Problem 27 - Essential Calculus

Question

In Exercises 21-30, sketch the region whose area is given by the definite integral. Then use a geometric formula to evaluate the integral \((a>0, r>0)\) $$ \int_{-1}^{1}(1-|x|) d x $$

Step-by-Step Solution

Answer

The area under the curve of the function \(1-|x|\) from -1 to 1 is 1 square unit.

Step 1: Sketching the graph

To start off, plot the function \(1-|x|\) via its absolute value properties splitting the interval into two parts: \(1-x\) for \(x \geq 0\) and \(1+x\) for \(x<0\). This results in two lines intersecting at the point (0,1): a line sloping downward to the right for \(x<0\) and another line sloping downwards to the left for \(x \geq 0\).

Step 2: Identify the Geometry

After you draw the graph of the integrand function, the region bounded by the graph of the function and the interval (-1,1) forms a triangle with base from -1 to 1 on the x-axis (length of 2 units) and height 1 unit (from x-axis up to the point (0,1)). Hence, you can realize that this is an isosceles triangle.

Step 3: Calculate the area

To compute the area of the isosceles triangle which occupies the definite integral, apply the formula for the area of a triangle which is \(\frac{1}{2} \times \text{base} \times \text{height}\). Substituting the values found in Step 2 the base of length 2 and height 1, it yields the result of \(\frac{1}{2} \times 2 \times 1 = 1\) square units.

Key Concepts

Geometric Interpretation

When dealing with definite integrals, we often interpret them as the area under a curve, between the curve and the x-axis, over a specific interval. This area gives us a practical way to understand what the integral represents. For our specific problem, the function given is shaped by the absolute value of x, creating a V-like graph. Imagine drawing this on a piece of paper: you would see a symmetrical shape across the y-axis. The area we're interested in is located between \(-1\ ext{ and }1\) on the x-axis. In this particular case, it forms a recognizable geometric shape — a triangle. This makes the calculation of the integral straightforward because we can rely on simple geometric principles.

Absolute Value Function

The absolute value function, denoted by \(|x|\), presents uniqueness in equations by being piecewise-defined. It means that \(|x|\) outputs the non-negative value of x, regardless of whether x itself is negative or positive. This adds an extra layer to visualizing its graph.

• For \(x \geq 0\), the absolute value function operates normally, essentially acting as the function \(x\) itself.
• For \(x < 0\), \(|x|\) converts any negative input into a positive, represented graphically as \(-x\).

When you apply an operation like \(1 - |x|\), as shown in this exercise, it influences the y-values, leading to two distinct line segments. Together, these segments give the graph its inverted-V shape, pivotal in understanding why the area under this curve forms a triangle.

Triangle Area Calculation

To compute the area of the triangle formed under this curve, we use a basic geometric formula. Since the region is a perfect isosceles triangle, we can employ the standard triangle area formula: \(\frac{1}{2} \times \text{base} \times \text{height}\). Here, the base is straightforward as it spans from \(-1\ ext{ to }1\) equating to 2 units. The height is determined by the peak of the triangle, lying at the intersection (0,1) on the graph.
Evaluate these measurements into the area formula:

• The base measures 2 units.
• The height stands at 1 unit.
• Plugging into the formula yields \(\frac{1}{2} \times 2 \times 1 = 1\) square unit.

This area calculation matches the value of the given definite integral, confirming our geometric interpretation and demonstrating the utility of visualizing integrals in this manner.