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 \((a>0, r>0)\) $$ \int_{-1}^{1}(1-|x|) d x $$

Short answer

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

Step-by-step solution

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 \(x \geq 0\).

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.

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.