Problem 16 - Calculus Early Transcendentals

Question

Use symmetry to evaluate the following integrals. $$\int_{-1}^{1}(1-|x|) d x$$

Step-by-Step Solution

Answer

Based on the step-by-step solution, the integral of the function \((1-|x|)\) from -1 to 1 is equal to 1.

Step 1: Visualize the function

Before we start evaluating the integral, it is important to have a clear understanding of the function. The provided function \(f(x) = 1 - |x|\) can be graphed as a triangle with vertices at \((-1,0), (0,1), \text{ and } (1,0)\). As stated in the analysis, this function is symmetric with respect to the y-axis.

Step 2: Break the integral into two parts

Since the function is symmetric about the y-axis, we can break the integral into two parts. For \(x \in [-1, 0]\), we have \(|x| = -x\). So, for this range, the function becomes \(1 - |x| = 1 - (-x) = 1+x\). For \(x \in [0,1]\), we have \(|x| = x\). Therefore, in this range, the function is \(1 - |x| = 1 - x\). Now we can write the integral as: $$\int_{-1}^{1}(1-|x|) d x = \left[\int_{-1}^{0}(1+x) d x\right] + \left[\int_{0}^{1}(1-x) d x\right]$$

Step 3: Evaluate the integral for \(x \in [-1, 0]\)

To evaluate the integral for the first part, between \(-1\) and \(0\), we can go step by step. $$\int_{-1}^0 (1+x) dx = \left[\frac{1}{2}x^2 + x\right]_{-1}^0 = \left[\frac{1}{2}(0)^2 + (0)\right] - \left[\frac{1}{2}(-1)^2 + (-1)\right] = 0 - (-\frac{1}{2}) = \frac{1}{2}$$

Step 4: Evaluate the integral for \(x \in [0, 1]\)

Now, let's evaluate the integral for the second part, between \(0\) and \(1\). $$\int_{0}^1 (1-x) dx = \left[x - \frac{1}{2}x^2\right]_{0}^1 = \left[(1) - \frac{1}{2}(1)^2\right] - 0 = 1 - \frac{1}{2} = \frac{1}{2}$$

Step 5: Add the two parts together

We have now evaluated the integral over both parts of the interval. To find the total value, we add the results of the two integrals together. $$\int_{-1}^{1}(1-|x|) d x = \frac{1}{2} + \frac{1}{2} = 1$$ So, the integral of the given function over the given interval is \(1\).