Problem 2 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int 3 x \cos x d x $$

Step-by-Step Solution

Answer

\( \int 3x \cos x \, dx = 3x \sin x + 3 \cos x + C \).

Step 1: Identify Parts for Integration by Parts

To apply integration by parts, we need to identify parts of the integral that can be expressed in the form \ \( \int u \, dv = uv - \int v \, du \). Let's choose \ \( u = 3x \) which makes \ \( du = 3 \, dx \), and \ \( dv = \cos x \, dx \) which makes \ \( v = \sin x \).

Step 2: Apply Integration by Parts Formula

With \ \( u = 3x, \, du = 3 \, dx, \, v = \sin x, \, dv = \cos x \, dx \), use the formula \ \( \int u \, dv = uv - \int v \, du \). Substituting these gives: \[ \int 3x \cos x \, dx = 3x \sin x - \int \sin x \, (3 \, dx) \].

Step 3: Simplify and Integrate Remaining Integral

Now simplify \[ 3x \sin x - 3 \int \sin x \, dx \]. The remaining integral is \[ \int \sin x \, dx \], whose antiderivative is \( - \cos x \). Thus, \[ 3x \sin x - 3 (- \cos x) = 3x \sin x + 3 \cos x \].

Step 4: Write the Final Expression with Constant of Integration

Finally, add the constant of integration \ \( C \) to give the general solution. So the integral evaluates to: \[ 3x \sin x + 3 \cos x + C \].

Key Concepts

Definite Integrals

Definite integrals are used to calculate the area under a curve for a specific interval. In contrast to indefinite integrals, which come with a constant of integration, definite integrals provide a specific value because they deal with exact intervals.

• The process involves determining both the upper and lower bounds of integration.
• The result of a definite integral is a single number representing the total area.
• This area can be positive, negative, or zero, depending on the position of the curve relative to the x-axis.

When using the integration by parts formula for definite integrals, the rule remains similar, but you must evaluate the final expression at the boundaries. If our earlier example were a definite integral, after integration, we would plug in the upper and lower limits into our result (e.g., \(3x \sin x + 3 \cos x\)) to find the precise value.

Antiderivative

An antiderivative is a function whose derivative is the original function given. Also known as indefinite integration, finding an antiderivative is akin to asking, "Which function, when differentiated, returns the current function?"

• The general form of an antiderivative includes a constant, usually written as \( C \), because differentiation removes constants.
• A simple way to find an antiderivative is to apply rules of integration to the function.

In our exercise, the antiderivative of \( \sin x \) is \( -\cos x \). In the solution, after integrating \( \sin x \), we added the constant \( C \) to express the most general form of the antiderivative. Every antiderivative of the equation \( 3x \cos x \) will be of the form \( 3x \sin x + 3 \cos x + C \). This demonstrates how integration by parts provides a methodical way to find antiderivatives even for seemingly complex expressions.

Calculus Problems

Calculus problems often require applying more than one calculus concept to obtain a solution. They might involve differentiating and integrating as well as employing specialized techniques.

• Integration by parts is one method in calculus used to simplify integrals where a direct application of basic rules isn't simple.
• These problems frequently call for a deep understanding of when and how to apply specific techniques like substitution or parts.
• Real-world problems can often be transcribed into a calculus form, solved mathematically, and then translated back to fit the context of the problem.

Addressing calculus problems with strategies like integration by parts requires practice and a clear comprehension of the fundamental theorems of calculus. By understanding when to use methods like substitution or integration by parts, students can tackle more challenging integrals efficiently. In the given exercise, appropriate selection of \( u \) and \( dv \) facilitated the solution using integration by parts, which is a testament to the multifaceted nature of calculus problems.