Question

Use the Interval Additive Property and linearity to evaluate \(\int_{0}^{4} f(x) d x .\) Begin by drawing a graph of \(f\). $$f(x)=3+|x-3|$$

Short answer

The integral evaluates to 17.

Step-by-step solution

Analyze the Function

The function given is \(f(x) = 3 + |x - 3|\). The absolute value \(|x-3|\) splits the function at \(x=3\), resulting in two linear pieces. We can express the function as follows: \(f(x) = 3 - (x-3) = 6-x\) for \(x < 3\) and \(f(x) = 3 + (x-3) = x\) for \(x \geq 3\). Use these pieces to sketch the graph of \(f(x)\).

Sketch the Graph

To sketch \(f(x)\), note that for \(x < 3\), \(f(x)\) is a line with a negative slope of \(-1\) intersecting the y-axis at \(y = 6\). For \(x \geq 3\), \(f(x)\) is a line with a positive slope of \(1\) intersecting the y-axis at \(y = 0\). Plot these lines within the interval \([0, 4]\).

Use Interval Additive Property

Since \(f(x)\) has a transition at \(x = 3\), evaluate the integral by splitting it at \(x = 3\). Thus: \(\int_{0}^{4} f(x) \, dx = \int_{0}^{3} (6-x) \, dx + \int_{3}^{4} x \, dx\).

Evaluate the First Integral

Calculate the integral \(\int_{0}^{3} (6-x) \, dx\). The antiderivative of \(6 - x\) is \(6x - \frac{x^2}{2}\). Evaluate this from 0 to 3: \((6(3) - \frac{3^2}{2}) - (6(0) - \frac{0^2}{2}) = 18 - 4.5 = 13.5.\).

Evaluate the Second Integral

Now, calculate the integral \(\int_{3}^{4} x \, dx\). The antiderivative of \(x\) is \(\frac{x^2}{2}\). Evaluate from 3 to 4: \(\frac{4^2}{2} - \frac{3^2}{2} = 8 - 4.5 = 3.5.\)

Combine Results

Add up the results of the two integrals: \(13.5 + 3.5 = 17.\) Thus, \(\int_{0}^{4} f(x) \, dx = 17.\)

Key concepts

Interval Additive Property

The interval additive property allows us to break up an integral over a large interval into smaller, more manageable pieces. This is particularly useful when dealing with integrals over functions that change behavior at certain points, such as piecewise functions. In simple terms:

• If you have an integral from point A to C, and there is a significant change at point B, you can split the integral at B.
• Instead of calculating \( \int_{A}^{C} f(x) \, dx \), you calculate \( \int_{A}^{B} f(x) \, dx \) and \( \int_{B}^{C} f(x) \, dx \), then add the results.

By applying this property, you can simplify complex integrals into easier components that are often straightforward to evaluate. This method is highly beneficial for functions with different expressions over different intervals, such as the given exercise where the function transitions at \(x = 3\).

Piecewise Linear Function

A piecewise linear function is a function that is defined by different linear expressions over different intervals. In our example, the given function is \(f(x) = 3 + |x-3|\), which simplifies to two linear pieces split by the point \(x = 3\). Here's how you can think of it:

• For \(x < 3\), the function becomes \(f(x) = 6 - x\). This represents a line with a negative slope that decreases as x increases.
• For \(x \geq 3\), the function changes to \(f(x) = x\). This is a simple line with a positive slope passing through the origin.

These pieces are essentially linear functions stitched together, creating a complete function that exhibits different behaviors depending on which interval of x you are examining. This method allows us to look at functions piece-by-piece, making analysis and integration easier.

Absolute Value Manipulation

Sometimes, an absolute value function can complicate how we analyze or integrate it. Fortunately, you can break it down into simpler linear expressions, especially in a piecewise function context. For the function given, \(f(x) = 3 + |x-3|\), you can evaluate the absolute value by considering two cases:

• If \(x < 3\), then \(|x-3| = -(x-3)\) because x is smaller than 3, flipping the sign of the expression inside the absolute value.
• If \(x \geq 3\), then \(|x-3| = x-3\) since the expression inside the absolute value is non-negative.

This approach helps you transform the original function into a simpler format that lends itself well to integration over each specific interval. Understanding and manipulating absolute values is key to piecewise integration.

Integration by Parts

Though not directly used in the solution, integration by parts is an essential calculus method. It's used when solving integrals of products of functions, derived from the product rule for differentiation. The formula is given by:

• \(\int u \, dv = uv - \int v \, du\)

Where \(u\) and \(dv\) are parts of the integrand you choose judiciously to simplify the problem. Select \(u\) to be the function that becomes simpler on differentiation, and \(dv\) to be the part that's easily integrable. Although it wasn't necessary for the problem in question, knowing integration by parts expands your tools for tackling a wide range of integral types efficiently.