Problem 105 - Thomas Calculus in SI Units

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. \(\int f(x) d x\) 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. \(x + C\), c. \(\sqrt{x} + C\), d. \(-x + C\), e. \(-\sqrt{x} + x + C\), f. \(-\sqrt{x} - x + C\)

Step 1: Find the derivative of 1-√x

To find \( f(x) \), we differentiate \( 1-\sqrt{x} \). The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \), so: \[ \frac{d}{dx}(1 - \sqrt{x}) = -\frac{1}{2\sqrt{x}} \] Thus, \( f(x) = -\frac{1}{2\sqrt{x}} \).

Step 2: Find the derivative of x+2

To find \( g(x) \), we differentiate \( x + 2 \). The derivative of \( x \) is 1 and the derivative of a constant (2) is 0, so: \[ \frac{d}{dx}(x + 2) = 1 \] Thus, \( g(x) = 1 \).

Step 3: Integrate f(x)

We now integrate \( f(x) = -\frac{1}{2\sqrt{x}} \). The integral of \( \frac{1}{\sqrt{x}} \) is \( 2\sqrt{x} \).\[ \int -\frac{1}{2\sqrt{x}} \, dx = -\sqrt{x} + C \]

Step 4: Integrate g(x)

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

Step 5: Integrate -f(x)

For \(-f(x)\), we have \(-(-\frac{1}{2\sqrt{x}})\), which is \(\frac{1}{2\sqrt{x}}\), so:\[ \int \frac{1}{2\sqrt{x}} \, dx = \sqrt{x} + C \]

Step 6: Integrate -g(x)

For \(-g(x)\), which is \(-1\):\[ \int -1 \, dx = -x + C \]

Step 7: Integrate the sum f(x) + g(x)

Now, we integrate \( f(x) + g(x) = -\frac{1}{2\sqrt{x}} + 1 \).\[ \int \left( -\frac{1}{2\sqrt{x}} + 1 \right) \, dx = -\sqrt{x} + x + C \]

Step 8: Integrate the difference f(x) - g(x)

Finally, we integrate \( f(x) - g(x) = -\frac{1}{2\sqrt{x}} - 1 \).\[ \int \left( -\frac{1}{2\sqrt{x}} - 1 \right) \, dx = -\sqrt{x} - x + C \]

Key Concepts

Definite Integrals

Definite integrals are a powerful tool in calculus that allow us to find the exact area under a curve from point a to point b on a graph. Unlike indefinite integrals, which deal with antiderivatives and include a constant of integration, definite integrals provide a specific numerical value.

• The general form of a definite integral can be represented as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the bounds of integration.
• A key property of definite integrals is that they can be used to calculate the total accumulation of a quantity, like distance traveled or the area of an irregular shape.
• Definite integrals can often be interpreted as the net area between the function and the x-axis over an interval \([a, b]\).

When solving problems involving definite integrals, it is important to evaluate the antiderivative of the function at both boundary points and then subtract. This creates a clear path from the more abstract idea of accumulation to obtaining actual numerical results.

Derivatives

The concept of derivatives is central to calculus as it measures the rate at which a quantity changes. Differentiation is the process of finding the derivative of a function. This operation is essential, as it helps determine the slope of a function at any given point.

• The derivative of a function \( f(x) \) is often denoted as \( f'(x) \) or \( \frac{df}{dx} \).
• A derivative provides a formula for the instantaneous rate of change, offering a precise way to calculate how one quantity reacts to changes in another.
• Standard rules for derivatives, such as the power rule, product rule, and chain rule, are utilized to differentiate complex functions.

For example, in the original exercise, the derivative of the function \( 1-\sqrt{x} \) was found by applying the power rule to \( \sqrt{x} \). Understanding derivatives allows us to work backwards when performing integration, a process known as finding antiderivatives.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are essentially the reverse of finding derivatives. If you know how a function changes (its derivative), you can figure out the original function using antiderivatives.

• To find an antiderivative of a function, you perform the integration, which is like 'undoing' the process of differentiation.
• The result of an indefinite integral provides a family of functions, each differing by a constant \( C \), because differentiation removes constant terms.
• Represented as \( \int f(x) \, dx = F(x) + C \), the process determines a function \( F(x) \) whose derivative is \( f(x) \).

In the exercise, determining the antiderivatives of given derivatives helped illustrate how calculus connects changes and accumulations over time. Remember, when integrating, you are reconstructing the function from the rate at which it changes.