MA 16500: Analytic Geometry and Calculus I

5.4. Working with Integrals

Goal: Use symmetry of even and odd functions, compute definite integrals efficiently, and understand the average value of a function and the Mean Value Theorem for integrals.

1. Even Functions

Definition:

For all \(x\),
\[ f \text{ is even } \iff \forall x, \ f(-x) = f(x) \]

Geometrically: symmetric with respect to the y-axis.

Integral Property: If \(f\) is even then

\[ \int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx \]

Example 1.1

\[ \int_{-2}^{2} x^4\,dx \]

Since \(x^4\) is even:

\[ \int_{-2}^{2} x^4\,dx = 2\int_{0}^{2} x^4\,dx = 2\left[\frac{x^5}{5}\right]_{0}^{2} = \frac{64}{5} \]

Example 1.2

\[ \int_{-\pi/2}^{\pi/2} \cos x\,dx = 2 \int_{0}^{\pi/2} \cos x\,dx = 2\left[\sin x\right]_{0}^{\pi/2} = 2 \]

2. Odd Functions

Definition:

\[ f \text{ is odd } \iff \forall x, \ f(-x) = -f(x) \]

Geometrically: symmetric with respect to the origin.

Integral Property: If \(f\) is odd then

\[ \int_{-a}^{a} f(x)\,dx = 0 \]

Example 2.1

\[ \int_{-2}^{2} 3x^3\,dx = 0 \]

Example 2.2

\[ \int_{-\pi/2}^{\pi/2} \sin x\,dx = 0 \]

Example 2.3

\[ \int_{-1}^{1} \frac{\tan x}{1+x^2+x^4}\,dx = 0 \]

3. Mean Value Theorem for Integrals

Definition (Average Value of a Function)

\[ f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b} f(x)\,dx \]

Statement (Mean Value Theorem for Integrals):

If \(f\) is continuous on \([a,b]\), then there exists \(c \in (a,b)\) such that:
\[ f(c) = \frac{1}{b-a}\int_{a}^{b} f(x)\,dx \]

Idea (Proof Sketch):

  • Define \(A(x) = \int_a^x f(t)\,dt\)
  • Then \(A'(x) = f(x)\)
  • Apply Mean Value Theorem to \(A(x)\)

Example:

Consider \( f(x) = 2x(1-x) \) on \([0,1]\). Compute the average value of the function on the interval. Find \( c \in (0,1) \) such that \( f(c) = f_{\text{avg}} \).

Step 1: Compute the average value

\[ f_{\text{avg}} = \frac{1}{1-0} \int_0^1 2x(1-x)\,dx = \int_0^1 (2x - 2x^2)\,dx \]
\[ = \left[x^2 - \frac{2}{3}x^3\right]_0^1 = 1 - \frac{2}{3} = \frac{1}{3} \]

Step 2: Solve \( f(c) = f_{\text{avg}} \)

\[ 2c(1-c) = \frac{1}{3} \]
Thus
\[ c = \frac{1 \pm \frac{\sqrt{3}}{3}}{2} \in (0,1). \]