Problem 34 - Calculus

Question

Use the Interval Additive Property and linearity to evaluate \(\int_{0}^{4} f(x) d x .\) Begin by drawing a graph off. $$ f(x)=\left\\{\begin{array}{ll} 1 & \text { if } 0 \leq x<1 \\ x & \text { if } 1 \leq x<2 \\ 4-x & \text { if } 2 \leq x \leq 4 \end{array}\right. $$

Step-by-Step Solution

Answer

The integral evaluates to 4.5.

Step 1: Understand the Definition of the Function

The function \( f(x) \) is defined piecewise: it is 1 for \( 0 \leq x < 1 \), \( x \) for \( 1 \leq x < 2 \), and \( 4-x \) for \( 2 \leq x \leq 4 \). Visualize this by sketching the graph on a coordinate plane with these intervals and functions.

Step 2: Apply the Interval Additive Property

The interval additive property states that we can break the integral over one interval \([0, 4]\) into the sum of integrals over smaller intervals: \([0, 1]\), \([1, 2]\), and \([2, 4]\). Therefore, \( \int_{0}^{4} f(x) \, dx = \int_{0}^{1} 1 \, dx + \int_{1}^{2} x \, dx + \int_{2}^{4} (4 - x) \, dx \).

Step 3: Evaluate the Integral Over \([0, 1]\)

For the interval \([0, 1]\), \( f(x) = 1 \), a constant. Thus, \( \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1 \).

Step 4: Evaluate the Integral Over \([1, 2]\)

For the interval \([1, 2]\), \( f(x) = x \). Evaluate the integral: \( \int_{1}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{(2)^2}{2} - \frac{(1)^2}{2} = 2 - 0.5 = 1.5 \).

Step 5: Evaluate the Integral Over \([2, 4]\)

For the interval \([2, 4]\), \( f(x) = 4 - x \). Evaluate the integral: \( \int_{2}^{4} (4-x) \, dx = [4x - \frac{x^2}{2}]_{2}^{4} = (4 \times 4 - \frac{4^2}{2}) - (4 \times 2 - \frac{2^2}{2}) = (16 - 8) - (8 - 2) = 8 - 6 = 2 \).

Step 6: Add the Evaluated Integrals

Sum the results of the integrals from each interval: \( 1 + 1.5 + 2 = 4.5 \).

Key Concepts

Piecewise Functions

Piecewise functions are mathematical expressions described by different formulas depending on the values of the independent variable. In the context of integrating piecewise functions, it involves dividing a function into distinct intervals, each with a separate expression. This kind of function is invaluable in real-life scenarios where a single formula is inadequate to define the function for its entire domain. Let's take the given function:

• From 0 to less than 1, the function is constant: \(f(x) = 1\).
• From 1 to less than 2, the function is defined by itself: \(f(x) = x\).
• From 2 to 4, the function decreases linearly from 4: \(f(x) = 4 - x\).

Understanding piecewise functions involves visualizing them. Drawing the graph for this function, you'll see a horizontal line at \(y = 1\) on the first interval, a diagonal line representing \(y = x\) on the next, and a sloping line going downward for \(y = 4 - x\). Each segment matches its domain.

By grasping these intervals, understanding integration later becomes more intuitive. It makes integrating functions over specified domains much simpler, allowing calculations to be precise and context-aware.

Interval Additive Property

The Interval Additive Property is a principle in calculus that helps break down the integration of piecewise functions. It states that if a function is being integrated over a broad interval, this wide interval can be divided into smaller intervals. We then evaluate the integral over these smaller intervals separately and sum the results. Mathematically, if we have a function \(f(x)\) over an interval \([a, c]\), it can be split into intervals \([a, b]\) and \([b, c]\). So, \(\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx\).

This property is crucial for dealing with functions defined piecewise. In our case, \(f(x)\) was integrated over \([0, 4]\), which breaks smoothly into \([0, 1]\), \([1, 2]\), and \([2, 4]\). Each sub-interval corresponds to a different piece of the function, allowing us to break down the integral in manageable chunks.

This approach is particularly advantageous because it transforms a complex task into simpler parts. It opens the way for applying different integration rules or techniques to each sub-function, which is often easier than attempting to integrate across a change in function expression all at once.

Linearity of Integration

Linearity of Integration is another important concept that assists in calculating definite integrals. This rule comes in two main parts:

• Additivity: This states that the integral of a sum of functions is the sum of their integrals. If you have \(\int f(x) + g(x) \, dx\), it equals \(\int f(x) \, dx + \int g(x) \, dx\).
• Scalar Multiplication: This states that if you multiply a function by a constant and then integrate, it's the same as multiplying the integral by that constant. Formally, \(\int k \, f(x) \, dx = k \int f(x) \, dx\).

This property is remarkably helpful while integrating each part of a piecewise function, ensuring calculations remain straightforward. In our exercise, each piece of the integral \(\int_{0}^{4} f(x) \, dx\) of \(f(x)\) was tackled independently based on this property, ensuring every segment was handled with appropriate techniques.

In practical terms, linearity allows for an easy breakdown of complex problems. It simplifies multidimensional or multifaceted integrals by dismantling complexity into components, which are more accessible to both understanding and computing.