Problem 66 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

a. Wrong, incorrect derivative. b. Wrong, incorrect derivative. c. Right, correct derivative.

Step 1: Analyze Formula a

The given integral formula is \( \int \sqrt{2x+1} \; dx = \sqrt{x^2+x+C} \). To check if this is correct, differentiate the right-hand side with respect to \( x \). The derivative is \( \frac{d}{dx} \sqrt{x^2+x+C} = \frac{(2x+1)}{2\sqrt{x^2+x+C}} \). This does not equal \( \sqrt{2x+1} \), thus the formula is incorrect.

Step 2: Analyze Formula b

The given integral formula is \( \int \sqrt{2x+1} \; dx = \sqrt{x^2+x}+C \). Differentiating, we have \( \frac{d}{dx} [\sqrt{x^2+x}] = \frac{(2x+1)}{2\sqrt{x^2+x}} \). This does not match with \( \sqrt{2x+1} \), so this formula is also incorrect.

Step 3: Analyze Formula c

The given integral formula is \( \int \sqrt{2x+1} \; dx = \frac{1}{3}(\sqrt{2x+1})^3 +C \). Differentiating, \( \frac{d}{dx} \left[ \frac{1}{3}(\sqrt{2x+1})^3 \right] = (2x+1)\sqrt{2x+1} \), which simplifies correctly to \( \sqrt{2x+1} \). Hence, this formula is correct.

Key Concepts

Definite Integrals

In integral calculus, definite integrals are used to calculate the area under a curve between two specific points along the x-axis. This type of integral provides a numerical value that represents the total accumulation or net displacement over an interval. Definite integrals are denoted by the integral sign with specified upper and lower limits. For example, to find the area under the function \( f(x) \) from \( x = a \) to \( x = b \), you write it as \( \int_{a}^{b} f(x) \, dx \).

• **Integration Limits**: Specify the start (lower limit) and end (upper limit) of integration.
• **Net Area Calculation**: Includes areas above the x-axis as positive and below as negative.
• **Integral Evaluation**: Usually involves finding an antiderivative first and then evaluating it at the upper and lower limits, using the formula \( F(b) - F(a) \).

While the concept of definite integrals applies for specific boundaries, remember that our formulas in the original exercise were indefinite, lacking these limits and focusing instead on finding an antiderivative.

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents an instantaneous rate of change, helping us understand how a function behaves at any point. In simpler terms, differentiation tells us how a function's output changes as the input changes.

• **Function Derivatives**: For example, if you have \( y = f(x) \), the derivative \( \frac{dy}{dx} \) or \( f'(x) \) describes how \( y \) changes with \( x \).
• **Rules of Differentiation**: Include product rule, quotient rule, and chain rule, which help compute derivatives of more complex functions.
• **Derivative Tests**: Useful for understanding function behavior, like locating maxima, minima, and points of inflection.

In the original exercise, differentiation was used to verify whether the integrated expressions for each formula were correct by comparing the results to the original integral expression.

Mathematical Formulas

Mathematical formulas in calculus play a crucial role in understanding relationships between functions, their derivatives, and integrals. Formulas provide a concise way to represent mathematical concepts and procedures necessary for solving complex problems.

• **Function Representation**: Formulas serve as statements showing relationships between variables, like equations defining derivatives or integrals.
• **Problem Solving**: They outline steps for computation, helping in tackling and simplifying calculus problems.
• **Verification**: In calculus, formulas are especially vital for validating results, as seen when differentiating to verify correctness in the provided exercise.

When analyzing the formulas in our exercise, differentiation was applied to check whether the integrated result corresponded with the original function under integration. The correctness of formula C was assured because the differentiated result matched the given integral function.

Antiderivatives

An antiderivative of a function is a new function whose derivative is the original function. Antiderivatives are central in solving indefinite integrals, helping simplify the process of finding general solutions to integrals.

• **General Solution**: When finding antiderivatives, include a constant \( C \) at the end to account for all possible derivatives leading to the same original function.
• **Indefinite Integrals**: Involve finding an antiderivative without specific limits, resulting in a function plus a constant \( C \).
• **Problem Solving**: Antiderivatives help reverse differentiation, returning to the original function or expression integrated.

In our exercise, identifying the proper antiderivative was key to determining which of the given formulas was correct. Only formula C successfully expressed the valid antiderivative for the function \( \sqrt{2x+1} \), showcasing the correct approach to finding an indefinite integral.