Problem 25 - Calculus

Question

Suppose that \(f(x)=f(-x), f(x) \leq 0, g(-x)=-g(x)\), \(\int_{0}^{2} f(x) d x=-4,\) and \(\int_{0}^{2} g(x) d x=5 .\) Evaluate each integral. (a) \(\int_{-2}^{2} f(x) d x\) (b) \(\int_{-2}^{2}|f(x)| d x\) (c) \(\int_{-2}^{2} g(x) d x\) (d) \(\int_{-2}^{2}[f(x)+f(-x)] d x\) (e) \(\int_{0}^{2}[2 g(x)+3 f(x)] d x\) (f) \(\int_{-2}^{0} g(x) d x\)

Step-by-Step Solution

Answer

(a) -8, (b) 8, (c) 0, (d) -16, (e) -2, (f) -5.

Step 1: Understanding Symmetry

We begin by identifying that since \( f(x) = f(-x) \), \( f(x) \) is an even function. This means \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \). Since \( g(-x) = -g(x) \), \( g(x) \) is an odd function, which means \( \int_{-a}^{a} g(x) \, dx = 0 \).

Step 2: Calculating Integral (a)

For \( \int_{-2}^{2} f(x) \, dx \), since \( f(x) \) is even, use symmetry: \( \int_{-2}^{2} f(x) \, dx = 2 \cdot \int_{0}^{2} f(x) \, dx = 2 \cdot (-4) = -8 \).

Step 3: Evaluating Absolute Value Integral (b)

Since \( f(x) \leq 0 \), \( |f(x)| = -f(x) \). Thus, \( \int_{-2}^{2} |f(x)| \, dx = - \int_{-2}^{2} f(x) \, dx = -(-8) = 8 \).

Step 4: Odd Function Integral (c)

Since \( g(x) \) is odd, \( \int_{-2}^{2} g(x) \, dx = 0 \).

Step 5: Calculating Function Sum (d)

For \( \int_{-2}^{2} [f(x) + f(-x)] \, dx \), since \( f(x) = f(-x) \), this becomes \( \int_{-2}^{2} 2f(x) \, dx = 2 \cdot \int_{-2}^{2} f(x) \, dx = 2 \cdot (-8) = -16 \).

Step 6: Evaluating Combination Integral (e)

For \( \int_{0}^{2} [2g(x) + 3f(x)] \, dx \), split and evaluate each integral: \( 2 \cdot \int_{0}^{2} g(x) \, dx + 3 \cdot \int_{0}^{2} f(x) \, dx = 2 \cdot 5 + 3 \cdot (-4) = 10 - 12 = -2 \).

Step 7: Computing Negative Range (f)

For \( \int_{-2}^{0} g(x) \, dx \), knowing \( g(x) \) is odd, use symmetry: \( \int_{-2}^{2} g(x) \, dx = \int_{-2}^{0} g(x) \, dx + \int_{0}^{2} g(x) \, dx = 0 \). Thus, \( \int_{-2}^{0} g(x) \, dx = -\int_{0}^{2} g(x) \, dx = -5 \).

Key Concepts

Symmetry in Functions

Symmetry in functions is an important concept in calculus as it allows us to simplify complex integrals by understanding the behavior of the function over a specific interval. Symmetry can be categorized into even or odd functions. Here's a look at each.

• An **even function** has reflective symmetry around the y-axis. If you fold the graph along the y-axis, the two sides will match perfectly. Mathematically, this means that for an even function, \(f(x) = f(-x)\).
• An **odd function** has rotational symmetry around the origin. If you rotate the graph 180 degrees about the origin, it looks the same. For an odd function, \(g(-x) = -g(x)\).

Understanding symmetry helps simplify the computation of definite integrals over symmetric intervals. For example, the integral of an even function over a symmetric interval about the y-axis can be doubled from zero to the endpoint, while the integral of an odd function over a symmetric interval results in zero. These properties lead to quick solutions to integral problems and highlight the beauty of mathematical symmetry.

Even and Odd Functions

Even and odd functions play a key role in determining the behavior of integrals especially when evaluating them over symmetric intervals.

**Even Functions**:
For even functions, the property that \(f(x) = f(-x)\) implies that the integral over a symmetric interval such as \([-a, a]\) is simply twice the integral from \[0\] to \[a\]. For example, \(\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx\). This property helps simplify calculations because you only need to compute the integral over half the domain.

**Odd Functions**:
Odd functions, where \(g(-x) = -g(x)\), exhibit a different symmetry. When integrated over a symmetric interval, the result is zero: \(\int_{-a}^{a} g(x) \, dx = 0\). This is due to how the negative and positive sides of the function cancel each other out over the symmetric interval. Identifying an odd function quickly leads to easy computation of the integral, saving time and effort.
In practice, determining if a function is odd or even is crucial for evaluating integrals efficiently and such insights can lead to significantly simpler calculations.

Definite Integrals

The definite integral is a fundamental concept in calculus that is used to find the area under a curve from one point to another on the x-axis. This calculation involves integrating a function within specific bounds, unlike indefinite integrals which focus on finding an antiderivative without limits.

• The **definite integral** \(\int_{a}^{b} f(x) \, dx\) represents the net area between the x-axis and the function \(f(x)\) from \(x = a\) to \(x = b\).
• For functions that lie entirely above the x-axis in the interval \([a, b]\), the definite integral gives the area directly. If a function dips below the x-axis, the integral computes the net area, which may include negative values.

Understanding definite integrals and their properties helps solve problems in physics, engineering, and beyond. By applying properties like summing areas of even functions or canceling areas of odd functions, we can evaluate the integral efficiently. This integral offers critical insights into the function's behavior over a specific interval.