Problem 14 - Calculus Single Variable

Question

A function \(f\) is defined piecewise on an interval \(I=[a, b] .\) Find the area of the region that is between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=\left\\{\begin{array}{cl} -x+3 & \text { if }-2 \leq x<0 \\ -x^{2}+3 x+3 & \text { if } 0 \leq x \leq 3 \end{array} \quad I=[-2,3]\right. $$

Step-by-Step Solution

Answer

The total area between the function and the x-axis is 17.5 square units.

Step 1: Understand the Function Segments

The function \( f(x) \) is given as a piecewise function with two segments. The first segment, \( f(x) = -x + 3 \), applies for \(-2 \leq x < 0\). The second segment, \( f(x) = -x^2 + 3x + 3 \), applies for \(0 \leq x \leq 3\). To find the total area, we'll need to integrate each segment separately.

Step 2: Set Up the Integral for the First Segment

For the first segment, \(-2 \leq x < 0\), set up the integral to find the area between the curve \( f(x) = -x + 3 \) and the x-axis. The area is given by the integral: \[ \int_{-2}^{0} (-x + 3) \; dx \].

Step 3: Evaluate the First Integral

Calculate the integral \( \int_{-2}^{0} (-x + 3) \; dx \). This results in: \[ \left[ -\frac{x^2}{2} + 3x \right]_{-2}^{0} = \left(-\frac{0^2}{2} + 3 \times 0\right) - \left( -\frac{(-2)^2}{2} + 3 \times (-2) \right). \] Simplifying gives \( 0 - (-2 + 6) = 4 \). Thus, the area for the first segment is 4.

Step 4: Set Up the Integral for the Second Segment

For the second segment, \(0 \leq x \leq 3\), set up the integral to find the area between the curve \( f(x) = -x^2 + 3x + 3 \) and the x-axis. The area is given by the integral: \[ \int_{0}^{3} (-x^2 + 3x + 3) \; dx \].

Step 5: Evaluate the Second Integral

Calculate the integral \( \int_{0}^{3} (-x^2 + 3x + 3) \; dx \). This results in: \[ \left[ -\frac{x^3}{3} + \frac{3x^2}{2} + 3x \right]_{0}^{3} = \left(-\frac{3^3}{3} + \frac{3 \times 3^2}{2} + 3 \times 3\right) - \left(-\frac{0^3}{3} + \frac{3 \times 0^2}{2} + 3 \times 0 \right). \] Simplifying gives \(-9 + 13.5 + 9 = 13.5\), so the area for the second segment is 13.5.

Step 6: Add Up the Areas of Both Segments

Add the areas of both segments to find the total area between the function \( f \) and the x-axis from \( x = -2 \) to \( x = 3 \). This is \( 4 + 13.5 = 17.5 \).

Key Concepts

Definite Integrals

Definite integrals are a core concept in calculus that help us determine the accumulated value of a function over a specific interval. When you integrate a function, you're essentially summing up all the infinitesimal parts over the desired range. For the function discussed in our example, we compute two separate definite integrals because it is defined piecewise. One integral calculates the area under the curve from -2 to 0, while the other calculates from 0 to 3.

The notation for a definite integral, such as \( \int_{a}^{b} f(x) \, dx \), indicates that we're finding the total accumulation of the function \( f(x) \) from \( x = a \) to \( x = b \). This process results in a numerical value representing the area under the curve, assuming the curve is above the x-axis. It is important to account for each section separately when dealing with piecewise functions.

Area Under a Curve

Finding the area under a curve is one of the primary applications of integration in calculus. It gives us the total area between the function and the x-axis within a specified interval. To visualize it, imagine the curve representing a function tracing out a landscape—integrating provides the 'ground' area covered by the original landscape curve over the x-axis.

In the exercise example, to find the total area under the piecewise function, we must calculate two separate areas. One from \( x = -2 \) to \( x = 0 \) and another from \( x = 0 \) to \( x = 3 \). By evaluating these integrals, we determine how much 'space' is occupied under the function. This method becomes crucial when the curve dips below the x-axis; calculus allows us to consider areas as negative or positive, based on their position relative to the axis.

Calculus Problem Solving

Solving calculus problems often requires breaking down complex tasks into manageable steps. This is especially true for integration problems involving piecewise functions. Here is a simple approach to problem-solving with calculus:

• Understand the Function: Recognize how the function behaves in different sections. For example, a piecewise function might have entirely different expressions on separate intervals.
• Setup the Problem: Identify the bounds and set up individual integrals for each segment of the function.
• Perform Calculations: Use integration techniques to evaluate each integral separately, ensuring correct calculations for each interval.
• Combine Results: Add together the results from each integral to solve the overall problem, recognizing the total effect or area under the curve.

By following these structured steps, tackling calculus problems becomes a more systematic and straightforward task.

Piecewise Functions

Piecewise functions are functions defined by multiple sub-functions, each applying to a certain interval of the domain. They allow us to describe situations where a single equation would not suffice. For instance, let's say you want to capture distinct behaviors of a system under different conditions—piecewise functions can model this variability.

In our exercise, the function \( f(x) \) is piecewise with two distinct `segments`: \(-x + 3\) for \(-2 \leq x < 0\) and \(-x^2 + 3x + 3\) for \(0 \leq x \leq 3\). Each segment has its rules for how the function behaves over its specific part of the range. When integrating or looking for areas under piecewise functions, handling each sub-function independently is crucial, as each describes different behavior in its respective interval.