Evaluate The Definite Integral Of (x/(1+x²))³ From 0 To Infinity.

In this article, we will delve into the fascinating world of calculus and explore the evaluation of a definite integral. Specifically, we aim to determine the value of the integral of the function (x/(1+x²))³ over the interval from 0 to infinity. This problem showcases the power of integration techniques and provides a valuable exercise in applying the fundamental principles of calculus. The journey will involve strategic substitutions, careful manipulation of limits, and a keen understanding of the behavior of functions as they approach infinity. Let's embark on this mathematical adventure and uncover the solution together.

Understanding the Integral

Before we dive into the solution, let's take a moment to understand the integral we're dealing with. We're asked to evaluate the definite integral:

∫[0 to ∞] (x/(1+x²))³ dx

This integral represents the area under the curve of the function f(x) = (x/(1+x²))³ from x = 0 to x = ∞. The function itself is a rational function, where the numerator is x and the denominator is the cube of (1+x²). As x increases, the denominator grows much faster than the numerator, suggesting that the function approaches zero as x approaches infinity. This is a crucial observation that will guide our solution process. To successfully evaluate this integral, we'll employ a technique called u-substitution, which allows us to simplify the integrand and make it easier to integrate. This method is particularly effective when we can identify a function and its derivative within the integral. Recognizing these patterns is key to mastering integration. Further, we will also need to understand how to deal with improper integrals, those with infinite limits of integration. This involves taking limits as the variable of integration approaches infinity, adding a layer of complexity to the problem. However, with a systematic approach and a clear understanding of the underlying concepts, we can successfully navigate these challenges and arrive at the correct answer. The process will not only provide us with the numerical value of the integral but also enhance our understanding of the behavior of functions and the power of calculus as a problem-solving tool.

Applying U-Substitution

The key to solving this integral lies in using the u-substitution method. This technique allows us to simplify complex integrals by replacing a part of the integrand with a new variable, 'u', and its differential 'du'. The goal is to transform the integral into a simpler form that we can readily integrate. In this case, a strategic choice for 'u' is:

u = 1 + x²

The reason for this choice becomes clear when we find the differential of 'u':

du = 2x dx

Notice that we have 'x dx' in our original integral, which is a multiple of 'du'. This is precisely what we need for the substitution to work. Now, we can rewrite the integral in terms of 'u'. First, we solve for 'x dx' from the 'du' equation:

x dx = (1/2) du

Next, we need to change the limits of integration. Our original limits were in terms of 'x' (0 and ∞), but now we need limits in terms of 'u'. When x = 0:

u = 1 + (0)² = 1

And when x approaches ∞:

u = 1 + (∞)² = ∞

So, our new limits of integration are 1 and ∞. Now we can rewrite the entire integral in terms of 'u':

∫[0 to ∞] (x/(1+x²))³ dx = ∫[1 to ∞] (1/u³) (1/2) du = (1/2) ∫[1 to ∞] u⁻³ du

We have successfully transformed the integral into a much simpler form. The integrand is now simply a power of 'u', which we can easily integrate using the power rule. The constant factor of (1/2) is also pulled out of the integral, further simplifying the expression. This transformation is a testament to the power of u-substitution, which allows us to tackle complex integrals by strategically changing the variable of integration. By carefully choosing 'u' and rewriting the integral in terms of 'u' and 'du', we have made the problem significantly more manageable. The next step is to apply the power rule of integration and evaluate the resulting expression at the new limits of integration. This will lead us to the final solution of the integral.

Integrating with Respect to u

Now that we have the integral in terms of 'u', we can proceed with the integration. We have:

(1/2) ∫[1 to ∞] u⁻³ du

To integrate u⁻³, we use the power rule for integration, which states:

∫xⁿ dx = (x^(n+1))/(n+1) + C, where n ≠ -1

Applying this rule to our integral, we get:

(1/2) [u⁻²/(-2)] [from 1 to ∞]

Simplifying, we have:

(-1/4) [1/u²] [from 1 to ∞]

Now we need to evaluate this expression at the limits of integration, ∞ and 1. This involves substituting these values for 'u' and finding the difference. When dealing with an infinite limit, we need to consider the limit as 'u' approaches infinity. This is a crucial step in evaluating improper integrals. We are essentially finding the value that the expression approaches as 'u' becomes infinitely large. The term 1/u² plays a key role here. As 'u' approaches infinity, 1/u² approaches zero. This is because dividing a constant by an increasingly large number results in a value that gets closer and closer to zero. This behavior is fundamental to understanding the convergence of improper integrals. If the expression approaches a finite value as the limit of integration approaches infinity, the integral converges. If it does not, the integral diverges. In our case, the fact that 1/u² approaches zero as 'u' approaches infinity is a good indication that the integral will converge. The next step is to carefully substitute the limits of integration and evaluate the resulting expression. This will give us the final numerical value of the integral.

Evaluating the Limits

To evaluate the limits, we substitute the upper and lower limits of integration into the expression we obtained after integration:

(-1/4) [1/u²] [from 1 to ∞] = (-1/4) [lim (u→∞) (1/u²) - 1/(1)²]

As we discussed earlier, the limit of 1/u² as u approaches infinity is 0. Therefore, we have:

(-1/4) [0 - 1]

Simplifying, we get:

(-1/4) [-1] = 1/4

Therefore, the value of the integral is 1/4. This result represents the area under the curve of the function f(x) = (x/(1+x²))³ from x = 0 to x = ∞. The fact that we obtained a finite value indicates that the integral converges. This means that the area under the curve is finite, even though the interval of integration extends to infinity. The convergence of this integral is a consequence of the function approaching zero sufficiently quickly as x approaches infinity. The denominator (1+x²)³ grows much faster than the numerator x, causing the function to decay rapidly. This rapid decay ensures that the area under the curve remains finite. The result of 1/4 is a precise and elegant solution to the problem. It demonstrates the power of calculus to solve problems involving continuous functions and infinite intervals. The process of u-substitution, integration, and limit evaluation has allowed us to determine the exact area under the curve, providing a valuable insight into the behavior of the function.

Conclusion

In conclusion, we have successfully evaluated the integral from 0 to infinity of the function (x/(1+x²))³. By employing the u-substitution technique, we transformed the integral into a simpler form that was readily integrable. We carefully changed the limits of integration to correspond to the new variable 'u' and then applied the power rule of integration. The evaluation of the limits involved considering the behavior of the function as 'u' approached infinity, which led us to the final result of 1/4. This problem showcases the elegance and power of calculus in solving problems involving continuous functions and infinite intervals. The techniques we used, such as u-substitution and limit evaluation, are fundamental tools in the arsenal of any mathematician or scientist. They allow us to tackle complex problems and gain a deeper understanding of the behavior of functions. The result of 1/4 represents the area under the curve of the function (x/(1+x²))³ from 0 to infinity. This finite value demonstrates the convergence of the integral, which is a testament to the rapid decay of the function as x approaches infinity. The process of solving this integral has not only provided us with a numerical answer but has also reinforced our understanding of key calculus concepts and techniques. The journey from the initial problem statement to the final solution has been a valuable exercise in mathematical problem-solving and highlights the beauty and utility of calculus in exploring the world around us.