Problem 95 - Calculus Early Transcendentals

Question

Find the area of the region \(R\) bounded by the graph of \(f\) and the \(x\) -axis on the given interval. Graph \(f\) and show the region \(R\) $$f(x)=2-|x| \text { on }[-2,4]$$

Step-by-Step Solution

Answer

Based on the step by step solution provided, here is a short answer question: Question: Determine the area of the region R bounded by the \(x\)-axis and the graph of the function \(f(x) = 2 - |x|\) on the interval \([-2, 4]\). Answer: The area of the region R is 4.

Step 1: Finding the point where the function changes its sign

First, we need to find the point where the function \(|x|\) changes its sign. We know that if \(x\) is greater or equal to 0, \(|x|=x\), and when \(x<0\), \(|x|=-x\). Set \(|x|=0\). The only point within the given interval \([-2,4]\) when this happens is \(x=0\).

Step 2: Writing the piecewise function

Now, we will write the function \(f(x)\) as a piecewise function, which would give us: $$ f(x)= \begin{cases} 2+x, & \text{if } x<0 \\ 2-x, & \text{if } x\geq 0 \end{cases} $$

Step 3: Finding the area over \([-2,0)\)

To find the area over the interval \([-2,0)\), we need to integrate the function \(f(x) = 2 + x\) over this interval: $$ A_1 = \int_{-2}^{0} (2+x) \, dx $$

Step 4: Calculating \(A_1\)

Let's now calculate the integral: $$ A_1 = \left[2x+\frac{1}{2}x^2\right]_{-2}^{0} = (2(0)+\frac{1}{2} \cdot 0^2) - (2(-2) + \frac{1}{2} \cdot (-2)^2) = 0 - (-4) = 4 $$

Step 5: Finding the area over \([0,4]\)

To find the area over the interval \([0,4]\), we need to integrate the function \(f(x) = 2 - x\) over this interval: $$ A_2 = \int_{0}^{4} (2-x) \, dx $$

Step 6: Calculating \(A_2\)

Let's now calculate the integral: $$ A_2 = \left[2x-\frac{1}{2}x^2\right]_{0}^{4} = (2(4)-\frac{1}{2} \cdot 4^2) - (2(0) - \frac{1}{2} \cdot 0^2) = 8 - 8 = 0 $$

Step 7: Calculating the total area

Finally, let's add up the areas \(A_1\) and \(A_2\) to find the total area of the region R: $$ A = A_1 + A_2 = 4 + 0 = 4 $$ The area of the region R bounded by the graph of \(f\) and the \(x\)-axis on the given interval \([-2,4]\) is 4.

Key Concepts

Area under curve

The area under a curve refers to the space enclosed by the curve and the x-axis over a given interval. To determine this area, we calculate the integral of the function defining the curve. This integral provides the accumulated "total area" between the curve and the x-axis from the starting point to the endpoint of the interval.

For instances where the curve dips below the x-axis, we typically take the absolute value to consider only the non-negative areas. In the exercise above, the function is split into two parts over the interval a single side:

• From o to -2 , the function is 2..+x
• From 0 to 4, the function is 2−x

To find the total area under the curve between these points, we calculate the definite integrals separately for each piece of the function and then sum the results.

Piecewise functions

Piecewise functions are functions defined by different expressions over different intervals on the x-axis. These are helpful in modeling situations where a function has distinct parts, such as changes in rate or direction.

In our exercise, the function \(f(x) = 2 - |x|\) changes its form at the point \(x = 0\). Here it becomes useful to express it as a piecewise function:

• \(f(x) = 2 + x\) when \(x < 0\)
• \(f(x) = 2 - x\) when \(x \geq 0\)

These equations represent the linear segments of the original function in the specified intervals. Graphing such a function resembles separate connected line segments, each portion adhering to its specific formula.

Definite integrals

A definite integral computes the net area under a curve between two specified limits on the x-axis, resulting in a numerical value representing this area. With definite integrals, we can calculate the accumulated sums of–in this case–area, while considering any overhead paths above or below the x-axis.

For the given intervals:

• The integral \(\int_{-2}^{0} (2+x) \, dx\) computes the area where the function is above the x-axis between \(-2\) and \(0\).
• The integral \(\int_{0}^{4} (2-x) \, dx\) computes the area from \(0\) to \(4\).

After evaluating the respective integrals, the areas gained are summed to find the total area of the region \(R\). Understanding how to manipulate and evaluate these integrals is essential to finding solutions in calculus related to areas under curves.