Problem 109 - Thomas Calculus

Question

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

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: Calculate Derivative f(x)

First, we find the derivative of the function given in \(f(x) = \frac{d}{dx}(1 - \sqrt{x})\). The function can be rewritten as \(1 - x^{1/2}\). Taking the derivative:\[\frac{d}{dx}(1 - x^{1/2}) = 0 - \frac{1}{2}x^{-1/2} = -\frac{1}{2\sqrt{x}}\] Therefore, \(f(x) = -\frac{1}{2\sqrt{x}}\).

Step 2: Calculate Derivative g(x)

The function \(g(x) = \frac{d}{dx}(x + 2)\) is a straightforward derivative of a linear expression. Calculate as follows:\[\frac{d}{dx}(x + 2) = 1 + 0 = 1\] Therefore, \(g(x) = 1\).

Step 3: Integrate f(x)

We now integrate \(f(x) = -\frac{1}{2\sqrt{x}}\) with respect to \(x\):\[\int -\frac{1}{2\sqrt{x}} \, dx = -\int \frac{1}{2}x^{-1/2} \, dx = -\left[x^{1/2}\right] + C = -\sqrt{x} + C\] where \(C\) is the constant of integration.

Step 4: Integrate g(x)

Integrate \(g(x) = 1\) with respect to \(x\):\[\int 1 \, dx = x + C\] where \(C\) is the constant of integration.

Step 5: Integrate [-f(x)]

Integrate \([-f(x)] = \frac{1}{2\sqrt{x}}\):\[\int \frac{1}{2\sqrt{x}} \, dx = \left[\sqrt{x}\right] + C = \sqrt{x} + C\] where \(C\) is the constant of integration.

Step 6: Integrate [-g(x)]

Integrate \([-g(x)] = -1\) with respect to \(x\):\[\int -1 \, dx = -x + C\] where \(C\) is the constant of integration.

Step 7: Integrate [f(x) + g(x)]

Integrate \(f(x) + g(x) = -\frac{1}{2\sqrt{x}} + 1\):\[\int \left(-\frac{1}{2\sqrt{x}} + 1\right) \, dx = -\sqrt{x} + x + C\] where \(C\) is the constant of integration.

Step 8: Integrate [f(x) - g(x)]

Integrate \(f(x) - g(x) = -\frac{1}{2\sqrt{x}} - 1\):\[\int \left(-\frac{1}{2\sqrt{x}} - 1\right) \, dx = -\sqrt{x} - x + C\] where \(C\) is the constant of integration.

Key Concepts

Derivative Calculation

In calculus, derivatives represent the rate at which a function changes with respect to a variable. For example, the derivative of a function expressed as \( f(x) = \frac{d}{dx}(1 - \sqrt{x}) \) can be calculated using the power rule for derivatives.
The power rule states that \( \frac{d}{dx}(x^n) = nx^{n-1} \). Therefore, we rewrite \( \sqrt{x} \) as \( x^{1/2} \) and differentiate using the rule:
\( \frac{d}{dx}(1 - x^{1/2}) = 0 - \frac{1}{2}x^{-1/2} = -\frac{1}{2\sqrt{x}} \).
This result tells us how the function \( 1 - \sqrt{x} \) changes as \( x \) changes. Calculating the derivative of simple linear functions like \( g(x) = x + 2 \) is straightforward since the derivative of a constant is zero and the derivative of \( x \) is \( 1 \). So, \( \frac{d}{dx}(x + 2) = 1 + 0 = 1 \). This indicates a consistent rate of change.

Integral of Functions

Integration is essentially the process of finding the 'anti-derivative' of a function. When you integrate a function, you are finding the area under its curve. For instance, integrating \( f(x) = -\frac{1}{2\sqrt{x}} \) involves finding its antiderivative:
\[ \int - \frac{1}{2\sqrt{x}} \, dx = -\int \frac{1}{2}x^{-1/2} \, dx = -\left[x^{1/2}\right] + C = -\sqrt{x} + C \]
where \( C \) is the constant of integration. Similarly, integrating a constant function like \( g(x) = 1 \) with respect to \( x \) is straightforward. The integral becomes:
\[ \int 1 \, dx = x + C \]. This represents the simplest type of integration, a constant slope across the x-axis.

Integration Constants

The integration constant, denoted by \( C \), is a crucial component in indefinite integrals. Since differentiation eliminates constants, when reversing the process via integration, we need to account for any possible constant that may have been present.
This is why after integrating a function, we add \( C \) to indicate that any constant number might have been part of the original function.
For example, when integrating \( f(x) = -\frac{1}{2\sqrt{x}} \), we arrive at the result \( -\sqrt{x} + C \). The \( + C \) captures the potential lost constant from the differentiation process.

Piecewise Integration

Piecewise integration is necessary when dealing with functions that are combined or altered, such as adding, subtracting, or multiplying together. In the exercises given, we encountered integration over combined functions like \( f(x) + g(x) \) and \( -f(x) \).
For example, integrating \( f(x) + g(x) = -\frac{1}{2\sqrt{x}} + 1 \) results in:
\[ \int \left(-\frac{1}{2\sqrt{x}} + 1\right) \, dx = -\sqrt{x} + x + C \].
This is done by integrating each part individually and then combining results. Similarly, for \( -f(x) \), the integration becomes \( \int \frac{1}{2\sqrt{x}} \, dx \), with the result \( \sqrt{x} + C \). Each piece is integrated separately, keeping the process manageable and correct. This technique is vital for handling more complex functions efficiently.