Problem 16 - 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-2 & \text { if } 0 \leq x \leq 2 \\ \sin (\pi x) & \text { if } 2

Step-by-Step Solution

Answer

The total area is 2.

Step 1: Identify the Function Segments

The function given is piecewise. For the interval \(0 \leq x \leq 2\), the function is \(f(x) = x - 2\). For the interval \(2 < x \leq 4\), the function is \(f(x) = \sin(\pi x)\).

Step 2: Set Up the Integral for the First Segment

To find the area between the graph and the \(x\)-axis for the first segment, we need to evaluate \(\int_0^2 (x - 2) \, dx\). This integral represents the area between the line \(x - 2\) and the \(x\)-axis from \(x = 0\) to \(x = 2\).

Step 3: Calculate the First Integral

Compute the integral \(\int_0^2 (x - 2) \, dx\). This becomes:\[ \int_0^2 (x - 2) \, dx = \left[ \frac{x^2}{2} - 2x \right]_0^2 \]Calculating gives \((1/2)(2^2) - 2(2) - ((1/2)(0^2) - 2(0)) = 2 - 4 = -2\). Since we're dealing with the area, take the absolute value: 2.

Step 4: Set Up the Integral for the Second Segment

Now set up the integral for the second segment over \(2 < x \leq 4\), which is \(\int_2^4 \sin(\pi x) \, dx\). This integral represents the area between the curve \(\sin(\pi x)\) and the \(x\)-axis from \(x = 2\) to \(x = 4\).

Step 5: Calculate the Second Integral

Compute the integral \(\int_2^4 \sin(\pi x) \, dx\). Use the substitution \(u = \pi x\), so \(du = \pi \, dx\) or \(dx = du/\pi\):\[ \int \sin(u) \, du = -\cos(u)\] Evaluating gives:\[ -\left. \frac{\cos(\pi x)}{\pi} \right|_2^4 = -\frac{\cos(4\pi)}{\pi} + \frac{\cos(2\pi)}{\pi} = -\frac{1}{\pi} + \frac{1}{\pi} = 0\]

Step 6: Combine the Results for Total Area

Add the absolute areas from both segments: The area from the first interval is 2, and the area from the second interval is 0. Thus, the total area is 2 + 0 = 2.

Key Concepts

Piecewise Functions

A piecewise function is a type of function defined by different expressions, each valid over a specific interval. Think of it as stitching together several simple functions to make a more complex one. In the provided problem, the function is defined in two parts:

• For the interval between 0 and 2, the function is defined as \( f(x) = x - 2 \).
• For the interval between 2 and 4, it changes to \( f(x) = \sin(\pi x) \).

Such functions are quite common in real-world problems where different conditions apply over different ranges. It's crucial to separate the function into its distinct segments to properly analyze and solve these kinds of problems.

Integration

Integration is a fundamental concept in calculus. It allows us to find the total accumulation of a quantity, such as area under a curve, from discrete rates of change.
In this problem, integration is used to find the area under each segment of the piecewise function. When you integrate a function, you are essentially summing up infinitesimally small slices of area to find the total area under a curve.
Remember, the integral of a function \( f(x) \) from \( a \) to \( b \) is generally represented as:\[ \int_a^b f(x) \, dx \]This calculation gives the overall contribution of the function value, \( f(x) \), over the interval \([a, b]\). This is vital in computing areas, as seen in our exercise.

Definite Integral

A definite integral computes the net area under a curve for a given interval, taking into consideration both positive and negative contributions.
In the exercise, we used definite integrals for:

• The first segment: \( \int_0^2 (x - 2) \, dx \). Upon calculation, the result was -2, but we take the absolute value when considering physical areas.
• The second segment: \( \int_2^4 \sin(\pi x) \, dx \). The evaluation resulted in 0, implying that this segment contributes no net area when considered from the x-axis.

The combination of these integrals provides the total net area between the function and the x-axis over the specified intervals, crucially featuring absolute values to account for areas below the x-axis.

Area Under a Curve

Finding the area under a curve is a practical application of calculus, often used to measure quantities that accumulate over some interval. In this exercise, this area signifies how much space the curve encloses above the x-axis.
For piecewise functions, you must calculate the area of each segment separately and then sum them up. This approach involves:

• Computing the definite integral for each segment.
• Considering absolute values for areas beneath the x-axis.

Thus, you get a total area of 2, combining the contributions from both segments. Calculating the area under a curve helps solve many real-world problems from physics to economics, where understanding the "total change" or accumulation is necessary.