Problem 127 - University Calculus: Early Transcendentals

Question

Suppose that $$ f(x)=\frac{d}{d x}(1-\sqrt{x}) \text { and } g(x)=\frac{d}{d x}(x+2) $$ Find: \(\mathbf{a} . \int f(x) d x\) b. \(\int g(x) d x\) c. \(\int[-f(x)] d x\) d. \(\int[-g(x)] d x\) e. \(\int[f(x)+g(x)] d x\) f. \(\int[f(x)-g(x)] d x\)

Step-by-Step Solution

Answer

a. \(-\sqrt{x} + C\); b. \(x + C_1\); c. \(\sqrt{x} + C_2\); \n d. \(-x + C_3\); e. \(-\sqrt{x} + x + C_4\); f. \(-\sqrt{x} - x + C_5\).

Step 1: Find the derivative for f(x)

Given that \( f(x) = \frac{d}{dx}(1-\sqrt{x}) \), compute the derivative. The derivative of \( 1 \) is \( 0 \), and the derivative of \( -\sqrt{x} \) using the power rule is \( -\frac{1}{2}x^{-1/2} \). Thus, \( f(x) = -\frac{1}{2}x^{-1/2} \).

Step 2: Find the derivative for g(x)

Given that \( g(x) = \frac{d}{dx}(x+2) \), compute the derivative. The derivative of \( x \) is \( 1 \) and the derivative of \( 2 \) is \( 0 \). Therefore, \( g(x) = 1 \).

Step 3: Solve \( \int f(x) \, dx \)

We need to find \( \int -\frac{1}{2}x^{-1/2} \, dx \). This integral can be solved using the reverse power rule. The antiderivative of \( x^{-1/2} \) is \( 2x^{1/2} \), so \( \int f(x) \, dx = -\frac{1}{2} \cdot 2x^{1/2} + C = -\sqrt{x} + C \), where \( C \) is the constant of integration.

Step 4: Solve \( \int g(x) \, dx \)

We need to integrate \( g(x) = 1 \). The integral of \( 1 \, dx \) is \( x + C_1 \), where \( C_1 \) is the constant of integration.

Step 5: Solve \( \int [-f(x)] \, dx \)

This is the negative of the integral of \( f(x) \), which was found in Step 3. Therefore, \( \int [-f(x)] \, dx = \int \frac{1}{2}x^{-1/2} \, dx = \sqrt{x} + C_2 \), where \( C_2 \) is the constant.

Step 6: Solve \( \int [-g(x)] \, dx \)

This is the negative of the integral of \( g(x) \). Therefore, \( \int [-g(x)] \, dx = \int -1 \, dx = -x + C_3 \), where \( C_3 \) is the constant of integration.

Step 7: Solve \( \int [f(x)+g(x)] \, dx \)

First, combine \( f(x) + g(x) = -\frac{1}{2}x^{-1/2} + 1 \). Integrate using the sum of the integrals rule: \( \int [f(x)+g(x)] \, dx = \int -\frac{1}{2}x^{-1/2} \, dx + \int 1 \, dx = -\sqrt{x} + x + C_4 \), where \( C_4 \) is the constant of integration.

Step 8: Solve \( \int [f(x)-g(x)] \, dx \)

First, combine \( f(x) - g(x) = -\frac{1}{2}x^{-1/2} - 1 \). Integrate using the sum of the integrals rule: \( \int [f(x)-g(x)] \, dx = \int -\frac{1}{2}x^{-1/2} \, dx - \int 1 \, dx = -\sqrt{x} - x + C_5 \), where \( C_5 \) is the constant of integration.

Key Concepts

Derivative

The derivative is a fundamental concept in calculus. It measures the rate at which a function changes at any given point. In simpler terms, it shows us how a function's output value will change with a tiny change in the input. In mathematical terms, if you have a function \( y = f(x) \), the derivative of this function is denoted as \( \frac{dy}{dx} \). The process of finding the derivative is called differentiation.

For example, in our exercise, the function \( f(x) \) is derived step by step. The derivative of \( 1 \) is \( 0 \) because constants do not change, while the derivative of \( -\sqrt{x} \) is computed using the power rule, resulting in \( -\frac{1}{2}x^{-1/2} \). Similarly, \( g(x) \) simplifies to \( 1 \) since the derivative of \( x \) is \( 1 \), and that of another constant \( 2 \) again results in \( 0 \).

This highlights how understanding derivatives helps us break down and analyze functions step by step.

Reverse Power Rule

The reverse power rule is a technique used for integrating functions that involve power expressions. It's the opposite of the power rule used in differentiation. When we integrate a function like \( x^n \), the reverse power rule tells us to add \( 1 \) to the exponent and then divide by the new exponent. The formula is as follows:

• If \( n eq -1 \), \( \int x^n \; dx = \frac{x^{n+1}}{n+1} + C \)
• If \( n = -1 \), the integral is \( \ln|x| + C \)

In our exercise, we used the reverse power rule to solve \( \int -\frac{1}{2}x^{-1/2} \, dx \). We increased \( -1/2 \) by \( 1 \) to obtain \( 1/2 \), and then divided, resulting in \( -2x^{1/2} \) or \( -\sqrt{x} \). This technique elegantly transforms complex expressions into manageable integrated forms.

Constant of Integration

When integrating a function, we must always add the constant of integration, denoted by \( C \). This constant is crucial because integration is essentially finding antiderivatives, and any constant added to a function before it is derived is lost in the process.

Thus, when we reverse the process, we have to add a constant to cover all possibilities. Essentially, the constant of integration ensures the completeness of our solution. For example, in our integrals such as \( \int f(x) \, dx = -\sqrt{x} + C \), \( C \) represents any constant value which can be added without affecting the derivative.

Neglecting this constant would result in an incomplete and potentially incorrect solution, overlooking crucial aspects of integration.

Power Rule

The power rule is one of the most straightforward rules in calculus for finding derivatives. It helps us differentiate functions or expressions that are in the form of \( x^n \). The power rule states that if \( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \). Simply multiply by the current power and reduce the power by one.

In our example, we apply the power rule to differentiate \( -\sqrt{x} \), treating it as \( -x^{1/2} \). The differentiation using the power rule gives us \( -\frac{1}{2}x^{-1/2} \). The simplicity of the power rule allows us to quickly and easily handle various polynomial expressions.

It's a fundamental concept that underpins many more advanced topics in calculus, making it essential for anyone learning integration and differentiation.