Question

In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{-1}^{1} \frac{5 r}{\left(4+r^{2}\right)^{2}} d r$$

Short answer

The value of the integral is \(0\).

Step-by-step solution

Choose a substitution

Let \(u = 4 + r^2\). The benefit of this choice is that the denominator \((4 + r^2)^2\) of the rational function will be simplified to \(u^2\) after substitution. The differential \(d u\) is given by \(d u = 2 r d r\).

Express the original integral in terms of \(u\)

Divide both sides of the differential by \(2\) to get \(rdr = \frac{1}{2} du\). Then substitute \(u = 4 + r^2\) and \(rdr = \frac{1}{2} du\) into the integral: it becomes $$\int_{-1}^{1} \frac{5 r}{u^{2}} \cdot \frac{1}{2} du = \frac{5}{2} \int_{u(-1)}^{u(1)} \frac{1}{u^{2}} du$$ The limits of the integral are also transformed from \([-1, 1]\) to \([u(-1), u(1)] = [4 + (-1)^2, 4 + 1^2] = [5, 5]\).

Perform the actual integration

The integral $$\frac{5}{2} \int_{5}^{5} \frac{1}{u^{2}} du$$, though, is \(0\) because the limits of integration are exactly the same. This means the area under the curve of \(1/u^2\), from \(u = 5\) to \(u = 5\), is \(0\).

Key concepts

Definite Integrals

Understanding definite integrals is crucial in calculus, as they represent the accumulation of quantities, such as area under the curve, between two points on a graph. In the context of the exercise provided, a definite integral has fixed limits, from \( r = -1 \) to \( r = 1 \). When we think about the geometric interpretation, we are essentially looking for the area enclosed by the curve of the function \( \frac{5r}{(4 + r^2)^2} \) and the x-axis, bound by the vertical lines \( r = -1 \) and \( r = 1 \).

However, as revealed in the solution, when the substitution method (u-substitution) is applied, the transformed limits become the same (from \( u = 5 \) to \( u = 5 \)). This singular scenario leads to an integral with a value of zero, because there is no 'distance' between the upper and lower limits. It's as if we're asking for the area under a curve from a point to itself, which intuitively would indeed be nothing. This is an essential takeaway for students; the power of definite integrals lies in their ability to neatly express the accumulation of quantities between two distinct bounds.

Integration Techniques

The field of integration is rich with various techniques that are essential for solving calculus problems, and u-substitution is one of the most fundamental and widely used methods. In the given exercise, u-substitution helps transform a complicated integral into a simpler one, making it easier to handle.

To effectively utilize u-substitution, one must identify a function within the integral whose derivative is also present, or at least closely resembles a portion of the integrand. Once the substitution is made, not only the function is replaced by a simpler algebraic form \( u \), but also the differential \(dr\) is replaced by \(du\), and the limits of integration are adjusted accordingly. This exercise demonstrates a pivotal aspect of u-substitution—the need to convert every part of the original integral into the new variable, including the limits of integration, to avoid any mix-up between variables. When done correctly, this technique can unravel integrals that may at first seem intractable.

Calculus Problem Solving

Calculus problem-solving requires a strategic approach that usually involves selecting the right technique for the task at hand. The selected problem exemplifies the importance of critically evaluating the problem before jumping in. Here, the choice of substitution, \( u = 4 + r^2 \), is not arbitrary; it's influenced by the form of the integrand and the realization that it will significantly simplify the expression.

Furthermore, problem-solving in calculus often includes validating the integration process and results. For instance, after transforming the limits of integration for \(u\), it's essential to notice the equivalence of the upper and lower limits—leading to the insight that the integral's value must be zero without executing further integration steps. This level of scrutiny and reflection is part of building a deep understanding and a strategic approach to calculus problems. Students should remember to contemplate the implications of each step in their solution, ensuring that their method is logically sound and their answer makes sense within the context of the question.