Problem 1 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x \cos x d x $$

Step-by-Step Solution

Answer

The solution is \( \int x \cos x \, dx = x \sin x + \cos x + C \).

Step 1: Choose Functions for Integration by Parts

To apply integration by parts, we first need to identify the parts of the formula: \( \int u \, dv = uv - \int v \, du \). For this integral, choose \( u = x \) (since its derivative becomes simpler) and \( dv = \cos x \, dx \) (since it integrates easily).

Step 2: Differentiate and Integrate the Chosen Parts

Differentiate \( u = x \) to get \( du = dx \). Integrate \( dv = \cos x \, dx \) to obtain \( v = \sin x \).

Step 3: Apply the Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula: \( \int x \cos x \, dx = x \sin x - \int \sin x \, dx \).

Step 4: Integrate the Remaining Integral

Now, find \( \int \sin x \, dx \), which is \( -\cos x \). Thus, the equation becomes \( x \sin x + \cos x + C \).

Step 5: Write the Final Solution

The evaluated integral is: \( \int x \cos x \, dx = x \sin x + \cos x + C \), where \( C \) is the constant of integration.

Key Concepts

Definite Integral

A definite integral evaluates the area under a curve within a specific interval. It results in a single numerical value. When working with definite integrals, the integration process follows similar steps as with indefinite integrals. However, after finding the antiderivative of the function,
you need to evaluate it at the upper and lower bounds of the interval.
To compute a definite integral, you:

• Determine the antiderivative, \( F(x) \).
• Evaluate \( F(x) \) at the upper limit and subtract the value of \( F(x) \) at the lower limit: \( F(b) - F(a) \).

When calculating a definite integral using integration by parts, the process is comparable, but ensure the final solution applies to the interval's limits. This methodology allows you to calculate exact areas beneath a curve,
providing significant applications in physics and engineering.

Indefinite Integral

An indefinite integral represents the collection of antiderivatives of a function. It does not evaluate over an interval unlike the definite integral. Instead, it includes a constant of integration, \( C \), representing an infinite family of functions.
The indefinite integral is expressed with the integral symbol followed by the function and a differential, such as \( \int f(x) \, dx \). In our specific problem, applying integration by parts, we find the indefinite integral of \( x \cos x \).
Integration by parts relies on choosing components, \( u \) and \( dv \), within the function, which help in breaking it down into simpler parts.

After applying integration by parts:

• Substitute \( u = x \) and \( dv = \cos x \, dx \).
• Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
• Plug these into the formula: \( \int x \cos x \, dx = x \sin x - \int \sin x \, dx \).
• Finally, integrate \( \sin x \) to result in \( -\cos x \), arriving at the solution: \( x \sin x + \cos x + C \).

This final expression contains a \(+ C\), accounting for all possible vertical shifts of the antiderivative.

Derivative and Antiderivative

The derivative and antiderivative are fundamental concepts in calculus, intricately related through integration and differentiation.
A derivative \( f'(x) \) describes the rate at which a function changes at any point, while an antiderivative is the reverse process. It involves recovering the original function from its derivative.
The antiderivative is what we aim to find during integration. In other terms, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).

Consider the process of integration by parts, especially in reference to solving \( \int x \cos x \, dx \):

• By differentiating \( u = x \to du = dx \) and integrating \( dv = \cos x \, dx \to v = \sin x \), you bridge the connection between derivatives and antiderivatives.
• This process alters and aids in tackling a seemingly challenging integral by leveraging known derivatives and antiderivatives.

Understanding these interactions deepens comprehension of how integration and differentiation are inverses, playing crucial roles across various fields like physics, engineering, and economics.