Problem 31 - Calculus

Question

Recall that \([x]\) denotes the greatest integer less than or equal to \(x\). Calculate each of the following integrals. You may use geometric reasoning and the fact that \(\int_{0}^{b} x^{2} d x=b^{3} / 3 .\) (The latter is shown in Problem 34.) (a) \(\int_{-3}^{3}[x \rrbracket d x\) (b) \(\int_{-3}^{3}[x]^{2} d x\) (c) \(\int_{-3}^{3}(x-[x]) d x\) (d) \(\int_{-3}^{3}(x-[x])^{2} d x\) (e) \(\int_{-3}^{3}|x| d x\) (f) \(\int_{-3}^{3} x|x| d x\) (g) \(\int_{-1}^{2}|x|[x] d x\) (h) \(\int_{-1}^{2} x^{2}[x] d x\)

Step-by-Step Solution

Answer

(a) -3, (b) 19, (c) 3, (d) 2, (e) 9, (f) 18, (g) 0, (h) 8/3.

Step 1: Understanding the Greatest Integer Function

The greatest integer function \[x\rrbracket\] is known as the floor function. It maps a real number to the largest integer less than or equal to it. For example, \[3.7\rrbracket = 3\], \[-2.3\rrbracket = -3\]. This understanding is crucial for breaking the integral into segments where \[x\rrbracket\] is constant.

Step 2: Solve Integral (a)

For \int_{-3}^{3}[x] d x\, observe that the floor function \[[x]\] changes at integer boundaries. Split the integral at each integer from -3 to 3: \[\int_{-3}^{3}[x] \, dx = \sum_{n=-3}^{2} \int_{n}^{n+1} n \, dx = \sum_{n=-3}^{2} n\cdot(1) = (-3) + (-2) + (-1) + 0 + 1 + 2 = -3.\] Hence, \(\int_{-3}^{3}[x] \, dx = -3\).

Step 3: Solve Integral (b)

For \int_{-3}^{3}[x]^2 \, dx\, evaluate similarly: \[\int_{-3}^{3}[x]^2 \cdot dx = \sum_{n=-3}^{2} \int_{n}^{n+1} n^2 \, dx = \sum_{n=-3}^{2} n^2 = 9 + 4 + 1 + 0 + 1 + 4 = 19.\] Consequently, \(\int_{-3}^{3}[x]^2 \, dx = 19\).

Step 4: Solve Integral (c)

For \int_{-3}^{3}(x - [x]) \, dx\, compute the integral over each interval \( n, n+1 \): \[\int_{n}^{n+1} (x-n) \, dx = \left[ \frac{(x-n)^2}{2} \right]_n^{n+1} = \frac{1}{2}.\] Summing from -3 to 2 gives: \[6 \cdot \frac{1}{2} = 3.\] Thus, \(\int_{-3}^{3}(x - \llbracket x \rrbracket) \, dx = 3\).

Step 5: Solve Integral (d)

For \int_{-3}^{3}(x-[x])^2 \, dx\, calculate \[\int_{n}^{n+1} (x-n)^2 \, dx = \left[ \frac{(x-n)^3}{3} \right]_n^{n+1} = \frac{1}{3}\] for each interval \(-3, 2\). Therefore, \[6 \cdot \frac{1}{3} = 2.\] So, \(\int_{-3}^{3}(x - [x])^2 \, dx = 2\).

Step 6: Solve Integral (e)

For \int_{-3}^{3}|x| \, dx\, utilize symmetry: \[2 \cdot \int_{0}^{3} x \, dx = 2 \cdot \left[ \frac{x^2}{2} \right]_0^3 = 2 \cdot \frac{9}{2} = 9.\] Hence, \(\int_{-3}^{3}|x| \, dx = 9\).

Step 7: Solve Integral (f)

For \int_{-3}^{3} x|x| \, dx\, consider the regions where \(x\) is negative and positive. Split: \[\int_{-3}^{0} x^2 \, dx + \int_{0}^{3} x^2 \, dx.\] Compute each: \[2 \cdot \int_{0}^{3} x^2 \, dx = 2 \cdot \left[ \frac{x^3}{3} \right]_0^3 = 2 \cdot 9 = 18.\] Thus, \(\int_{-3}^{3} x|x| \, dx = 18\).

Step 8: Solve Integral (g)

Calculate \[\int_{-1}^{2} |x|[x] \, dx\]. First, break down to segments: from -1 to 0; from 0 to 1; and from 1 to 2. Solve these: \[|x| = -x \text{ for } -1

Step 9: Solve Integral (h)

For \int_{-1}^{2} x^2[x] \, dx\, evaluate in segments: - From -1 to 0: \[\int_{-1}^{0} x^2 [x] \, dx = 0;\] - From 0 to 1: \[\int_{0}^{1} x^2 \, dx = \frac{1}{3};\]- From 1 to 2: \[\int_{1}^{2} x^2 \cdot 1 \, dx = \frac{7}{3};\] The sum: \(0 + \frac{1}{3} + \frac{7}{3} = \frac{8}{3}.\)

Key Concepts

Greatest Integer Function

The Greatest Integer Function is a mathematical function that rounds down a real number to its nearest integer. It is also often referred to as the floor function. For a given real number \(x\), the greatest integer function, denoted as \(\lfloor x \rfloor\), returns the largest integer less than or equal to \(x\).

For example:

• If \(x = 3.7\), then \(\lfloor 3.7 \rfloor = 3\) because 3 is the largest integer less than 3.7.
• If \(x = -2.3\), then \(\lfloor -2.3 \rfloor = -3\) because -3 is the largest integer less than -2.3.

Understanding this function is essential, especially when evaluating integrals that involve \(\lfloor x \rfloor\) because the value of \(\lfloor x \rfloor\) stays constant over certain intervals, specifically between consecutive integers. This property allows you to break down integrals into manageable segments where the function's value doesn't change abruptly.

Floor Function

The floor function, usually represented as \(\lfloor x \rfloor\), and the greatest integer function are essentially the same concept. These terms can be used interchangeably.

Characteristics of the floor function include:

• It maps any real number to an integer value.
• The output is always less than or equal to the original number.
• The function is constant between any two consecutive integers.

For example, within the interval between 1 and 2, the floor of any number will be 1. This trait simplifies the evaluation of many integrals, as it turns complex functions into piecewise constant ones, at least over each unit interval. Understanding its behavior over different intervals is key when applying it in definite integrals.

Definite Integrals

Definite integrals enable us to calculate the area under a curve between two points on the x-axis. The notation \(\int_{a}^{b} f(x) \, dx\) is used to represent the definite integral of a function \(f(x)\) from \(x = a\) to \(x = b\).

Some key properties of definite integrals include:

• The integral \(\int_{a}^{b} f(x) \, dx\) can be split into multiple integrals across smaller intervals.
• If \(f(x)\) is constant across an interval, the integral calculation becomes straightforward: simply multiply the constant by the width of the interval.
• Definite integrals can be solved using geometric approaches by interpreting certain functions into areas and using symmetry when applicable.

These properties are useful for solving integrals involving step functions or ones that change characteristics at integer points, like those containing the floor function, as it requires splitting the integral at each integer boundary to take into account the constant function values.

Geometric Reasoning

Geometric reasoning involves leveraging the geometry of functions to simplify integral calculations. This approach is exceptionally useful in integral calculus when dealing with functions that can be represented as shapes of known area formulas, such as triangles, rectangles, or trapezoids.

For instance:

• The area under a constant function over an interval resembles a rectangle, facilitating straightforward area calculations.
• Symmetric functions around the y-axis can often simplify calculations by computing the integral over half of the interval and then doubling the result.

By recognizing these patterns, especially with floor and step functions, we avoid lengthy algebraic computations by reducing the problem to simple geometric shapes. This strategy not only economizes time and effort but also deepens our understanding of how integrals encapsulate areas under curves.

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Problem 30

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Problem 31

Other exercises in this chapter

Problem 30

Use the method of substitution to find each of the following indefinite integrals. $$ \int x^{6}\left(7 x^{7}+\pi\right)^{8} \sin \left[\left(7 x^{7}+\pi\right)

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Problem 30

30\. On her way to work, Teri noted her speed every 3 minutes. The results are shown in the table below. How far did she drive? \(\begin{array}{l}\text { Time }

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Problem 31

Show that if \(f(x)=\int_{2 x}^{5 x} \frac{1}{t} d t,\) then \(f\) is a constant function on \((0, \infty)\)

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Problem 31

Find the interval \((s)\) on which the graph of \(y=f(x), x \geq 0,\) is (a) increasing, and (b) concave up. $$ f(x)=\int_{1}^{x} \frac{1}{\theta} d \theta $$

View solution

Key Concepts

Greatest Integer Function

The Greatest Integer Function is a mathematical function that rounds down a real number to its nearest integer. It is also often referred to as the floor function. For a given real number \(x\), the greatest integer function, denoted as \(\lfloor x \rfloor\), returns the largest integer less than or equal to \(x\).

For example:

• If \(x = 3.7\), then \(\lfloor 3.7 \rfloor = 3\) because 3 is the largest integer less than 3.7.
• If \(x = -2.3\), then \(\lfloor -2.3 \rfloor = -3\) because -3 is the largest integer less than -2.3.

Understanding this function is essential, especially when evaluating integrals that involve \(\lfloor x \rfloor\) because the value of \(\lfloor x \rfloor\) stays constant over certain intervals, specifically between consecutive integers. This property allows you to break down integrals into manageable segments where the function's value doesn't change abruptly.

Floor Function

The floor function, usually represented as \(\lfloor x \rfloor\), and the greatest integer function are essentially the same concept. These terms can be used interchangeably.

Characteristics of the floor function include:

• It maps any real number to an integer value.
• The output is always less than or equal to the original number.
• The function is constant between any two consecutive integers.

For example, within the interval between 1 and 2, the floor of any number will be 1. This trait simplifies the evaluation of many integrals, as it turns complex functions into piecewise constant ones, at least over each unit interval. Understanding its behavior over different intervals is key when applying it in definite integrals.

Definite Integrals

Definite integrals enable us to calculate the area under a curve between two points on the x-axis. The notation \(\int_{a}^{b} f(x) \, dx\) is used to represent the definite integral of a function \(f(x)\) from \(x = a\) to \(x = b\).

Some key properties of definite integrals include:

• The integral \(\int_{a}^{b} f(x) \, dx\) can be split into multiple integrals across smaller intervals.
• If \(f(x)\) is constant across an interval, the integral calculation becomes straightforward: simply multiply the constant by the width of the interval.
• Definite integrals can be solved using geometric approaches by interpreting certain functions into areas and using symmetry when applicable.

These properties are useful for solving integrals involving step functions or ones that change characteristics at integer points, like those containing the floor function, as it requires splitting the integral at each integer boundary to take into account the constant function values.

Geometric Reasoning

Geometric reasoning involves leveraging the geometry of functions to simplify integral calculations. This approach is exceptionally useful in integral calculus when dealing with functions that can be represented as shapes of known area formulas, such as triangles, rectangles, or trapezoids.

For instance:

• The area under a constant function over an interval resembles a rectangle, facilitating straightforward area calculations.
• Symmetric functions around the y-axis can often simplify calculations by computing the integral over half of the interval and then doubling the result.

By recognizing these patterns, especially with floor and step functions, we avoid lengthy algebraic computations by reducing the problem to simple geometric shapes. This strategy not only economizes time and effort but also deepens our understanding of how integrals encapsulate areas under curves.