Problem 62 - Calculus Early Transcendentals

Question

Miscellaneous integrals Evaluate the following integrals. \(\int_{0}^{\pi} 2^{\sin x} \cos x d x\)

Step-by-Step Solution

Answer

Answer: The value of the definite integral is 0.

Step 1: Set up integration by substitution

For the substitution, we will choose \(u = \sin x\) so that \(du = \cos x dx\). It is also necessary to change the limits of integration accordingly. When \(x=0\), \(\sin(0) = 0\), so the lower limit becomes \(u=0\). When \(x=\pi\), \(\sin(\pi) = 0\), so the upper limit remains \(u=0\). The integral will now be in terms of \(u\): $$\int_{0}^{0} 2^u du$$ As the upper and lower limits are the same, the integral will have a value of zero.

Step 2: Conclusion

Applying integration by substitution and changing the limits of integration, the definite integral simplifies to zero: $$\int_{0}^{\pi} 2^{\sin x} \cos x d x = 0$$

Key Concepts

Definite Integrals

Definite integrals play a pivotal role in calculus, being one of the fundamental concepts students encounter in their studies. In simple terms, a definite integral represents the accumulated sum of function values, tightly packed, across an interval on the x-axis. Mathematically, it's expressed as \( \int_{a}^{b} f(x) \,dx \), where \(a\) and \(b\) are the lower and upper limits, respectively, and \(f(x)\) is the integrand.

When computing a definite integral, we are often finding the area under the curve of a graph of the function, between the given limits. This can interpret as calculating total distance, area, or other quantities that accumulate over a specific interval. One fascinating aspect of definite integrals is that they may result in zero, which occurs if the area above the x-axis is perfectly balanced by the area below it, or in the case of the integral from the problem above, when the upper and lower integration limits are the same.

U-Substitution

The Technique of U-Substitution

U-substitution is a method frequently used to simplify the process of integration by replacing a complicated function with a simpler one. The goal is to transform the integral into a form where standard integration rules can be easily applied. It is analogous to performing a change of variables in an algebraic equation to make it more tractable.

In essence, u-substitution is about selecting a part of the integrand to represent as \( u \), and expressing \(dx\) in terms of \(du\). After the substitution, the integral becomes \( \int g(u) \,du \), where the integration process hopefully becomes more straightforward. In the exercise provided, \( u \) was chosen as \( \sin x \), leading to an unexpected yet correctly computed result because the limits turned out to be equal.

Calculus

The Realm of Calculus

Calculus is a vast and essential branch of mathematics, focusing on change and motion. It's divided into two main parts: differential calculus, concerning rates of change and slopes of curves; and integral calculus, which focuses on accumulation of quantities and the areas under and between curves.

In the context of our exercise, we are engaging with integral calculus. The process of integration, inverse to differentiation, requires various techniques to solve problems effectively. One such technique is integration by substitution, which we’ve already touched upon. The ability to link an area with the definite integral of a function is a powerful tool in both pure and applied mathematics as well as in fields like physics and engineering.

Integration Limits

Understanding Integration Limits

The concept of integration limits can be quite intriguing. In a definite integral, the limits define the bounds of integration, marking the start and end points on the x-axis over which we evaluate our function. These limits can be numbers, as seen in most cases, or they can extend to infinity, employed in improper integrals.

When performing u-substitution, a critical step involves converting the original limits of integration to new limits that correspond to the variable \( u \). Skipping or incorrectly adjusting these limits can result in inaccurate answers. However, sometimes, after applying u-substitution to the integration limits, you may find that they are equal, as in the exercise provided. When this occurs, the value of the integral simplifies considerably—since the bounds are the same, there's essentially no distance between them for the function to accumulate any value, resulting in an integral value of zero.