Problem 50 - Calculus: Early Transcendentals

Question

Find \( \displaystyle \int^5_0 f(x) \, dx \) if \( f(x) = \left\\{ \begin{array}{ll} 3 & \mbox{if \) x < 3 \(}\\\ x & \mbox{if \) x \ge 3 \(} \end{array} \right.\)

Step-by-Step Solution

Answer

The integral is 17.

Step 1: Understand the Function

The function \( f(x) \) is piecewise: it is 3 when \( x < 3 \), and it is \( x \) when \( x \ge 3 \). We need to integrate this function from 0 to 5.

Step 2: Set Up the Integral

The integral from 0 to 5 can be split into two separate integrals to account for the piecewise definition. These are: \( \int_{0}^{3} 3 \, dx \) and \( \int_{3}^{5} x \, dx \).

Step 3: Evaluate the First Integral

For the integral \( \int_{0}^{3} 3 \, dx \), treat 3 as a constant. Integrating this from 0 to 3 gives:\[ 3x \Big|_0^3 = 3(3) - 3(0) = 9. \]

Step 4: Evaluate the Second Integral

For the integral \( \int_{3}^{5} x \, dx \), integrate \( x \) to get \( \frac{x^2}{2} \). Then, evaluate from 3 to 5:\[ \frac{x^2}{2} \Big|_3^5 = \frac{5^2}{2} - \frac{3^2}{2} = \frac{25}{2} - \frac{9}{2} = 8. \]

Step 5: Add the Results

Now, add the results from the two parts: \( 9 + 8 = 17 \).

Key Concepts

Piecewise Functions

A piecewise function is a type of function defined by different expressions depending on the value of the input. In our example, the function \( f(x) \) acts differently when \( x < 3 \) compared to when \( x \geq 3 \). This structure allows each part of the function to suit different intervals of \( x \).

To understand piecewise functions easily, think of them like a switch that flips between multiple behaviors based on the input's range. They are quite useful in describing real-world phenomena where a single rule cannot apply everywhere.

When dealing with piecewise functions in calculus, especially in integration, it's crucial to consider each piece separately and ensure you respect the intervals over which each piece is valid. In our example, the task was divided into two integrals because we had two pieces in the function, and each integral covered a different part of the domain.

Integration Techniques

Integration techniques are strategies used to find antiderivatives or areas under curves. In our problem, we used basic integration techniques.

It's helpful to familiarize yourself with some key techniques:

• Power Rule: When integrating a power of \( x \), such as \( x^n \), use the formula \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is an integration constant.
• Constant Rule: When integrating a constant, such as 3 in our first integral, the result is the constant times the variable of integration. So, \( \int k \, dx = kx + C \).

In our example, we applied the constant rule for the interval \( [0, 3] \) and power rule for the interval \( [3, 5] \).

It's important to handle each segment of a piecewise function within its applicable bounds to ensure the calculated integrals reflect the behavior of the function correctly across its domain.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects the concept of differentiation and integration, two core operations in calculus. It consists of two parts that practically allow us to evaluate definite integrals using antiderivatives.

The theorem states:

• Part 1: If \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then \( \int_a^b f(x)\, dx = F(b) - F(a) \).
• Part 2: If \( f \) is continuous on \( [a, b] \), then the derivative of the integral \( F(x) = \int_a^x f(t)\, dt \) is \( f(x) \).

In solving the integral \( \int_0^5 f(x) \, dx \), we used the first part. We found appropriate antiderivatives for each part of the piecewise function and applied their values at the boundaries specified by our limits.

This theorem simplifies calculating definite integrals, allowing us to find exact areas under curves efficiently and accurately.