Problem 76 - Calculus

Question

Find a nice formula for \(\int_{0}^{b}[x] d x, b>0 .\)

Step-by-Step Solution

Answer

The integral \\(\int_{0}^{b}[x] \, dx = nb - \frac{n(n+1)}{2}\\), with \\(n = \lfloor b \rfloor\\).

Step 1: Understanding the Floor Function

The floor function, denoted by \([x]\), is the greatest integer less than or equal to \((x)\). For example, for any number between 5 and 6, \([x] = 5\). This means that the function \(([x])\) will create a step-like behavior over the integration interval.

Step 2: Partitioning the Interval

Let's partition the integral \([0, b]\) into sub-intervals defined by consecutive integers: \([0, 1), [1, 2), \ldots, [n, b]\), where \([n, b]\) includes non-integer upper bound \((b)\) and \((n = \lfloor b \rfloor)\).

Step 3: Computing the Integral Over Each Interval

For each interval \([k, k+1)\), the floor function is constant and equal to \(k\). Hence, the integral over this interval is: \[ \int_{k}^{k+1}[x] \, dx = \int_{k}^{k+1} k \, dx = k \cdot (k+1 - k) = k \cdot 1 = k. \]

Step 4: Expressing the Integral from 0 to n

Sum the integrals from \([0, n)\): \[ \sum_{k=0}^{n-1} k = 0 + 1 + 2 + \ldots + (n-1). \] This is an arithmetic series, and its sum can be expressed as \(\frac{n(n-1)}{2}\).

Step 5: Calculating the Remaining Partial Interval

For the last interval \([n, b]\), where \(([x] = n)\), compute \[ \int_{n}^{b} [x] \, dx = \int_{n}^{b} n \, dx = n(b - n). \]

Step 6: Combining the Results

Combine the sums from Step 4 and Step 5 to form the complete integral: \[ \int_{0}^{b}[x] \, dx = \frac{n(n-1)}{2} + n(b-n). \] Simplifying gives: \[ \int_{0}^{b}[x] \, dx = \frac{n^2 - n}{2} + nb - n^2. \] The final simplified form is: \[ \int_{0}^{b}[x] \, dx = nb - \frac{n(n+1)}{2}. \] where \(n = \lfloor b \rfloor\).

Key Concepts

Floor Function

The floor function is a fundamental concept in mathematics, often symbolized as \([x]\). It represents the greatest integer less than or equal to a given number \(x\). For instance, if \(x\) lies between 5 and 6, \([x]\) would yield 5. Essentially, the floor function "rounds down" any non-integer value to the nearest whole number.

• Step-like Behavior: When you visualize the floor function, imagine a staircase, as it creates a step-like behavior on a graph, moving "up" at each integer value.
• Application in Definite Integrals: When integrating over an interval using the floor function, it’s important to note that it remains constant within sub-intervals between consecutive integers.

Understanding the floor function is key to analyzing problems involving definite integrals where the function value changes at integer points. It is widely used in computer science and discrete mathematics to manipulate numbers precisely when fractional components are involved.

Arithmetic Series

An arithmetic series is a sequence of numbers in which the difference between consecutive terms remains constant. It’s crucial in mathematical analysis and number theory. When dealing with the definite integral of the floor function over an interval like \([0, b]\), identifying and summating an arithmetic series comes into play.

• Finding the Sum: In the context of an integral, the arithmetic series sums up the values of the floor function across intervals. For instance, \(0 + 1 + 2 + \ldots + (n-1)\) is the series we've seen in the problem.
• Sum Formula: The sum of the first \(n\) integers is given by the formula \(\frac{n(n-1)}{2}\). This formula simplifies the process of summing up sequential numbers quickly and efficiently.

Understanding arithmetic series enables one to decompose complex integral problems into simpler, manageable parts, making it easier to find solutions.

Partitioning Intervals

Partitioning intervals is a technique used extensively in calculus, particularly when working with integrals. It involves dividing an interval into smaller segments or sub-intervals, which makes complex calculations more manageable. In our exercise, partitioning plays a pivotal role.

• Sub-intervals: The interval \([0, b]\) is partitioned into segments like \([0, 1), [1, 2), \ldots, [n, b]\). Here, \(n\) represents the integer part of \(b\), specifically \(\lfloor b \rfloor\).
• Non-integer Bounds: The last partition, \([n, b]\), highlights how to handle intervals where \(b\) isn't an exact integer, ensuring that all parts of \([0, b]\) are accounted for in the integral.
• Constant Value within Intervals: Each sub-interval maintains a constant floor function value, simplifying the integration process as each can be individually evaluated.

Understanding partitioning is fundamental for comprehending how integrals are evaluated over non-standard intervals, making intricate problems simpler and more intuitive to solve.