Problem 13 - Thomas Calculus in SI Units

Question

Find an antiderivative for each function. Do as many as you can mentally. Check your answers by differentiation. a. \(\frac{1}{2} \sec ^{2} x\) b. \(\frac{2}{3} \sec ^{2} \frac{x}{3}\) c. \(-\sec ^{2} \frac{3 x}{2}\)

Step-by-Step Solution

Answer

a) \( \frac{1}{2} \tan x + C \) b) \( 2 \tan \left( \frac{x}{3} \right) + C \) c) \( -\frac{2}{3} \tan \left( \frac{3x}{2} \right) + C \)

Step 1: Identify the Basic Antiderivative

The basic antiderivative for \( \sec^2 x \) is known to be \( \tan x \) because the derivative of \( \tan x \) is \( \sec^2 x \). We will use this basic formula to find antiderivatives for the given functions.

Step 2: Solve Part a

For the function \( f(x) = \frac{1}{2} \sec^2 x \), we take the antiderivative of the function:\[ \int \frac{1}{2} \sec^2 x \, dx = \frac{1}{2} \int \sec^2 x \, dx = \frac{1}{2} \tan x + C \]Where \( C \) is the constant of integration.

Step 3: Solve Part b

For \( f(x) = \frac{2}{3} \sec^2 \frac{x}{3} \), we use substitution. Let \( u = \frac{x}{3} \), then \( du = \frac{1}{3} dx \) or \( dx = 3 \, du \).The integral becomes:\[ \int \frac{2}{3} \sec^2 u \, (3 \, du) = 2 \int \sec^2 u \, du = 2 \tan u + C = 2 \tan \left( \frac{x}{3} \right) + C \]

Step 4: Solve Part c

For \( f(x) = -\sec^2 \frac{3x}{2} \), similar use substitution. Let \( u = \frac{3x}{2} \), then \( du = \frac{3}{2} \, dx \) or \( dx = \frac{2}{3} \, du \).The integral becomes:\[ \int -\sec^2 u \, \left( \frac{2}{3} \, du \right) = -\frac{2}{3} \int \sec^2 u \, du = -\frac{2}{3} \tan u + C = -\frac{2}{3} \tan \left( \frac{3x}{2} \right) + C \]

Step 5: Verify by Differentiation

For each solution, differentiate to verify:- Differentiate \( \frac{1}{2} \tan x + C \) to get \( \frac{1}{2} \sec^2 x \).- Differentiate \( 2 \tan \left( \frac{x}{3} \right) + C \) to get \( \frac{2}{3} \sec^2 \frac{x}{3} \).- Differentiate \( -\frac{2}{3} \tan \left( \frac{3x}{2} \right) + C \) to get \( -\sec^2 \frac{3x}{2} \).Each differentiation confirms the correctness of the antiderivatives.

Key Concepts

Calculus

Calculus is a branch of mathematics focused on change and motion. It is divided into two primary areas: differentiation and integration. Differentiation is about understanding rates of change, such as velocity, while integration deals with finding areas under curves and accumulations. We use calculus to solve problems involving quantities that are changing. When we talk about finding an antiderivative, we refer to integration, which is the reverse process of differentiation. It's like asking where a function came from if we know its rate of change. Calculus plays a pivotal role in many fields, such as physics, engineering, economics, and beyond. It helps us model real-world situations, predict outcomes, and optimize results by understanding how different variables correlate and change magnitude over time.

Integration

Integration is a fundamental concept in calculus that involves finding a function given its derivative. This process helps determine the accumulated quantity, such as area or volume, under a curve.When we integrate a function, we're looking for all possible original functions that could produce the given derivative. This collection of functions is the set of antiderivatives. It's important to include a constant of integration, usually represented by \(C\), since differentiation of a constant yields zero, making multiple functions possible antiderivatives of the same derivative.In the exercise, integration is used to find the antiderivatives of trigonometric functions like \(\sec^2 x\). The integral of \(\sec^2 x\) is known to be \(\tan x + C\) because differentiating \(\tan x\) results in \(\sec^2 x\). Therefore, we apply this knowledge to find specific antiderivatives for given expressions.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, relate angles of a triangle to the lengths of its sides. These functions are crucial in calculus, serving as building blocks for more complex functions.In the context of integration, some trigonometric derivatives have commonly known antiderivatives. For example, the derivative of \(\tan x\) is \(\sec^2 x\), making \(\tan x\) the antiderivative of \(\sec^2 x\).Understanding how to manipulate these trigonometric expressions can simplify finding antiderivatives, especially when dealing with integrals involving secant squared \( (\sec^2 x) \). This understanding helps in solving integrals easily by substitution or using basic integral rules, essential for solving complex calculus problems involving trigonometric expressions.