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Chapter 7: Problem 63

Determine whether the following statements are true and give an explanation or counterexample. a. It is possible that a computer algebra system says \(\int \frac{d x}{x(x-1)}=\ln (x-1)-\ln x\) and a table of integrals says \(\int \frac{d x}{x(x-1)}=\ln \left|\frac{x-1}{x}\right|+C.\) b. A computer algebra system working in symbolic mode could give the result \(\int_{0}^{1} x^{8} d x=\frac{1}{9},\) and a computer algebra system working in approximate (numerical) mode could give the result \(\int_{0}^{1} x^{8} d x=0.11111111\)

Short Answer

Expert verified
a) A computer algebra system can give the result of the integral \(\int \frac{1}{x(x-1)} dx = \ln \left|\frac{x-1}{x}\right| + C\), while a table of integrals might give the more general result \(-\ln |x| + \ln |x-1| + C\). b) A computer algebra system can give the definite integral \(\int_0^1 x^8 dx\) in symbolic mode as \(\frac{1}{9}\), while in numerical mode, it might provide the result 0.11111111. Answer: Yes, both statement a and statement b are true.

Step by step solution

01

Integrate the given function using partial fraction decomposition

To integrate the function \(\frac{1}{x(x-1)}\), we first perform partial fraction decomposition. We can rewrite the function as: $$\frac{1}{x(x-1)} = \frac{A}{x} + \frac{B}{x-1},$$ where A and B are constants. To find the values of A and B, follow these steps: 1. Multiply both sides by \(x(x-1)\) to clear the denominators, resulting in: $$1 = A(x-1) + Bx$$ 2. Substitute x = 0 into the equation and solve for A: $$1 = A(0-1) + B(0) \implies A = -1$$ 3. Substitute x = 1 into the equation and solve for B: $$1 = (-1)(1-1) + B(1) \implies B = 1$$ Now the partial fraction decomposition becomes: $$\frac{1}{x(x-1)} = \frac{-1}{x} + \frac{1}{x-1}$$
02

Integrate the decomposed function

Now integrate each term separately: $$\int \frac{1}{x(x-1)} dx = \int \left( -\frac{1}{x}+\frac{1}{x-1}\right) dx = -\int \frac{1}{x} dx+\int \frac{1}{x-1} dx$$ Now, \(\int \frac{1}{x} dx = \ln |x| + C_1\) and \(\int \frac{1}{x-1} dx = \ln |x-1| + C_2\). Adding the results together, we have: $$\int \frac{1}{x(x-1)} dx = -\ln |x| + \ln |x-1| + C$$ At this point, we can use the properties of logarithms to simplify the result, and we obtain: $$\int \frac{1}{x(x-1)} dx = \ln \left|\frac{x-1}{x}\right| + C$$
03

Compare the results and provide an explanation for statement a

Upon comparing the results, we see that the table of integrals gave the correct result, while the computer algebra system did not. However, the computer algebra system only differs from the true answer in that it does not take the absolute value of the quotient and didn't include the constant C. Therefore, the statement a is considered true, as it is possible that a computer algebra system gives the answer without the absolute value, but it would not be the most general answer. #Statement b: Checking the Definite Integral#
04

Compute the definite integral symbolically

Integrate the given polynomial function \(x^8\) on the interval [0, 1]: $$\int_{0}^{1} x^8 dx = \left[\frac{1}{9}x^9\right]_0^1 = \frac{1}{9}(1^9 - 0^9) = \frac{1}{9}$$ As the statement said, the result in symbolic mode is indeed \(\frac{1}{9}\).
05

Compare the numerical results and provide an explanation for statement b

Using the numerical result given in statement b, \(\int_{0}^{1} x^8 dx = 0.11111111\). This is an approximation, and when we convert the symbolic result, \(\frac{1}{9}\), into a decimal, we obtain: $$\frac{1}{9} \approx 0.11111111$$ So the statement b is also true, as a computer algebra system working in approximate mode can provide a result similar to the decimal approximation of the symbolic result, and this small error might be due to the numerical method used or the number of decimal places considered.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Fraction Decomposition
Partial fraction decomposition is a technique used to simplify the integration of rational functions, which are ratios of two polynomials. This method decomposes a single complex rational expression into simpler, easier-to-integrate terms. Here's how it works:
  • First, express the given rational function (e.g., \(\frac{1}{x(x-1)}\)) as a sum of simpler fractions like \(\frac{A}{x} + \frac{B}{x-1}\), where \(A\) and \(B\) are constants to be determined.
  • Clear the denominators by multiplying across by the original denominator function; this simplifies comparison and calculation.
  • To find \(A\) and \(B\), substitute strategic values for \(x\) to solve the resulting equations. This often involves values that nullify part of the equation, making it easier to isolate terms.
If done correctly, the process turns a complex integrand into simpler, separate fractions that can be individually integrated using basic calculus techniques. Using partial fraction decomposition can greatly simplify integral solving, especially when dealing with non-linear denominators.
Definite Integral
The definite integral is a fundamental concept in calculus representing the accumulation of quantities, such as areas under curves over a specified interval. Unlike indefinite integrals, which are represented with the constant \(C\), definite integrals result in a specific numerical value.
  • The definite integral of a function \(f(x)\) from \(a\) to \(b\) is denoted as \( \int_{a}^{b} f(x) \, dx \).
This integral essentially accumulates quantities represented by \(f(x)\) starting from \(x = a\) to \(x = b\). For example, given \(\int_{0}^{1} x^8 \, dx = \frac{1}{9}\), this calculation involves the symbolic integration of \(x^8\) and applying boundary values. This produces a numerical result representing total accumulation over the interval from 0 to 1.
Definite integrals are used extensively in various fields, including physics for calculating mass, volume, and other measurable properties.
Computer Algebra System
A Computer Algebra System (CAS) is a software program designed to perform symbolic mathematics on a computer. CAS can manipulate and solve mathematical expressions, providing numeric and symbolic solutions to various mathematical problems. Here鈥檚 how it can be especially useful:
  • It can handle complex expressions, performing operations like integration, differentiation, and algebraic simplification quickly and accurately.
  • CAS typically offers both symbolic and numerical computation modes, with symbolic computation delivering exact answers using mathematical symbols, while numerical mode provides decimal approximations.
The CAS might occasionally provide solutions that appear different from textbook answers, such as when it omits constants of integration or absolute value signs. It's essential to interpret CAS results with an understanding of the context and potential simplifications. In the exercise, the CAS omitted an absolute value and constant, but these elements are, in reality, very important for a general solution.
Approximate and Symbolic Calculation
Approximate and symbolic calculations are two modes in which mathematic computations can be performed, and they each serve different purposes.
  • **Symbolic Calculation**: This method focuses on deriving exact expressions in terms of symbols. For instance, integrating \(x^8\) symbolically from 0 to 1 gives \(\frac{1}{9}\). Symbolic calculations provide exact solutions whenever possible, maintaining mathematical equations in their algebraic form.
  • **Approximate Calculation**: This is used when a precise numerical answer is needed. Instead of focusing on the form, it outputs a decimal representation, which can be useful in real-world scenarios where approximations suffice. As seen in the solution, \(\int_{0}^{1} x^8 \approx 0.11111111\) is an approximate numerical result of \(\frac{1}{9}\).
Both types of calculations are valuable. Symbolic methods validate a theoretical framework, while approximate calculations provide practical, usable data in numerical form. Generally, even with approximate calculations, understanding the level of precision is crucial to interpreting results accurately.

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Most popular questions from this chapter

The cycloid is the curve traced by a point on the rim of a rolling wheel. Imagine a wire shaped like an inverted cycloid (see figure). A bead sliding down this wire without friction has some remarkable properties. Among all wire shapes, the cycloid is the shape that produces the fastest descent time. It can be shown that the descent time between any two points \(0 \leq a \leq b \leq \pi\) on the curve is $$\text { descent time }=\int_{a}^{b} \sqrt{\frac{1-\cos t}{g(\cos a-\cos t)}} d t$$ where \(g\) is the acceleration due to gravity, \(t=0\) corresponds to the top of the wire, and \(t=\pi\) corresponds to the lowest point on the wire. a. Find the descent time on the interval \([a, b]\) by making the substitution \(u=\cos t\) b. Show that when \(b=\pi\), the descent time is the same for all values of \(a ;\) that is, the descent time to the bottom of the wire is the same for all starting points.

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=\cos a t \longrightarrow F(s)=\frac{s}{s^{2}+a^{2}}$$

Exact Simpson's Rule Prove that Simpson's Rule is exact (no error) when approximating the definite integral of a linear function and a quadratic function.

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?

An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x}$$
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