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Chapter 5: Problem 1

If \(f\) is an odd function, why is \(\int_{-a}^{a} f(x) d x=0 ?\)

Short Answer

Expert verified
Question: Prove that for an odd function \(f(x)\), the definite integral over the interval \([-a, a]\) is equal to 0. Answer: To prove that \(\int_{-a}^{a} f(x) d x = 0\) for an odd function \(f(x)\), we used the properties of odd functions and the definition of the definite integral. We showed that the integral can be rewritten in terms of another variable, and when added together, the result is the integral of 0 over the given interval. By the definition of the integral, the integral of 0 over any interval is 0. Thus, for an odd function, \(\int_{-a}^{a} f(x) d x = 0\).

Step by step solution

01

Understanding Odd Functions

An odd function is a function that has the following property: \(f(-x)=-f(x)\) for all \(x\) in the domain of \(f\). Graphically, this means that the function is symmetric with respect to the origin, if you rotate the graph 180 degrees around the origin, it will overlap itself.
02

Defining the Definite Integral

The definite integral of a function \(f(x)\) over an interval \([a, b]\) is defined as the signed area between the curve of the function, the x-axis, and the vertical lines at \(x=a\) and \(x=b\). Mathematically, this can be represented as: \(\int_{a}^{b} f(x) d x\).
03

Evaluating the Integral of an Odd Function

Let's evaluate the definite integral of the odd function \(f(x)\) over the interval \([-a, a]\): \(\int_{-a}^{a} f(x) d x\). Since \(f(x)\) is an odd function, we have that: \[\int_{-a}^{a} f(x) d x = \int_{-a}^{a} (-f(-x)) d x\] Now, we will make a substitution. Let \(u=-x\). Then \(-u = x\) and \(-du = dx\).
04

Substituting Variables

With our substitution (\(u=-x\)), we can rewrite the integral as: \[\int_{-a}^{a} (-f(-x)) d x = \int_{a}^{-a} f(u) (-du)\] Notice that the limits of integration switched places. To fix this, we can change the limits back to their original order (\(-a\) to \(a\)) by simply multiplying the integral by -1.
05

Evaluating the New Integral and Adding

Now, we have: \[-\int_{a}^{-a} f(u) (-du) = \int_{-a}^{a} f(u) du\] Since the integrals are now identical, except for the variable used for integration, we can say that: \[ \int_{-a}^{a} f(x) d x= \int_{-a}^{a} f(u) du \] Now, we will add the two integrals together: \[ \int_{-a}^{a} f(x) d x + \int_{-a}^{a} f(u) du = \int_{-a}^{a} (f(x) + f(u)) d x = \int_{-a}^{a} 0 d x \] By the definition of the integral, the integral of 0 over any interval is 0.
06

Conclusion

Therefore, for an odd function \(f(x)\), \(\int_{-a}^{a} f(x) d x = 0\).

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Most popular questions from this chapter

Use a calculator and right Riemann sums to approximate the area of the region described. Present your calculations in a table showing the approximations for \(n=10,30,60,\) and 80 subintervals. Comment on whether your approximations appear to approach a limit. The region bounded by the graph of \(f(x)=2^{x}\) and the \(x\) -axis on the interval [1,2]

Gateway Arch The Gateway Arch in St. Louis is \(630 \mathrm{ft}\) high and has a 630 -ft base. Its shape can be modeled by the parabola $$ y=630\left[1-\left(\frac{x}{315}\right)^{2}\right] $$ Find the average height of the arch above the ground.

Does the right Riemann sum underestimate or overestimate the area of the region under the graph of a positive decreasing function? Explain.

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