Problem 101 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

a. \(-\sqrt{x} + C\); b. \(x + C\); c. \(\sqrt{x} + C\); d. \(-x + C\); e. \(-\sqrt{x} + x + C\); f. \(-\sqrt{x} - x + C\)

Step 1: Differentiate the Functions

First, we need to find the derivatives, as given:Differentiating \( f(x) = \frac{d}{dx}(1-\sqrt{x}) \), we get \( f(x) = 0 - \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x}} \).Differentiating \( g(x) = \frac{d}{dx}(x+2) \), we get \( g(x) = 1 \).

Step 2: Integrate \( f(x) \)

Now, we integrate \( f(x) = -\frac{1}{2\sqrt{x}} \):\[ \int f(x) \ dx = \int -\frac{1}{2} x^{-1/2} \ dx = -\frac{1}{2} \cdot \frac{x^{1/2}}{1/2} + C = -\sqrt{x} + C \]

Step 3: Integrate \( g(x) \)

Next, we integrate \( g(x) = 1 \):\[ \int g(x) \ dx = \int 1 \ dx = x + C \]

Step 4: Integrate \(-f(x)\)

Integrating \( -f(x) = \frac{1}{2\sqrt{x}} \):\[ \int -f(x) \ dx = \int \frac{1}{2} x^{-1/2} \ dx = \frac{1}{2} \cdot \frac{x^{1/2}}{1/2} + C = \sqrt{x} + C \]

Step 5: Integrate \(-g(x)\)

Integrating \(-g(x) = -1 \):\[ \int -g(x) \ dx = \int -1 \ dx = -x + C \]

Step 6: Integrate \(f(x) + g(x)\)

Integrate the sum \( f(x) + g(x) = -\frac{1}{2\sqrt{x}} + 1 \):\[ \int [f(x) + g(x)] \ dx = \int \left(-\frac{1}{2} x^{-1/2} + 1\right) \ dx = -\sqrt{x} + x + C \]

Step 7: Integrate \(f(x) - g(x)\)

Integrate the difference \( f(x) - g(x) = -\frac{1}{2\sqrt{x}} - 1 \):\[ \int [f(x) - g(x)] \ dx = \int \left(-\frac{1}{2} x^{-1/2} - 1\right) \ dx = -\sqrt{x} - x + C \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any given point. Think of it as a way to determine how quickly or slowly something is happening. In mathematical terms, the process of differentiation involves taking the derivative of a function. This derivative tells us the slope of the tangent line to the function at any point, giving an idea of its steepness.

For example, in this exercise, we start by differentiating the given functions. For function \( f(x) = 1 - \sqrt{x} \), differentiation yields \( f'(x) = -\frac{1}{2\sqrt{x}} \). This result indicates how the function \( 1 - \sqrt{x} \) changes with respect to \( x \). Similarly, differentiating \( g(x) = x + 2 \) gives us \( g'(x) = 1 \), a constant that tells us the function increases at a steady rate of 1.

Integration

Integration is the process of finding the integral of a function, which is essentially the reverse of differentiation. Whereas differentiation looks for the rate of change, integration looks to accumulate quantities, much like summing up slices to find the whole. When integrating a function, we aim to find what function could have produced the given derivative. This can also be visualized as the area under a curve.

In this exercise, we integrate the derivatives we found earlier. For \( f(x) = -\frac{1}{2\sqrt{x}} \), the integral is \( -\sqrt{x} + C \), where \( C \) is the constant of integration. This result represents the original function whose rate of change is given by \( f(x) \). Integrating \( g(x) = 1 \), we get \( x + C \), indicating the accumulation of a steady rate of increase over \( x \).

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is one of the key principles that ties together the concepts of differentiation and integration. It suggests that differentiation and integration are inverse processes. This theorem makes it incredibly handy to solve complex problems by switching between derivatives and integrals.

The first part of the theorem states that if a function is continuous over an interval, then it has an antiderivative over that interval. In simpler terms, if you can differentiate a function to get \( f(x) \), then integrating \( f(x) \) should give you back to the function from which \( f(x) \) came. The exercise demonstrates this by finding the integral of the derivatives that were differentiated initially, bringing us back to the original functions within constant terms.

Definite and Indefinite Integrals

In calculus, integrals can be classified into two types: definite and indefinite. An indefinite integral, like those in this exercise, gives a family of functions and includes a constant \( C \). This constant represents any number since integration can be thought of as reversing differentiation, which can differ by a constant.

Conversely, definite integrals have limits and provide a specific numerical value representing the area under the curve between two points. For example, starting from \( x = a \) to \( x = b \), integrating determinate functions will give the precise accumulation from \( a \) to \( b \). The exercise primarily deals with indefinite integrals, providing a general form \( + C \), where \( C \) is the constant of integration. It shows how finding indefinite integrals recovers the original functions(modulo a constant) from their derivatives.