Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Evaluate the integrals. $$ \int x \sin x d x $$

Step-by-Step Solution

Answer

The integral \(\int x \sin x d x = -x \cos x + \sin x + C\).

Step 1: Choose \(u\) and \(dv\)

Usually, we choose \(u\) to be a function that becomes simpler once we differentiate it. So we'll choose \(u = x\). Then we must choose \(dv\) to be everything else, so \(dv = \sin x dx\). Then we differentiate \(u\) and integrate \(dv\) to get \(du\) and \(v\) respectively, which gives: \(du = dx\), \(v = -\cos x\).

Step 2: Apply the integration by parts formula

Now we'll apply the integration by parts formula: \(\int u dv = uv - \int v du \)This gives:\(\int x \sin x dx = uv - \int v du = x(-\cos x) - \int -\cos x dx\)

Step 3: Simplify the integral and evaluate

The integral simplifies to:\(-x \cos x + \int \cos x dx\)Next, we evaluate the remaining integral of \(\cos x\), which is \( \sin x\).So, the final answer is \(-x \cos x + \sin x + C\) where \(C\) is the constant of integration.

Key Concepts

Indefinite Integrals

Understanding indefinite integrals is crucial as they represent the opposite process of differentiation, known as antidifferentiation. An indefinite integral, shown symbolically as \(\int f(x)dx\), is essentially a function \(F(x)\) whose derivative is equal to the original function \(f(x)\). That is, if \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\), and \(F(x) + C\), including the constant of integration \(C\), represents the indefinite integral. This means that the integral includes a family of functions that differ only by a constant.

When you find an indefinite integral, you are looking for all possible functions that could be differentiated to get back to your original equation. For example, in the exercise above, the indefinite integral of \(x \sin(x) dx\) yields a result that includes \(C\), recognizing the multitude of possible antiderivatives that could produce the function \(x \sin(x)\).

Integral Calculus

Integral calculus is one of the two main branches of calculus, with the other being differential calculus. It primarily deals with the accumulation of quantities, such as areas under curves, volumes of solids, and more abstract concepts like net change. This branch of mathematics is not only about finding antiderivatives but also encompasses various applications such as computing lengths, areas, and volumes--all essential tasks in engineering, physics, economics, and more.

In the context of our exercise, integral calculus helps us determine the function, or the 'accumulated quantity,' generated as we integrate \(x \sin(x) dx\). This process computes the area under the curve represented by the function \(x \sin(x)\), leading us to the general form of the antiderivative plus the constant of integration.

U-Substitution

U-substitution is a technique that simplifies certain integrals by transforming them into a simpler form that is easier to evaluate. It's analogous to a change of variables in algebra. The goal is to identify a part of the integral, usually referred to as \(u\), that when differentiated, appears somewhere else in the integral.

However, not all problems require u-substitution, especially when simpler techniques, like integration by parts as seen in our example, can be applied. Nonetheless, u-substitution remains a cornerstone strategy, particularly when encountering integrals that involve chain-rule derivatives or complex compositions of functions.

Integration Techniques

Several methods can be employed to solve different types of integrals, known as integration techniques. These include the power rule, substitution method, integration by parts, partial fractions, trigonometric integrals, and others. The choice of technique largely depends on the form of the integral we're trying to solve.

For instance, in our exercise related to \(\int x \sin(x) dx\), integration by parts is chosen as the appropriate method because it efficiently handles products of functions where one function is easily differentiated, and the other easily integrated. Each technique has specific scenarios where it excels, and learning when to apply which is key to mastering integral calculus. Furthermore, practice improves recognization of patterns and selection of the most effective integration technique for a given problem.