Question

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

Short answer

The integral \(\int_{0}^{4} f(x) \, dx\) evaluates to 10.

Step-by-step solution

Visualize the Function

First, we need to understand how the function \( f(x) \) behaves on the interval from \(0\) to \(4\). For \(0 \leq x < 2\), \(f(x) = 2\), which is a constant function represented by a horizontal line at \(y=2\). For \(2 \leq x \leq 4\), \(f(x) = x\), which is a linear function represented by a line with slope 1. Draw these pieces on a graph to see their behavior.

Apply the Interval Additive Property

Using the Interval Additive Property, we can split the integral \(\int_{0}^{4} f(x) \, dx\) into two separate integrals: \(\int_{0}^{2} f(x) \, dx\) and \(\int_{2}^{4} f(x) \, dx\). This allows us to evaluate each section of \(f(x)\) separately.

Evaluate the First Integral (\(0 \leq x < 2\))

For \(0 \leq x < 2\), \(f(x) = 2\). The integral becomes \(\int_{0}^{2} 2 \, dx\). To solve this, compute the antiderivative of 2, which is \(2x\), and evaluate it from 0 to 2:\[\int_{0}^{2} 2 \, dx = \left[2x\right]_{0}^{2} = 2(2) - 2(0) = 4\]

Evaluate the Second Integral (\(2 \leq x \leq 4\))

For \(2 \leq x \leq 4\), \(f(x) = x\). The integral becomes \(\int_{2}^{4} x \, dx\). To solve this, compute the antiderivative of \(x\), which is \(\frac{x^2}{2}\), and evaluate it from 2 to 4:\[\int_{2}^{4} x \, dx = \left[\frac{x^2}{2}\right]_{2}^{4} = \frac{4^2}{2} - \frac{2^2}{2} = 8 - 2 = 6\]

Combine the Results

Add the results from the two integrals to find the value of the original integral:\[\int_{0}^{4} f(x) \, dx = \int_{0}^{2} 2 \, dx + \int_{2}^{4} x \, dx = 4 + 6 = 10\]

Key concepts

Interval Additive Property

The Interval Additive Property is a useful concept in calculus that allows us to break down a complex integral into smaller, more manageable parts.

When faced with evaluating an integral over a composite interval, like from 0 to 4 in this problem, we can split it into integrals over simpler intervals.

For example, for the function given, we divide the integral \( \int_{0}^{4} f(x) \, dx \) into two parts: \( \int_{0}^{2} f(x) \, dx \) for the constant part of the function, and \( \int_{2}^{4} f(x) \, dx \) for the linear part.

This separation makes solving the integral easier and allows us to apply different techniques on each individual section. By adding the results of these parts, we obtain the final value of the original integral.

Step-by-Step Solutions

Providing solutions in a step-by-step manner aids greatly in understanding.

In this exercise, we started by graphing the function piecewise to visualize its behavior over the different intervals.

This setup gives a clear understanding of how the function behaves across its domain. Once visualized, we then took deliberate steps to evaluate each integral section.

• First, the integral from 0 to 2 was evaluated by recognizing the function as constant.
• Next, the integral from 2 to 4 considered the variable part of the function.

Breaking it down like this ensures that nothing is overlooked, and each part of the function is handled appropriately, culminating in the straightforward addition of results from the two intervals.

Piecewise Functions

Piecewise functions are functions defined by multiple sub-functions, each applying to a particular interval in the domain.

In this exercise, the function \( f(x) \) was defined separately for \( 0 \leq x < 2 \) and \( 2 \leq x \leq 4 \).

• The interval from \( 0 \leq x < 2 \) was governed by a constant value, \( f(x) = 2 \).
• For the interval \( 2 \leq x \leq 4 \), \( f(x) = x \), a linearly increasing function.

By acknowledging these pieces, we see how different rules apply depending on the interval, impacting how the integral is solved.

This concept ensures we pay attention to the boundaries and transitions within the domain, a key skill for tackling more advanced calculus problems.

Integration Techniques

Various integration techniques are employed to handle different types of functions.

For the given piecewise function, two key techniques were employed: evaluating the integral of a constant function and evaluating the integral of a linear function.

• For the constant function \( f(x) = 2 \), the integration technique involves finding the antiderivative, which is a straightforward application of multiplying the constant by the width of the interval.
• For the linear function \( f(x) = x \), we find the antiderivative, \( \frac{x^2}{2} \), by using polynomial integration rules, and evaluate it over its respective interval.

These methods demonstrate how different approaches need to be selected based on the function behavior within each interval. Understanding when and how to apply these techniques is crucial for successful integration.