Problem 65 - Thomas Calculus

Question

Right, or wrong? Say which for each formula and give a brief reason for each answer. \begin{equation}\begin{array}{l}{\text { a. } \int(2 x+1)^{2} d x=\frac{(2 x+1)^{3}}{3}+C} \\ {\text { b. } \int 3(2 x+1)^{2} d x=(2 x+1)^{3}+C} \\\ {\text { c. } \int 6(2 x+1)^{2} d x=(2 x+1)^{3}+C}\end{array}\end{equation}

Step-by-Step Solution

Answer

Part (a) is wrong; parts (b) and (c) are correct.

Step 1: Analyze Part (a)

Let's find the integral of \((2x + 1)^2\). To solve \(\int(2x + 1)^2 \,dx\), use substitution. Let \(u = 2x + 1\), hence \(du = 2\,dx\) or \(\frac{du}{2} = dx\).Now the integral becomes:\[ \int u^2 \cdot \frac{1}{2} \,du = \frac{1}{2} \cdot \frac{u^3}{3} + C = \frac{(2x + 1)^3}{6} + C \]Compare \(\frac{(2x + 1)^3}{6} + C\) with the given expression \(\frac{(2x + 1)^3}{3} + C\). The given expression is incorrect because the actual coefficient of the answer should be \(\frac{1}{6}\), not \(\frac{1}{3}\).

Step 2: Analyze Part (b)

Now, we'll solve \(\int 3(2x + 1)^2 \, dx\). Again, using substitution \(u = 2x + 1\) gives \(du = 2 \, dx\) or \(dx = \frac{du}{2}\).The integral becomes:\[ 3 \int (u^2) \cdot \frac{1}{2} \,du = \frac{3}{2} \cdot \frac{u^3}{3} + C \]\[ = \frac{u^3}{2} + C = \frac{(2x + 1)^3}{2} + C \]Comparing this with \((2x + 1)^3 + C\), the expression given is correct. \(\frac{(2x + 1)^3}{2}\) simplifies perfectly to \((2x + 1)^3\) since they divided by a common factor of 2 and combined with the constant \(C\). Therefore, the given expression is correct.

Step 3: Analyze Part (c)

Now, check \(\int 6(2x + 1)^2 \, dx\). Using substitution \(u = 2x + 1\) gives \(du = 2 \, dx\) or \(dx = \frac{du}{2}\).The integral becomes:\[ 6 \int (u^2) \cdot \frac{1}{2} \,du = 3 \cdot \frac{u^3}{3} + C \]\[ = u^3 + C = (2x + 1)^3 + C \]Thus, the expression \((2x + 1)^3 + C\) is correct.

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used to evaluate integrals where the standard rules of integration might not be directly applicable. It resembles the chain rule used in differentiation, where one part of the expression is substituted to simplify the integral.

Here's how it works:

• Identify a portion of the integrand that can be replaced with a simpler variable, commonly labeled as \( u \).
• Calculate the differential \( du \), which replaces the \( dx \) in the integral, by differentiating \( u \) with respect to \( x \).
• Express \( dx \) in terms of \( du \) by rearranging the derived expression, and substitute back into the integral.
• The integral often transforms into a simple polynomial which can be easily integrated.
• After integration, substitute back the original expressions for \( u \) to get the final answer.

This method is essential in solving complex problems, like those involving polynomial expressions raised to a power, for instance, \( (2x + 1)^2 \), simplifying it to an easier integration of \( u^2 \).

It ultimately assists in transforming a difficult integral into a basic one.

Definite and Indefinite Integrals

Integrals are divided into two main types: definite and indefinite integrals. The difference lies primarily in their usage and the results they produce.

• Indefinite Integrals represent a family of functions and include a constant \( C \), since they consider all possible anti-derivatives of a given function. They take the form \( \int f(x) \, dx = F(x) + C \).
• Definite Integrals calculate the net area under a curve between two specified limits \( a \) and \( b \). The notation \( \int_{a}^{b} f(x) \, dx \) is used and provides a specific numerical result based on the limits.

Both types are foundational in calculus:

• Indefinite integrals are useful for finding general formulas and solving differential equations.
• Definite integrals are crucial for applications involving accumulated values, like total distance or area.

In the given exercise, we primarily deal with indefinite integrals where a constant \( C \) is included in each solution to reflect the family of possible functions.

Antiderivatives

Antiderivatives, closely linked to integrals, involve finding a function whose derivative matches a given function. In essence, taking the antiderivative is the inverse operation of differentiation.

Key points about antiderivatives include:

• Finding the antiderivative involves determining \( F(x) \) when given \( f(x) \), such that \( F'(x) = f(x) \).
• The antiderivative is not uniquely defined; it contains an arbitrary constant \( C \) due to the indefinite integration process.
• This constant \( C \) reflects the vertical shift in the graph of the function and is crucial when solving problems involving initial conditions.

Understanding antiderivatives is essential to solving calculus problems like those presented in your exercise, where evaluating integrals requires determining antiderivative formulas and including the constant \( C \) to denote the family of all possible solutions.