite integrals is a crucial skill in calculus with applications in various fields, including physics, engineering, and economics. By mastering the techniques discussed in this article, you'll be well-equipped to tackle a wide range of integration problems. Remember, practice is key, so try working through more examples to solidify your understanding. Each problem you solve will build your confidence and proficiency in integral calculus.
For further exploration and a deeper understanding of calculus concepts, consider visiting Khan Academy's Calculus section. It's a fantastic resource for learning and practicing calculus.
