Question

Find the area covered by the curve \(y=\max .\\{2-x, 2,1+x\\}\) with \(x\)-axis from \(x=-1\) to \(x=1\) is (1) \(1 / 2\) (2) \(5 / 2\) (3) \(7 / 2\) (4) \(9 / 2\)

Short answer

The area covered by the curve is \( \frac{7}{2} \).

Step-by-step solution

- Identify the Function Segments

The given function is defined as the maximum of three functions: \[ y = \text{max}\{2-x, 2, 1+x\} \]For different values of x, we'll find out which of these functions has the highest value in the range from x = -1 to x = 1.

- Evaluate the Functions at Key Points

At x = -1:\[ 2-x = 3 \]\[ 2 = 2 \]\[ 1+x = 0 \]So, the maximum value at x = -1 is 3.At x = 0:\[ 2-x = 2 \]\[ 2 = 2 \]\[ 1+x = 1 \]So, the maximum value at x = 0 is 2.At x = 1:\[ 2-x = 1 \]\[ 2 = 2 \]\[ 1+x = 2 \]So, the maximum value at x = 1 is 2.

- Determine the Maximum Function Segment

From the evaluations, the function segments can be determined:For x from -1 to 0, the function y = \(2-x\) gives the highest value.For x from 0 to 1, the function y = 2 gives the highest value. Both functions touch at x = 0.

- Integrate to Find the Area

Now, integrate each segment separately and sum them up:For \(-1 \leq x < 0\):\[ \text{Area}_1 = \int_{-1}^{0} (2-x) \, dx \]For \(0 \leq x \leq 1\):\[ \text{Area}_2 = \int_{0}^{1} 2 \, dx \]

- Calculate the Area for Each Segment

Calculate \( \text{Area}_1 \):\[ \text{Area}_1 = \int_{-1}^{0} (2-x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{-1}^{0} = 0 - \left(-2 + \frac{1}{2}\right) = \frac{3}{2} \]Calculate \( \text{Area}_2 \):\[ \text{Area}_2 = \int_{0}^{1} 2 \, dx = \left[ 2x \right]_{0}^{1} = 2 - 0 = 2 \]

- Sum the Areas

Combine the areas from the calculated segments:\[ \text{Total Area} = \text{Area}_1 + \text{Area}_2 = \frac{3}{2} + 2 = \frac{7}{2} \]

Key concepts

integration

Integration is a fundamental concept in calculus. It helps us find the area under a curve. There are different types of integrals, but we focus on definite integrals.

A definite integral computes the accumulation of quantities, such as area, between two bounds. We use the notation \(\int_{a}^{b} f(x) \, dx \) to indicate the integral of \( f(x) \) from \( a \) to \( b \).

To solve an integral:

• Identify the bounds \( a \) and \( b \).
• Determine the function \( f(x) \).
• Integrate the function and apply the bounds.

We subtract the value of the antiderivative at \( a \) from its value at \( b \). This gives us the accumulated area between \( a \) and \( b \).

piecewise functions

Piecewise functions consist of different expressions based on the input's value. They are written as multiple sub-functions, each with a specific domain.

The exercise involves a piecewise function defined as \( y = \text{{max}}\{2-x, 2, 1+x\} \). This 'max' function selects the maximum value among its components for each \( x \).
For instance, within a specific range:

• For \( -1 \leq x < 0 \), the function \( y = 2 - x \) provides the highest value.
• For \( 0 \leq x \leq 1 \), the function \( y = 2 \) provides the highest value.

This segmentation helps us identify the correct function to integrate over each interval.

definite integral

A definite integral calculates the area under a curve between two specific points on the x-axis. We denote it with two limits, \( a \) and \( b \), as in \(\int_{a}^{b} f(x) \, dx \).

In our exercise, we split the integral into segments based on the dominant piecewise function.

• For \( -1 \leq x < 0 \), we integrate \( 2 - x \).
• For \( 0 \leq x \leq 1 \), we integrate \( 2 \).

Each integral gives the area of its segment. Summing these areas provides the total area under our piecewise-defined curve within the given range.

For example, integrating \( 2 - x \) from \( -1 \) to \( 0 \) and \( 2 \) from \( 0 \) to \( 1 \) leads to:
\( \text{{Total Area}} = \int_{-1}^{0} (2 - x) \, dx + \int_{0}^{1} 2 \, dx = \frac {3}{2} + 2 = \frac{7}{2} \). This final value represents the total area covered by the curve.