Math Problem Statement

Use upper and lower sums to show that −a −b f (x) dx = ∫ b a f (−x) dx What do these mean geometrically? (Assume that a < b, and that all the above integrals exist)

Solution

To prove the equation using upper and lower sums, we need to analyze the two integrals:

  1. abf(x)dx\int_a^b f(-x) \, dx — The integral of the function f(x)f(-x) over the interval [a,b][a, b].
  2. abf(x)dx\int_{-a}^{-b} f(x) \, dx — The integral of the function f(x)f(x) over the interval [b,a][-b, -a].

Step 1: Change of Variable in abf(x)dx\int_a^b f(-x) \, dx

To simplify, we will use a substitution. Let’s define a change of variable:

  • Let u=xu = -x, so that du=dxdu = -dx.

Now, we change the limits of integration:

  • When x=ax = a, then u=au = -a,
  • When x=bx = b, then u=bu = -b.

Thus, the integral abf(x)dx\int_a^b f(-x) \, dx becomes: abf(x)dx=abf(u)(du)=abf(u)du.\int_a^b f(-x) \, dx = \int_{-a}^{-b} f(u) \, (-du) = -\int_{-a}^{-b} f(u) \, du.

This matches the form of the second integral: abf(u)du=baf(x)dx.-\int_{-a}^{-b} f(u) \, du = \int_{-b}^{-a} f(x) \, dx.

Step 2: Relationship Between the Integrals

Now, notice that we have shown: abf(x)dx=baf(x)dx.\int_a^b f(-x) \, dx = \int_{-b}^{-a} f(x) \, dx.

This shows that abf(x)dx\int_a^b f(-x) \, dx is equal to the integral baf(x)dx\int_{-b}^{-a} f(x) \, dx, which is the second integral in the original equation.

Geometrical Interpretation

  • Upper and Lower Sums: These represent approximations of the area under the curve f(x)f(x) and f(x)f(-x). Geometrically, the upper and lower sums approximate the area under the curve on either side of the y-axis.
  • When we make the substitution u=xu = -x, we reflect the function across the y-axis. This reflection changes the limits of integration, but the area under the curve (represented by the integral) remains the same.
  • Essentially, the integrals on the left and right-hand sides of the equation measure the same area but on different intervals. The change of variable corresponds to a reflection, preserving the overall area under the curve.

Conclusion

Using the change of variable u=xu = -x, we have shown that: abf(x)dx=baf(x)dx.\int_a^b f(-x) \, dx = \int_{-b}^{-a} f(x) \, dx. Geometrically, this reflects the idea that reversing the direction of integration (by switching the limits and substituting f(x)f(-x)) does not change the total area under the curve but just reflects it across the y-axis.


Would you like any further clarification or details on any specific part? Here are some related questions to expand on the concepts:

  1. How do upper and lower sums help approximate the area under a curve?
  2. How does a change of variable in integration relate to transformations of the graph of a function?
  3. What happens when you perform a substitution involving a linear function of xx?
  4. How do Riemann sums compare with upper and lower sums in approximating integrals?
  5. What is the significance of the symmetry of functions in definite integrals?

Tip: When working with definite integrals, always check for possible symmetries (such as even or odd functions), as these can simplify the computation or lead to intuitive insights about the area under the curve.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Definite Integrals
Change of Variables in Integration
Geometric Interpretation of Integrals
Upper and Lower Sums

Formulas

Change of variable formula: ∫a^b f(-x) dx = -∫-a^-b f(u) du
Relationship of definite integrals: ∫a^b f(-x) dx = ∫-b^-a f(x) dx

Theorems

Fundamental Theorem of Calculus
Properties of Definite Integrals

Suitable Grade Level

Undergraduate Level (Calculus I or II)