Problem 4 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int 3 x \cos (1-x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \[-3x \sin(1-x) + 3\cos(1-x) + C\].

Step 1: Choose u and dv

For integration by parts, select a function to differentiate (\(u\)) and another to integrate (\(dv\)). Here, let \(u = 3x\) and \(dv = \cos(1-x) \, dx\).

Step 2: Differentiate u to get du

Differentiate \(u\) to find \(du\). Since \(u = 3x\), then \(du = 3 \, dx\).

Step 3: Integrate dv to get v

Integrate \(dv\) to find \(v\). The integral of \(\cos(1-x)\) can be approached by substitution. Let \(w = 1-x\) so \(dw = -dx\). Then the integral becomes \(-\int \cos(w) dw = -\sin(w) + C\), or \(-\sin(1-x) + C\).

Step 4: Apply Integration by Parts Formula

Use the integration by parts formula \(\int u \, dv = uv - \int v \, du\). Substitute \(u = 3x\), \(v = -\sin(1-x)\), and \(du = 3 \, dx\) into the formula: \(\int 3x \cos(1-x) \, dx = (3x)(-\sin(1-x)) - \int (-\sin(1-x))(3 \, dx)\).

Step 5: Simplify and Solve the Remaining Integral

Simplify the expression: \[-3x \sin(1-x) - 3 \int -\sin(1-x) \, dx\]Integrate the second term: \[\int \sin(1-x) \, dx = -\int \sin(w) \frac{1}{-1} \, dw\],where \(w = 1-x\), becomes \(-\cos(w) + C = -\cos(1-x) + C\).

Step 6: Final Answer

Combine results from previous steps: \[-3x \sin(1-x) - 3(-\cos(1-x)) + C\ \ = -3x \sin(1-x) + 3\cos(1-x) + C\].This is the final evaluated integral.

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on the concept of integration, the process of finding the antiderivative or integral of a function.Integration is essentially the reverse process of differentiation and is used to calculate areas, volumes, and other quantities.The basic idea is to add up infinitely small pieces, sometimes called "differentials," to find a total quantity. It is crucial in solving problems involving the total quantity or overall change across an interval.There are several methods to integrate functions:

• Substitution: or the "reverse chain rule," simplifies integrals by changing variables.
• Integration by parts: useful when integrals involve products of functions, applying \(\int u \, dv = uv - \int v \, du\).
• Partial fractions: breaking down complex rational functions to simpler parts.

Integration is involved in analyzing physical systems and determining accumulated values over time, making it a crucial tool in physics, engineering, and other sciences.

Trigonometric Integration

Trigonometric integration is a specific technique in integral calculus that deals with integrating functions involving trigonometric functions like sine, cosine, or tangent.These integrals appear frequently in problems involving periodic or oscillating behaviors, such as wave motion or circular motion.For example, the integral \(\int \cos(1-x) \, dx\), from the original problem, requires understanding the behavior of cosine and its antiderivative.Some tips for tackling trigonometric integrals include:

• Using trigonometric identities to simplify expressions (e.g., \(\sin^2(x) + \cos^2(x) = 1\)).
• Recognizing known integral results, such as \(\int \cos(x) \, dx = \sin(x) + C\).
• Applying techniques like substitution or integration by parts when functions involve complex arguments.

Solving these types of integrals might seem tricky, but building familiarity with trigonometric properties and identities can greatly aid in integration.

Substitution Method

The substitution method in calculus is a technique used to simplify the process of integration.Sometimes referred to as "u-substitution," it resembles the reverse chain rule from differentiation.The idea is to change variables to transform a complicated integral into a simpler one.Typically, this involves:

• Identifying a portion of the integral to substitute with a new variable (e.g., \(w = f(x)\)).
• Determining the differential of the new variable (e.g., \(dw\)).
• Replacing all instances of the original variables in the integral with the new substitution.
• Integrating with respect to the new variable, then reversing the substitution to return to the original variable.

For example, in the original problem, the substitution \(w = 1-x\) and thus \(dw = -dx\) translates the integral \(\int \cos(1-x) \, dx\) to \(-\int \cos(w) \, dw\), simplifying the computation.This method is particularly useful when dealing with composite functions or when integrals involve a series of nested functions.Understanding this technique opens up solving more complex integrals efficiently.