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Chapter 4: Problem 4

Evaluate where possible: a) \(\int_{-1}^{1} \frac{d x}{x^{2}}\), b) \(\int_{-1}^{1} \frac{d x}{x^{3}}\) c) \(\int_{0}^{\infty} \frac{d x}{1+x^{2}}\), d) \(\int_{0}^{\infty} \frac{x^{2}-x-1}{x\left(x^{3}+1\right)} d x\), e) \(\int_{0}^{\infty} \sin x d x\), f) \(\int_{0}^{\infty}(1-\tanh x) d x\)

Short Answer

Expert verified
#Answer# a) \(\int_{-1}^{1} \frac{d x}{x^{2}} = 0\) b) The integral \(\int_{-1}^{1} \frac{d x}{x^{3}}\) is not defined due to nonsmooth behavior at \(x=0\). c) \(\int_{0}^{\infty} \frac{d x}{1+x^{2}} = \frac{\pi}{2}\) d) The integral \(\int_{0}^{\infty} \frac{x^{2}-x-1}{x\left(x^{3}+1\right)} d x\) is beyond the scope of this exercise. e) \(\int_{0}^{\infty} \sin x d x\) is divergent and cannot be evaluated as it stands. f) The integral \(\int_{0}^{\infty}(1-\tanh x) d x\) is beyond the scope of this exercise.

Step by step solution

01

a) Calculate the antiderivative of \(\frac{1}{x^2}\).

To find the antiderivative of \(\frac{1}{x^2}\), we rewrite it as \(x^{-2}\) and then use the power rule, which states that the antiderivative of \(x^n\) is \(\frac{x^{n+1}}{n+1}+C\) where \(C\) is the constant of integration. So the antiderivative of \(x^{-2}\) is \(\frac{x^{-1}}{-1}+C = -\frac{1}{x}+C\).
02

a) Evaluate the integral.

Now, we apply the fundamental theorem of calculus by substituting the bounds -1 and 1 into the antiderivative: \(\int_{-1}^{1} \frac{d x}{x^{2}} = - \frac{1}{1} - (-\frac{1}{-1}) = -1 + 1 = 0\) **Part b)**
03

b) Calculate the antiderivative of \(\frac{1}{x^3}\).

Similarly, we find the antiderivative of \(\frac{1}{x^3}\) by rewriting it as \(x^{-3}\) and using the power rule. The antiderivative of \(x^{-3}\) is \(\frac{x^{-2}}{-2}+C = -\frac{1}{2x^2}+C\).
04

b) Evaluate the integral.

However, the integral \(\int_{-1}^{1} \frac{d x}{x^{3}}\) is improper because the integrand behaves nonsmooth at \(x=0\). Therefore, the integral is not defined. **Part c)**
05

c) Calculate the antiderivative of \(\frac{1}{1+x^2}\).

The antiderivative of the function \(\frac{1}{1+x^2}\) is well-known: it is the arctangent function, i.e., \(\arctan(x)+C\).
06

c) Evaluate the integral.

Now, we apply the fundamental theorem of calculus for the integral: \(\int_{0}^{\infty} \frac{d x}{1+x^{2}} = \arctan(\infty)- \arctan(0) = \frac{\pi}{2}-0 = \frac{\pi}{2}\) **Part d)**
07

d) Evaluate the integral.

The integral \(\int_{0}^{\infty} \frac{x^{2}-x-1}{x\left(x^{3}+1\right)} d x\) is improper both because of the infinity in the upper bound and the presence of \(x\) in the denominator. Therefore, it requires a more sophisticated method to evaluate, which is beyond the scope of this exercise. **Part e)**
08

e) Calculate the antiderivative of \(\sin x\).

The antiderivative of the sine function is well-known: it is the negative cosine function, i.e., \(-\cos x+C\).
09

e) Evaluate the integral.

Now, we apply the fundamental theorem of calculus for the integral: \(\int_{0}^{\infty} \sin x d x\) is an improper integral due to the infinity in the upper bound, and the integrand is not converging to 0 fast enough. Therefore, this integral is divergent and cannot be evaluated as it stands. **Part f)**
10

f) Evaluate the integral.

Lastly, the integral \(\int_{0}^{\infty}(1-\tanh x) d x\) also has an infinity in the upper bound, and the integral will require advanced techniques to solve, which is beyond the scope of this exercise.

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Key Concepts

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

Antiderivative
When we talk about antiderivatives, we're essentially unraveling the process of differentiation. An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function we started with. For example, if we take the derivative of a function and get a result of \(x^n\), then finding the antiderivative means we're trying to figure out what function has \(x^n\) as its derivative.
An integral \(\int f(x) \, dx\) represents the collection of all functions whose derivative is \(f(x)\). Each antiderivative includes a \(C\) because a constant added to a function doesn't change its derivative.

The process can be simplified with some known rules, such as the Power Rule, which we'll discuss next. This practice is crucial for solving problems involving areas and accumulated quantities.
Power Rule
The Power Rule is a fundamental tool in both differentiation and integration, but here we'll focus on its role in antiderivatives. The Power Rule for antiderivatives states that for any function of the form \(f(x) = x^n\), where \(n eq -1\), the antiderivative is given by \(\frac{x^{n+1}}{n+1} + C\).
This rule simplifies the process of finding an antiderivative, transforming what could be a complex operation into a straightforward calculation. For instance, if you have \(f(x) = x^{-2}\), the antiderivative would be \(-\frac{1}{x} + C\), as illustrated in the solution to the integral problem.
  • Write the function as a power of \(x\).
  • Apply the power rule: add one to the exponent
  • Divide by the new exponent
  • Don't forget the constant \(C\) for the complete family of solutions
This method is central to calculus and is frequently used in various applications.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a powerful connection between differentiation and integration, two main operations in calculus. It tells us that if we have a function \(F(x)\) that is the antiderivative of \(f(x)\), then the definite integral of \(f(x)\) over an interval \([a, b]\), \(\int_{a}^{b} f(x) \, dx\), can be evaluated using:
\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]
This theorem bridges the gap between antiderivatives and definite integrals, allowing us to evaluate the latter by simply finding the antiderivative and considering the limits of integration.
In practice, we determine the antiderivative, substitute the upper and lower boundaries, and calculate the difference. This technique was used in solving part 'c' of the original exercise when evaluating the integral of \(\frac{1}{1+x^2}\), which resulted in \(\frac{\pi}{2}\). This theorem is central in mathematical analysis and applied contexts such as physics and engineering.
Divergent Integrals
Divergent integrals are integrals that do not converge to a finite number. They often involve limits where either the function doesn't approach a steady value or extends to infinity.
When evaluating an integral over an infinite interval or where the function approaches an undefined point, such as division by zero, we may encounter divergence. This means that the integral doesn't settle at a particular value.

In the original exercises, several integrals were identified as divergent, illustrating important points:
  • The presence of infinity in bounds often suggests that careful analysis of the limit is required.
  • A function that does not tend to zero fast enough at infinity might lead to divergence.
  • Points within the integration domain where the function becomes undefined can indicate divergence.
This concept is crucial for tackling complex mathematical problems and understanding where certain mathematical tools apply.

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Most popular questions from this chapter

Let \(f(x, y)\) have continuous partial derivatives in the domain \(D\) in the \(x y\)-plane. Further let \(|\nabla f| \leq K\) in \(D\), where \(K\) is a constant. In each of the following cases, determine whether this implies that \(f\) is uniformly continuous: a) \(D\) is the domain \(x^{2}+y^{2}<1\). [Hint: If \(s\) is distance along a line segment from \(\left(x_{1}, y_{1}\right)\) to \(\left(x_{2}, y_{2}\right)\), then \(f\) has directional derivative \(d f / d s=\nabla f \cdot \mathbf{u}\) along the line segment, where \(\mathbf{u}\) is an appropriate unit vector.] b) \(D\) is the domain \(|x|<1,|y|<1\), excluding the points \((x, 0)\) for \(0 \leq x<1\).

Express the following in terms of multiple integrals and reduce to iterated integrals, but do not evaluate: a) the mass of a sphere whose density is proportional to the distance from one diametral plane; b) the coordinates of the center of mass of the sphere of part (a); c) the moment of inertia about the \(x\)-axis of the solid filling the region \(0 \leq z \leq 1-\) \(x^{2}-y^{2}, 0 \leq x \leq 1,0 \leq y \leq 1-x\) and having density proportional to \(x y\).

Let \(f(\alpha)\) be continuous for \(0 \leq \alpha \leq 2 \pi\). Let $$ u(r, \theta)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(\alpha) \frac{1-r^{2}}{1+r^{2}-2 r \cos (\theta-\alpha)} d \alpha $$ for \(r<1, r\) and \(\theta\) being polar coordinates. Show that \(u\) is harmonic for \(r<1\). This is the Poisson integral formula. [Hint: Write \(w=1+r^{2}-2 r \cos (\theta-\alpha)\) and \(v(r, \theta, \alpha)=\) \(\left(1-r^{2}\right) w^{-1}\). Use Leibnitz's Rule to conclude that \(\nabla^{2} u=(2 \pi)^{-1} \int_{0}^{2 \pi} f(\alpha) \nabla^{2} v d \theta\), where \(\nabla^{2} v=v_{r r}+r^{-2} v_{\theta \theta}+r^{-1} v_{r}\) as in Eq. (2.138) in Section 2.17. Show that \(v_{r}=\) \(-2 r w^{-1}-\left(1-r^{2}\right) w^{-2} w_{r}\) etc. and finally $$ \begin{aligned} \nabla^{2} v=&-4 w^{-1}+\left(5 r-r^{-1}\right) w w_{r}+\left(r^{2}-1\right)\left(w^{-2} w_{r r}-2 w^{-3} w_{r}^{2}\right.\\\ &\left.+r^{-2} w^{-2} w_{\theta \theta}-2 r^{-2} w^{-3} w_{\theta}^{2}\right) . \end{aligned} $$ Multiply both sides by \(r^{2} w^{3}\), insert the proper expressions for \(w, w_{r}, \ldots\) on the right and collect terms in powers of \(r\left(r^{6}, r^{5} \ldots\right)\) to verify that \(r^{2} w^{3} \nabla^{2} v=0\) and hence \(\nabla^{2} u=0\). ]

Consider a 1-dimensional fluid motion, the flow taking place along the \(x\) axis. Let \(v=\) \(v(x, t)\) be the velocity at position \(x\) at time \(t\), so that if \(x\) is the coordinate of a fluid particle at time \(t\), one has \(d x / d t=v\). If \(f(x, t)\) is any scalar associated with the flow (velocity, acceleration, density,...), one can study the variation of \(f\) following the flow with the aid of the Stokes derivative: $$ \frac{D f}{D t}=\frac{\partial f}{\partial x} \frac{d x}{d t}+\frac{\partial f}{\partial t} $$ [see Problem 12 following Section 2.8]. A picce of the fluid occupying an interval \(a_{0} \leq\) \(x \leq b_{0}\) when \(t=0\) will occupy an interval \(a(t) \leq x \leq b(t)\) at time \(t\), where \(\frac{d a}{d t}=\) \(v(a, t), \frac{d b}{d t}=v(b, t)\). The integral $$ F(t)=\int_{a(t)}^{b(t)} f(x, t) d x $$ is then an integral of \(f\) over a definite piece of the fluid, whose position varies with time; if \(f\) is density, this is the mass of the piece. Show that $$ \frac{d F}{d t}=\int_{a(t)}^{b(t)}\left[\frac{\partial f}{\partial t}(x, t)+\frac{\partial}{\partial x}(f v)\right] d x=\int_{a(t)}^{b(t)}\left(\frac{D f}{D t}+f \frac{d v}{d x}\right) d x, $$ This is generalized to arbitrary 3-dimensional flows in Section 5.15.

a) Show that the integral $$ \iint_{R} \frac{1}{r^{p}} d x d y d z, \quad r=\sqrt{x^{2}+y^{2}+z^{2}}, $$ i. over the spherical region \(x^{2}+y^{2}+z^{2} \leq 1\) converges for \(p<3\) and find its value. For \(p=1\) this is the Newtonian potential of a uniform mass distribution over the solid sphere, evaluated at the origin. b) For the integral of part (a), let \(R\) be the exterior region \(x^{2}+y^{2}+z^{2} \geq 1\). Show that the integral converges for \(p>3\) and find its value.
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