Question

Area versus net area Graph the following functions. Then use geometry (not Riemann sums) to find the area and the net area of the region described. The region between the graph of \(y=1-|x|\) and the \(x\) -axis, for \(-2 \leq x \leq 2\)

Short answer

Question: Find the area and the net area of the region between the graph of \(y=1-|x|\) and the x-axis for \(-2 \leq x \leq 2\). Solution: The area and the net area of the region between the graph of \(y=1-|x|\) and the x-axis for \(-2 \leq x \leq 2\) are both equal to 14 square units.

Step-by-step solution

Graph the function and separate into two functions

In order to find the area and the net area of the region described, we must first graph the given function and separate it into two different functions on the interval \([-2, 2]\). The function \(y=1-|x|\) can be broken down into two functions \(y=1-x\) and \(y=1+x\) for \(x<0\) and \(x \geq 0\), respectively.

Integrate both functions

Now we will find the definite integral of both functions, \(1-x\) and \(1+x\), over the intervals \([-2,0]\) and \([0,2]\). This will give us the area under the curve for the respective functions. For \(y=1-x\) over the interval \([-2, 0]\): $$\int_{-2}^0 (1-x)dx$$ For \(y=1+x\) over the interval \([0, 2]\): $$\int_{0}^2 (1+x)dx$$

Calculate the definite integrals and areas

Next, we calculate the definite integrals and find the areas for both functions. For \(y=1-x\): $$\int_{-2}^0 (1-x)dx = [(1-x)x]_{-2}^0 = (1(0) - 0) - (1(-2) - 2^2) = 0 + 6 = 6$$ For \(y=1+x\): $$\int_{0}^2 (1+x)dx = [(1+x)x]_{0}^2 = (1(2) + 2^2) - (1(0) - 0) = 4 + 4 = 8$$

Calculate the area and net area

We now have the individual areas for the two separate functions. To find the total area, add the areas of both functions. Total area: $$6 + 8 = 14 \text{sq units}$$ For the net area, we have the absolute value of each integral, which does not affect the total area in this case since both integrals are positive. Net area: $$|6| + |8| = 6 + 8 = 14 \text{sq units}$$ Thus, both the area and the net area of the region between the graph of \(y=1-|x|\) and the \(x\)-axis for \(-2 \leq x \leq 2\) are equal to 14 square units.

Key concepts

Area Under the Curve

The area under the curve of a function like \(y = 1 - |x|\) represents the total space enclosed between the curve and the \(x\)-axis over a given interval. It's a visual representation of the quantity or magnitude the function provides over its domain. In our problem, the function \(y = 1 - |x|\) is graphed over the interval \([-2, 2]\).

The area is computed by splitting the absolute value function into its two defining linear segments:

• \(y = 1 - x\) for \(x < 0\)
• \(y = 1 + x\) for \(x \geq 0\)

Both need to be integrated over their respective intervals. The absolute value ensures that each segment of the function is non-negative, simplifying the calculation to a geometry problem rather than a complicated integration method.

This approach uses simple shapes—triangles in this case—to compute areas without summing infinite small values like in Riemann sums. This makes working with linear pieces straightforward, allowing the use of basic area formulas for geometric shapes.

Definite Integrals

Definite integrals help us find the area under a curve over a specific interval. They give the total accumulated value from integrating from one boundary of an interval to another. In our scenario, the definite integrals are calculated for \(y = 1 - x\) and \(y = 1 + x\).

We work through the calculations with the specific intervals:

• For \(y = 1 - x\) from \([-2, 0]\)
• For \(y = 1 + x\) from \([0, 2]\)

Each integral is evaluated by applying the limits of integration:\[\int_{-2}^0 (1-x)\,dx = 6\text{ square units}\]\[\int_{0}^2 (1+x)\,dx = 8\text{ square units}\]The definite integrals provide the exact area under each segment of the curve without adding or subtracting due to crossing the \(x\)-axis, as the function \(1 - |x|\) does not cross the \(x\)-axis within the interval. Thus, we simply compute these integrals directly, giving both area segments their respective positive values.

Net Area

The term "net area" usually refers to considering the algebraic signs during integration, accounting for areas above and below the \(x\)-axis. In this specific problem, though, all areas calculated between the segments of \(y = 1 - |x|\) and the \(x\)-axis are positive, hence area and net area are identical.

For functions that cross the \(x\)-axis within the interval, the net area would take into account the negative contributions below the axis. This means you would subtract those from the positive areas.

• Net area considers directionality, while total area does not.
• In scenarios with no crossing, as seen here, net area equals total area.

Thus, the net area concept importantly distinguishes between direction and magnitude, though our function's lack of crossing makes it straightforward. Here the net area sums the absolute values of each computed definite integral:\[|6| + |8| = 14\text{ square units}\]This ensures any negative values (if they existed) would be reinforced as positive, maintaining the true measure of 'space' between the curve and \(x\)-axis.