Problem 14 - Calculus

Question

Suppose that \(\int_{0}^{1} f(x) d x=2, \int_{1}^{2} f(x) d x=3, \int_{0}^{1} g(x) d x=-1\)and \(\int_{0}^{2} g(x) d x=4\). Use properties of definite integrals (linearity, interval additivity, and so on) to calculate each of the integrals. $$ \int_{1}^{1}[3 f(x)+2 g(x)] d x $$

Step-by-Step Solution

Answer

The integral evaluates to 0 because it is over a zero-length interval.

Step 1: Identify the Integral to Evaluate

We need to evaluate the integral \( \int_{1}^{1}[3f(x)+2g(x)]\,dx \). This is the integral of a function evaluated over a zero-length interval, from 1 to 1.

Step 2: Apply the Property of Zero-Length Interval

When integrating over an interval where the upper and lower limits are the same \( (a = b) \), the integral evaluates to zero. Mathematically, \( \int_{a}^{a} h(x) dx = 0 \).

Step 3: Conclude the Integral Value

Since the interval of integration is zero-length (from 1 to 1), the integral \( \int_{1}^{1} [3f(x) + 2g(x)] \,dx \) evaluates directly to zero.

Key Concepts

Linearity of Integrals

One of the core principles of definite integrals is their linearity. This means we can distribute an integral over the sum of functions and factor out constants. For example, suppose we have a function of the form \(c_1 \cdot f(x) + c_2 \cdot g(x)\). When we take the integral over this combination, we can split it like this:

• \( \int [c_1 \cdot f(x) + c_2 \cdot g(x)] \, dx = c_1 \int f(x) \, dx + c_2 \int g(x) \, dx \)

This property makes complex integrals easier to handle by breaking them into simpler parts. We can tackle each small part separately and recombine them at the end. This is especially useful when we already know the integral values of individual functions over the desired intervals.
Linearity helps in computations significantly by using known integral values, simplifying calculations, and verifying results easily.

Properties of Integrals

Definite integrals have several important properties that make them powerful and versatile.

• Additivity over Intervals: For a continuous function \(f(x)\), if \(a \leq b \leq c\), then \( \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \). This property lets us break down complicated integration over a large interval into manageable parts.
• Linearity: As discussed, scaling and adding functions are easy due to the linearity of integrals.
• Zero-Length Interval: A very useful property which states that the integral of any function from a to a is zero, i.e., \( \int_{a}^{a} h(x) \, dx = 0 \). This simplifies our computations when the limits are equal.
• Reversal of Limits: \( \int_{a}^{b} f(x) \, dx = - \int_{b}^{a} f(x) \, dx \). Simply put, reversing the limits changes the sign of the integral.

These properties simplify the complexity associated with integrals, providing tools to evaluate integrals efficiently.

Zero-Length Intervals

One practical property of definite integrals is the zero-length interval rule. When the upper and lower bounds of an interval are the same, the integral is zero. This is stated as:

• \( \int_{a}^{a} h(x) \, dx = 0 \)

The core idea here is that there is no "area" under the curve between two identical points. Therefore, regardless of the function \(h(x)\), the integral is zero.
In exercises and problems, identifying such intervals can greatly simplify the solution. For instance, as shown in the original exercise, the integral \( \int_{1}^{1} [3f(x) + 2g(x)] \, dx \) is evaluated immediately to zero due to its zero-length interval. This highlights how certain properties, although simple, have a big impact on solving mathematical challenges efficiently.