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Chapter 5: Problem 20

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

Short Answer

Expert verified
Question: Evaluate the following integral using symmetry: \(\int_{-1}^{1}(1-|x|) d x\). Solution: By taking advantage of the function's symmetry around the y-axis, we found that the integral is equal to \(\frac{1}{2} + \frac{1}{2} = 1\).

Step by step solution

01

Determine the cases for \(x\).

We need to evaluate the integral \(\int_{-1}^{1}(1-|x|) d x\). Since the function involves an absolute value, we need to split it into two cases: one for \(x \ge 0\) and another for \(x < 0\). Case 1: \(x \ge 0\). In this case, the function is \(f(x) = 1-x\) because \(|x| = x\). Case 2: \(x < 0\). In this case, the function is \(f(x) = 1+x\), because \(|x| = -x\).
02

Apply symmetry to the function.

Note that the domain of integration spans from \(-1\) to \(1\), which means the function is symmetric around the y-axis. Since the function is symmetric, we can rewrite the integral for each case to take advantage of the symmetry. For case 1 (x 鈮� 0), the integral is \(\int_{0}^{1}(1-x) d x\). For case 2 (x < 0), the integral is symmetric to case 1, so we have \(\int_{-1}^{0}(1+x) d x\).
03

Evaluate the integrals separately.

Now, we'll evaluate the integrals for each case. For case 1: $$\int_{0}^{1}(1-x) d x = \left[x - \frac{1}{2}x^2\right]_{0}^{1} = (1 - \frac{1}{2}) - (0) = \frac{1}{2}$$ For case 2: $$\int_{-1}^{0}(1+x) d x = \left[x + \frac{1}{2}x^2\right]_{-1}^{0} = (0) - (-1 + \frac{1}{2}) = \frac{1}{2}$$
04

Sum the results.

Now that we have the results for both cases, we can sum them up, as we are required to integrate over the interval \([-1, 1]\): $$\int_{-1}^{1}(1-|x|) d x = \int_{-1}^{0}(1+x) d x + \int_{0}^{1}(1-x) d x = \frac{1}{2} + \frac{1}{2} = 1$$ The result of the integral is \(1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Symmetry in integration
Symmetry can often make evaluating definite integrals simpler, especially when integrals are taken over symmetric intervals, such as from -a to a.
In such cases, if the function is an even function鈥攎eaning f(x) = f(-x)鈥攖he integral of f from -a to a can be simplified.
This is because the contributions of the function's positive and negative parts cancel out any odd behavior (where f(x) = -f(-x)), making the integral of only the positive part suffice.
For the exercise's function involving an absolute value, we recognized symmetry by splitting the integral into a piece for positive x and another for negative x.
This allowed the function to be integrated separately from 0 to 1 and from -1 to 0, then summed for the final result.
  • Even functions: You only need to calculate one half and double it.
  • Odd functions: The integral over symmetric limits cancels to zero.
  • Using symmetry reduces computational effort.
Absolute value
The absolute value function, denoted as \(|x|\), reflects how far a number is from zero on the real number line, regardless of direction.
Mathematically, the absolute value of x is x if x is non-negative and -x if x is negative.
In calculus, particularly for integration, dealing with absolute values requires separating the domain into regions where the expression inside the absolute value changes sign.
This is because each side of zero will have different linear expressions when the absolute value is resolved.
In this exercise, the function \(1 - |x|\) involves an absolute value of x and was split into two cases:
  • When \(x \ge 0\), \( |x| = x \, \text{so} \ 1 - |x| = 1 - x\).
  • When \(x < 0\), \( |x| = -x \, \text{so} \ 1 - |x| = 1 + x\).
By handling absolute value functions piecewise, we can apply appropriate limits and evaluate each part separately. This method leads us to evaluate the integral correctly over its entire domain.
Piecewise functions
Piecewise functions are an essential concept in calculus and analysis, allowing different formulas or expressions to be applied to different segments of a domain.
When a function behaves differently in different intervals, it is smart to express it in a piecewise format.
This means defining multiple expressions, each valid over a certain subdomain.
For the integral problem provided in the exercise, the function \(1 - |x|\) was expressed as part of a piecewise function:
  • For \(x \ge 0\), we have \(1 - x\).
  • For \(x < 0\), it switches to \(1 + x\).
Defining and using piecewise functions effectively requires:
* Understanding how the function behaves in each interval* Appropriately setting the integration limits for these intervals* Calculating each portion separately before summing them to find the total value of the integral
This strategy provides clarity and precision, especially when dealing with more complex functions or discontinuities. In calculus, mastering piecewise functions can make solving integrals and evaluating limits much more manageable.

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