Problem 2 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int 2 x \sin x d x $$

Step-by-Step Solution

Answer

\(-2x \cos x + 2 \sin x + C\)

Step 1: Identify parts for Integration by Parts

Integration by parts uses the formula \( \int u \cdot dv = uv - \int v \cdot du \). We need to identify parts, \( u \) and \( dv \), from the integral \( \int 2x \sin x \, dx \). Choose \( u = 2x \) (since its derivative simplifies) and thus \( dv = \sin x \, dx \).

Step 2: Differentiate and Integrate the Parts

Differentiate \( u \) to find \( du \): \( du = 2 \, dx \). Integrate \( dv \) to find \( v \): \( v = \int \sin x \, dx = -\cos x \).

Step 3: Apply the Integration by Parts Formula

Substitute \( u \), \( dv \), \( du \), and \( v \) into the integration by parts formula: \( \int 2x \sin x \, dx = 2x(-\cos x) - \int (-\cos x)(2) \, dx \). Simplify this to get \( -2x \cos x + \int 2 \cos x \, dx \).

Step 4: Integrate the Remaining Integral

Simplify and integrate the second part: \( \int 2 \cos x \, dx \) becomes \( 2 \sin x \). Thus, the expression becomes \( -2x \cos x + 2 \sin x \).

Step 5: Write the Final Solution with the Constant of Integration

Combine the terms and add the constant of integration \( C \): The integral evaluates to \( \int 2x \sin x \, dx = -2x \cos x + 2 \sin x + C \).

Key Concepts

Calculus

Calculus is a major branch of mathematics that focuses on change and motion, using derivatives and integrals as its central tools. There are two main types of calculus: differential calculus, which deals with rates of change and slopes of curves, and integral calculus, which focuses on the accumulation of quantities and areas under and between curves. In real-world applications, calculus helps to model and solve problems where change is a significant factor, such as physics, engineering, economics, statistics, and more.

Definite integral

The definite integral is a fundamental concept in calculus, serving as a tool to find the total change or area under a curve over a specified interval.

• The definite integral of a function from point \(a\) to point \(b\) is denoted as \(\int_{a}^{b} f(x) \, dx\).
• Geometrically, it represents the net area between the function curve \(y = f(x)\) and the x-axis, calculated from \(x = a\) to \(x = b\).
• This involves calculating the limit of a sum of areas of rectangles under the curve as the number of rectangles approaches infinity.

The importance of definite integrals lies in their ability to provide precise measurements of quantities that vary, such as distance traveled over time or the total energy consumed.

Integration techniques

Integration techniques are methods used to evaluate integrals, especially when they cannot be solved by simple antiderivatives.

• Integration by Parts: This is particularly useful for integrating products of functions. It is based on the product rule of differentiation and follows the formula: \( \int u \, dv = uv - \int v \, du \). In our problem, choosing \( u = 2x \) and \( dv = \sin x \, dx \) allowed us to break down the integral into simpler parts.
• Substitution: In cases where a change of variables simplifies the integral, substitution is the go-to technique, resembling the reverse process of the chain rule.
• Partial Fractions: This method is useful for integrating rational functions by expressing them as sums of simpler fractions.

Mastering these techniques allows you to handle a wide variety of integrals, advancing your problem-solving skills in calculus.