Question

Show that $$ \int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x $$ for all possible relative orderings of \(a, b,\) and \(c,\) provided that \(f\) is integrable on a closed interval containing them.

Short answer

Question: Prove that for any integrable function f, and for any real numbers a, b, and c: $$ \int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_{b}^{c} f(x) dx $$

Step-by-step solution

Case 1: a ≤ b ≤ c

For this case, we have \(\int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x\) directly.

Case 2: a ≤ c ≤ b

We can write it as \(\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\) = \(\int_{a}^{b} f(x) d x\). By rearranging, we get \(\int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\).

Case 3: b ≤ a ≤ c

We have \(\int_{a}^{c} f(x) d x=\int_{b}^{a} f(x) d x+\int_{a}^{c} f(x) d x\). Since \(\int_{b}^{a} f(x) d x = -\int_{a}^{b} f(x) d x\), we get \(\int_{a}^{c} f(x) d x = \int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x\).

Case 4: b ≤ c ≤ a

We have \(\int_{a}^{c} f(x) d x = -\int_{c}^{a} f(x) d x = -\int_{c}^{b} f(x) d x - \int_{b}^{a} f(x) d x\). Swapping a and c and rearranging, we get \(\int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x\).

Case 5: c ≤ a ≤ b

This case is equivalent to Case 3 when we swap a and c.

Case 6: c ≤ b ≤ a

This case is equivalent to Case 1 when we swap a and c. The property is now proven for all possible relative orderings of a, b, and c, and we have shown that $$ \int_{a}^{c} f(x) d x=\int_{a}^{b} f(x) d x+\int_{b}^{c} f(x) d x $$ for all possible relative orderings of \(a, b,\) and \(c,\) provided that \(f\) is integrable on a closed interval containing them.

Key concepts

Definite Integrals

A definite integral is a fascinating concept with a wealth of applications in mathematics and beyond. When we talk about definite integrals, we're referring to the integral of a function over a specific interval. This interval is usually denoted by \[ \int_{a}^{b} f(x) \, dx \] Here, the numbers \( a \) and \( b \) are the lower and upper limits of integration, and \( f(x) \) is the function we're integrating. The result of this operation gives us the net area under the curve \( f(x) \) from \( a \) to \( b \).

• Think of it like measuring the shade under a curve starting at \( x = a \) and ending at \( x = b \).
• This is different from an indefinite integral, which focuses on finding a general formula for all possible integrals of \( f(x) \).

Definite integrals are crucial because they provide precise measurements which can be applied in various fields such as physics, engineering, and economics. They're not limited to positive areas; they can account for negative areas too, which they subtract from the total. The examples you see in your textbooks, like the exercise we discussed, often explore how definite integrals behave across different intervals. By examining these properties, you gain a better understanding of how integrals can be useful in real-world calculations.

Integrable Functions

When we talk about integrable functions, we're focusing on functions that can be integrated over a specific interval. For a function \( f(x) \) to be integrable on an interval \([a, b]\), it must meet certain requirements. Primarily, it should be bounded (not going towards infinity) and should not have too many discontinuities.

• Bounded means there are real numbers that the function values don't exceed across the interval.
• For a function to be truly integrable in the context of Riemann integrals, any discontinuities must not add up to something too overwhelming.

Riemann integrability, which is what most high school and early college math courses focus on, requires that the function has a finite number of jump discontinuities. This means you can plot the function without encountering too much trouble with undefined behavior or endless stretches up or down a y-axis. This concept is important in exercises like the one you've seen, where we calculate integrals over intervals that might involve switching the start and end points. Even when we have complicated intervals or functions made up of several smaller pieces, as long as our function remains integrable, then we can apply the rules of integration seamlessly.

Properties of Integrals

Understanding the properties of integrals is key to unlocking their full potential. Integral properties make it easy to calculate complex expressions by breaking them down into simpler parts. The primary property discussed in the exercise is additivity over intervals:\[ \int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx \]This principle allows us to split an integral into parts, essentially summing up areas under a curve for different sections. Remarkably, this holds true regardless of the order of \( a, b, \) and \( c \). Whether \( a \) comes before \( b \) or is smaller than \( c \), the property remains valid.

• Properties like linearity mean you can break up an integral with additions and subtractions, making complex problems manageable.
• If you change the function's order, you swap signs, another important trait.

Each property is a valuable tool in simplifying and solving integration problems across various scenarios. By mastering these properties, you're better equipped to understand how integrals work and apply them efficiently in real-world contexts.