Question

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

Short answer

The value of \( \int_{0}^{4} f(x) \, dx \) is 4.5.

Step-by-step solution

Break Down the Integral

To evaluate \( \int_{0}^{4} f(x) \ dx \) using the Interval Additive Property, first notice that \( f(x) \) is a piecewise function. This means we need to integrate over each sub-interval and then add the results. The sub-intervals are \([0, 1), [1, 2),\) and \([2, 4]\.\) Therefore, we can express the integral as \( \int_{0}^{1} f(x) \ dx + \int_{1}^{2} f(x) \ dx + \int_{2}^{4} f(x) \ dx.\)

Evaluate the First Integral

The function \( f(x) = 1 \) on \([0, 1)\). Thus, the integral over this interval is \( \int_{0}^{1} 1 \, dx \). To evaluate this, use the formula for the integral of a constant: \( \int_{a}^{b} c \, dx = c(b-a) \). Here, \( c = 1 \), so the integral is \( 1 \times (1-0) = 1 \).

Evaluate the Second Integral

On the interval \([1, 2)\), \( f(x) = x \). Thus, the integral is \( \int_{1}^{2} x \, dx \). The antiderivative of \( x \) is \( \frac{x^2}{2} \). Evaluate this from 1 to 2: \( \left[ \frac{x^2}{2} \right]_{1}^{2} = \frac{2^2}{2} - \frac{1^2}{2} = \frac{4}{2} - \frac{1}{2} = 2 - 0.5 = 1.5 \).

Evaluate the Third Integral

For the interval \([2, 4]\), \( f(x) = 4 - x \). Therefore, the integral is \( \int_{2}^{4} (4 - x) \, dx \). The antiderivative of \( 4-x \) is \( 4x - \frac{x^2}{2} \). Evaluating from 2 to 4 gives: \( \left[ 4x - \frac{x^2}{2} \right]_{2}^{4} = \left( 4 \times 4 - \frac{4^2}{2} \right) - \left( 4 \times 2 - \frac{2^2}{2} \right) = (16 - 8) - (8 - 2) = 8 - 6 = 2 \).

Sum of Integrals

Now that we have all the individual integrals, use the interval additive property to sum them: \( \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx + \int_{2}^{4} f(x) \, dx = 1 + 1.5 + 2 = 4.5 \).

Key concepts

Integral Calculus

Integral calculus focuses on the concept of integration, which is essentially the process of finding the area under a curve. This is the opposite operation of differentiation, which determines the slope of a function at any given point. Integrals can be definite or indefinite. A definite integral includes limits of integration, giving a number that represents the total accumulation over an interval. It's written as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the limits of integration. Indefinite integrals, on the other hand, do not have specific limits and include an arbitrary constant \( C \). They represent a family of functions. The definite integral is used in various applications such as calculating areas, volumes, central points, and many other useful things.

Piecewise Functions

Piecewise functions are functions defined by multiple sub-functions, each applying to a certain interval of the main function's domain. The range of input values is broken up into intervals, and each interval has its own corresponding function. In the provided problem, the function \( f(x) \) is defined differently over three intervals. This requires a separate approach when performing operations like integration, as each part of the piecewise function must be handled according to its specific definition.

• The first piece applies to \([0, 1)\), where \( f(x) = 1 \).
• The second piece applies to \([1, 2)\), where \( f(x) = x \).
• The third piece applies to \([2, 4]\), where \( f(x) = 4 - x \).

These separate definitions require integrating each piece individually over their respective intervals and then combining the results to get the total area under the curve.

Interval Additive Property

The interval additive property of integrals states that if you divide the interval of a definite integral into smaller intervals, you can integrate each sub-interval separately and sum up the results to get the integral of the original interval.
This property is extremely useful in cases involving piecewise functions, where handling the pieces separately makes the process simpler. The additive property provides a logical way to handle separate segments of functions that combine to form a complete response.
In the example of \( f(x) \), this property allows us to compute the integral over \([0, 4]\) by adding up the integrals over \([0, 1)\), \([1, 2)\), and \([2, 4]\). This makes it easier to compute integrals for a function with multiple rules on different intervals.

Linearity of Integration

One fundamental property of integrals is their linearity. Linearity of integration means that if you have the integral of a sum of functions, it can be split into the sum of their individual integrals. The same rule applies for a constant multiple of a function, allowing the constant to be factored out of the integral.
Thus, for functions \( f(x) \) and \( g(x) \) and a constant \( c \), the properties can be stated as follows:

• \( \int (f(x) + g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx \)
• \( \int c \cdot f(x) \, dx = c \cdot \int f(x) \, dx \)

These properties make computations manageable by breaking down complex integrals into more straightforward pieces, allowing students to use simple arithmetic operations to find solutions. In the example, linearity simplifies the process by confirming that the sum of the integrals of individual intervals gives you the integral over the whole range.