Evaluate ∫₋₁.₉² 7xe^(5x) Dx Using Integration By Parts
Introduction
In the realm of calculus, evaluating definite integrals is a fundamental skill. Among the various techniques available, integration by parts stands out as a powerful method for handling integrals involving the product of two functions. This article delves into the application of integration by parts to evaluate the definite integral of 7xe^(5x) from -1.9 to 2. We will explore the underlying theory, step-by-step calculations, and the final result, accurate to three decimal places. Understanding integration by parts is crucial for solving a wide range of problems in mathematics, physics, and engineering. This method allows us to break down complex integrals into simpler forms that can be readily evaluated. The key idea is to choose appropriate parts of the integrand as 'u' and 'dv', such that the integral becomes more manageable. We'll walk through the selection process and the subsequent steps to arrive at the solution. Moreover, this technique highlights the interplay between differentiation and integration, showcasing the elegance and power of calculus. As we progress, we'll emphasize the practical application of the formula and address common challenges encountered in its use. By the end of this discussion, you'll have a solid grasp of how to apply integration by parts effectively, enhancing your ability to tackle a variety of integral problems with confidence. This integral is a classic example that showcases the effectiveness of integration by parts when dealing with a product of a polynomial and an exponential function. The process involves careful selection of the 'u' and 'dv' components, followed by the application of the integration by parts formula, which transforms the original integral into a more manageable form. The application of the formula is iterative, and in some cases, it may need to be applied multiple times to reach a solvable integral. However, in this specific case, a single application suffices to bring us to a straightforward integral that we can easily evaluate. The final step involves substituting the limits of integration and calculating the definite value, rounding the result to the required decimal places for accuracy. This method not only provides the solution but also enhances our understanding of the underlying concepts and techniques in calculus.
The Theory Behind Integration by Parts
At its core, integration by parts is derived from the product rule of differentiation. If we have two functions, u(x) and v(x), the product rule states:
d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
Integrating both sides with respect to x, we get:
∫ d/dx [u(x)v(x)] dx = ∫ u'(x)v(x) dx + ∫ u(x)v'(x) dx
This simplifies to:
u(x)v(x) = ∫ v(x) du + ∫ u(x) dv
Rearranging the terms, we arrive at the integration by parts formula:
∫ u dv = uv - ∫ v du
The effectiveness of this method hinges on the judicious choice of 'u' and 'dv'. The goal is to select 'u' and 'dv' such that the integral ∫ v du is simpler than the original integral ∫ u dv. A common guideline for this selection is the acronym LIATE, which helps prioritize the functions for 'u' as follows:
- Logarithmic functions
- Inverse trigonometric functions
- Algebraic functions
- Trigonometric functions
- Exponential functions
The function appearing earlier in the list is generally a good candidate for 'u'. In our specific problem, we have the product of an algebraic function (7x) and an exponential function (e^(5x)). Following the LIATE rule, we should choose 7x as 'u' and e^(5x) dx as 'dv'. This choice is critical because differentiating 7x results in a simpler expression (7), while integrating e^(5x) results in an expression that is still manageable (1/5 e^(5x)). This strategic selection is what allows us to simplify the integral and eventually solve it. The process of selecting 'u' and 'dv' is not always straightforward, and sometimes it may require trying different combinations to find the one that works best. However, the LIATE rule provides a useful starting point and often leads to the correct choice. Understanding the underlying theory and applying the rule thoughtfully are essential for mastering integration by parts and tackling a wide range of integral problems.
Applying Integration by Parts to the Integral
Now, let's apply integration by parts to the definite integral:
∫₋₁.₉² 7xe^(5x) dx
Following the LIATE rule, we choose:
- u = 7x
- dv = e^(5x) dx
Differentiating u, we get:
- du = 7 dx
Integrating dv, we get:
- v = ∫ e^(5x) dx = (1/5)e^(5x)
Now, we apply the integration by parts formula:
∫ u dv = uv - ∫ v du
Substituting our values, we have:
∫₋₁.₉² 7xe^(5x) dx = [7x * (1/5)e^(5x)]₋₁.₉² - ∫₋₁.₉² (1/5)e^(5x) * 7 dx
This simplifies to:
∫₋₁.₉² 7xe^(5x) dx = [(7/5)xe^(5x)]₋₁.₉² - (7/5) ∫₋₁.₉² e^(5x) dx
Now, we need to evaluate the remaining integral:
∫ e^(5x) dx = (1/5)e^(5x)
So, our expression becomes:
∫₋₁.₉² 7xe^(5x) dx = [(7/5)xe^(5x)]₋₁.₉² - (7/5) * [(1/5)e^(5x)]₋₁.₉²
This further simplifies to:
∫₋₁.₉² 7xe^(5x) dx = [(7/5)xe^(5x) - (7/25)e^(5x)]₋₁.₉²
This step-by-step application of the integration by parts formula demonstrates how the original complex integral is transformed into a form that is much easier to evaluate. By carefully selecting 'u' and 'dv', we were able to simplify the integral into a manageable expression. The remaining steps involve substituting the limits of integration and calculating the final value. This process highlights the power and elegance of integration by parts as a technique for solving integrals involving the product of functions. The careful application of the formula, along with a clear understanding of the underlying theory, is crucial for achieving accurate results.
Evaluating the Definite Integral
Now, we substitute the limits of integration:
[(7/5)xe^(5x) - (7/25)e^(5x)]₋₁.₉² = [(7/5)(2)e^(52) - (7/25)e^(52)] - [(7/5)(-1.9)e^(5*(-1.9)) - (7/25)e^(5*(-1.9))]
Let's calculate each part separately:
- [(7/5)(2)e^(10) - (7/25)e^(10)] = (14/5)e^(10) - (7/25)e^(10) = (70/25)e^(10) - (7/25)e^(10) = (63/25)e^(10)
- [(7/5)(-1.9)e^(-9.5) - (7/25)e^(-9.5)] = (-13.3/5)e^(-9.5) - (7/25)e^(-9.5) = (-66.5/25)e^(-9.5) - (7/25)e^(-9.5) = (-73.5/25)e^(-9.5)
So, the expression becomes:
(63/25)e^(10) - (-73.5/25)e^(-9.5) = (63/25)e^(10) + (73.5/25)e^(-9.5)
Now, we can approximate the values:
- e^(10) ≈ 22026.4658
- e^(-9.5) ≈ 0.00007485
Substituting these values:
(63/25) * 22026.4658 + (73.5/25) * 0.00007485 ≈ 55496.654 + 0.00022005 ≈ 55496.65422
Rounding to three decimal places, we get 55496.654.
This final calculation brings us to the numerical value of the definite integral. The process involves careful substitution of the limits of integration and approximation of the exponential terms. The result, rounded to three decimal places, provides a precise value for the integral. This step underscores the importance of accuracy in numerical computations and highlights the practical application of calculus in real-world scenarios. The combination of analytical techniques, such as integration by parts, with numerical approximation allows us to solve complex problems and obtain meaningful results.
Final Answer
Therefore, the definite integral ∫₋₁.₉² 7xe^(5x) dx, accurate to three decimal places, is approximately 55496.654.
Conclusion
In conclusion, we have successfully evaluated the definite integral ∫₋₁.₉² 7xe^(5x) dx using integration by parts. This method allowed us to transform a complex integral into a simpler form, which we could then evaluate. The key steps involved choosing appropriate parts for 'u' and 'dv', applying the integration by parts formula, and substituting the limits of integration. The final result, accurate to three decimal places, is 55496.654. Mastering integration by parts is essential for anyone studying calculus and its applications. This technique is not only a powerful tool for solving integrals but also provides insights into the fundamental relationships between differentiation and integration. The ability to strategically choose 'u' and 'dv', along with a clear understanding of the formula, allows us to tackle a wide range of integral problems with confidence. The detailed walkthrough presented here serves as a valuable resource for students and professionals alike, reinforcing the principles of calculus and enhancing problem-solving skills. The application of these techniques extends beyond the classroom, finding relevance in various fields such as physics, engineering, and economics, where integral calculus plays a crucial role in modeling and analyzing complex systems. By understanding and mastering integration by parts, we equip ourselves with a valuable tool for tackling mathematical challenges in both academic and real-world contexts. This method, when applied thoughtfully and strategically, opens doors to solving a vast array of problems that would otherwise be intractable, underscoring its significance in the broader landscape of mathematical analysis. The combination of analytical rigor and numerical approximation, as demonstrated in this example, exemplifies the power of calculus as a versatile and indispensable tool for problem-solving.
