Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

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) \(\int_{-3}^{3}[x] d x\) (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) 0, (g) 2, (h) \(\frac{5}{3}\)

Step 1: Understanding the Floor Function

The floor function \([x]\) denotes the greatest integer less than or equal to \(x\). For example, \([x] = -1\) for \(-1 \leq x < 0\), \([x] = 0\) for \(0 \leq x < 1\), and so on. We need to calculate integrals involving \([x]\) and other expressions.

Step 2: Integral (a) \\( \\int_{-3}^{3}[x] \, dx \\)

Break the integral \(\int_{-3}^{3}[x] \, dx\) into sections where \([x]\) is constant over each interval: - \([-3,-2)\), \([-2,-1)\), \([-1,0)\), \([0,1)\), \([1,2)\), \([2,3)\).For example, on \([-3,-2)\), \([x]=-3\), and the length of each interval is 1. We have six intervals, each with steps of 1 and a height corresponding to the \([x]\) term.Calculate the integral over each interval and sum them: - \([-3,-2)\): \(-3 \times 1 = -3\)- \([-2,-1)\): \(-2 \times 1 = -2\)- \([-1,0)\): \(-1 \times 1 = -1\)- \([0,1)\): \(0 \times 1 = 0\) - \([1,2)\): \(1 \times 1 = 1\) - \([2,3)\): \(2 \times 1 = 2\)Summing: \(-3 + (-2) + (-1) + 0 + 1 + 2 = -3\).

Step 3: Integral (b) \\( \\int_{-3}^{3}[x]^{2} \, dx \\)

Use the same intervals as above:- \([-3,-2)\): \((-3)^2 \times 1 = 9\)- \([-2,-1)\): \((-2)^2 \times 1 = 4\)- \([-1,0)\): \((-1)^2 \times 1 = 1\)- \([0,1)\): \(0^2 \times 1 = 0\)- \([1,2)\): \(1^2 \times 1 = 1\)- \([2,3)\): \(2^2 \times 1 = 4\)Sum them: \(9 + 4 + 1 + 0 + 1 + 4 = 19\).

Step 4: Integral (c) \\( \\int_{-3}^{3}(x-[x]) \, dx \\)

Notice that \(x - [x]\) is the fractional part of \(x\), which ranges from 0 to just below 1 over each unit interval. Therefore, integrating over intervals from \(-3\) to \(3\):- The function \(x - [x]\) on any interval like \([-3, -2)\) or \([2, 3)\) contributes 0.5 (since its average over this unit size interval is 0.5).- This amounts to \(0.5 \times 6\) (there are 6 unit intervals between \(-3\) and \(3\)).Thus, the integral equals 3.

Step 5: Integral (d) \\( \\int_{-3}^{3}(x-[x])^{2} \, dx \\)

Evaluate \(x - [x]\) over \([-3,-2)\), \([-2,-1)\), \([-1,0)\), \([0,1)\), \([1,2)\), \([2,3)\) where \(x-[x]\) ranges from 0 to 1. - \((x - [x])^2\) curve over each unit interval contributes \(\frac{1}{3}\) as this is the integral of \(x^2\) from 0 to 1.- Since \(0 \leq (x - [x]) < 1\) always repeats 6 times (the intervals are regular), the integral equals \(\frac{6}{3} = 2\).

Step 6: Integral (e) \\( \\int_{-3}^{3}|x| \, dx \\)

Since \(|x|\) is symmetric, consider only \([0,3]\) portion and double the result:- From \(-3\) to \(0\), the contribution is \(\frac{3^2}{2} = 4.5\)- From \(0\) to \(3\), the contribution is also \(\frac{3^2}{2} = 4.5\)Hence, doubled gives 9.

Step 7: Integral (f) \\( \\int_{-3}^{3} x|x| \, dx \\)

Notice \(x|x|\) is symmetric and odd over the range.- Because of symmetry and being odd \((-x|x|) = - (x|x|)\), the integral becomes 0 over \([-3, 3]\).

Step 8: Integral (g) \\( \\int_{-1}^{2}|x|[x] \, dx \\)

Split the integral from \([-1,0)\) and \([0,1)\), \([1,2)\):- From \([-1,0)\): \(|x| = -x\), \([x] = -1\), hence the integral contributes 0.5- From \([0,1)\): \(|x| = x\), \([x] = 0\), no contribution- From \([1,2)\): \(|x| = x\), \([x] = 1\), contributes 1.5Summing: \(0.5 + 0 + 1.5 = 2\).

Step 9: Integral (h) \\( \\int_{-1}^{2} x^{2}[x] \, dx \\)

Focus on pieces \([-1,0)\), \([0,1)\), and \([1,2)\):- On \([-1,0)\), \(x^2 \times (-1)\), contributes \(-\frac{1}{3}\) (as per the formula given).- On \([0,1)\), \(x^2 \times 0 = 0\), so nothing.- On \([1,2)\), \(x^2 \times 1\), contributes \(\frac{7}{3} - \frac{1}{3} = 2\)Summing: \(-\frac{1}{3} + 0 + 2 = \frac{5}{3}\).

Key Concepts

floor function

The floor function, often denoted as \([x]\), finds the greatest integer less than or equal to a real number \(x\). It acts like a downward step effect, where it reduces any decimal number to its integer part. For example:

• If \(x = 2.9\), then \([x] = 2\).
• If \(x = -1.1\), then \([x] = -2\).

This function is crucial in integral calculus for problems involving discontinuities. When calculating definite integrals involving \([x]\), it's often necessary to break the integral into separate intervals where the function remains constant.
For example, for the integral \( \int_{-3}^{3} [x] \, dx \), we calculate the area over each interval where \([x]\) remains the same value. Each interval spans one unit, making calculations straightforward by simply multiplying the floor value by the interval length (usually 1). This highlights how the floor function can create piecewise constant functions across its range, making it easier to calculate the integral by summing contributions from each part.

definite integrals

Definite integrals measure the area under a curve between two points. In calculus, they are generally denoted as \( \int_{a}^{b} f(x) \, dx \), where \(a\) and \(b\) are the limits of integration and \(f(x)\) is the function being integrated. They can help find:

• The total accumulation of a quantity.
• The net area between the x-axis and the curve.

When dealing with definite integrals, it's important to understand the behavior of the function over the interval. This can involve splitting the integral across subintervals where the behavior of the function makes the calculation simpler. For example, \[\int_{-3}^{3} |x| \, dx \]can be tackled by recognizing the absolute value function behaves differently on negative and positive domains. The area tends to be symmetric, allowing for clever calculations, like doubling the positive computation to cover both sides of the zero axis.
Moreover, for functions with floor values or piecewise definitions, breaking them into constant segments allows summing areas of simpler, often rectangular shapes, providing a practical way to integrate complex functions.

piecewise functions

Piecewise functions are defined by different expressions over various intervals of their domain. They essentially "piece" together multiple functions, allowing a complex overall function to behave differently across varying regions. This is especially useful when a function has abrupt changes or discontinuities.
For example, the floor function in the integral calculus exercise is itself a piecewise function, as it maintains constant integer values over different intervals. To integrate a piecewise function:

• First identify each piece where the function remains consistent.
• Calculate the integral over each interval separately.
• Sum the results to get the total value of the integral.

This approach is essential as it ensures accuracy by addressing how the function changes at each piece's boundaries. Such a strategy might seem laborious at first, but it prevents errors arising from ignoring function behavior changes within the integration limits.
Piecewise functions embody the adaptability of functions in modeling real-world situations where continuity is frequently disrupted, making them a staple of mathematical analysis.