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It should not be thought that the passing of a curve from one side of its tangent line to the other at a point is a sufficient condition for the point to be a point of inflection. It may, after all, happen that the curve does not have any constant convexity on either a left- or a right-hand neighborhood of the point. | Let\n\[ f\left( x\right) = \left\{ \begin{matrix} 2{x}^{3} + {x}^{3}\sin \frac{1}{{x}^{2}} & \text{ for }x \neq 0, \\ 0 & \text{ for }x = 0. \end{matrix}\right. \]\n\nThen \( {x}^{3} \leq f\left( x\right) \leq 3{x}^{3} \) for \( 0 \leq x \) and \( 3{x}^{3} \leq f\left( x\right) \leq {x}^{3} \) for \( x \leq 0 \), so th... | Yes |
Proposition 7. (Jensen’s inequality). \( {}^{19} \) If \( f : \rbrack a, b\lbrack \rightarrow \mathbb{R} \) is a convex function, \( {x}_{1},\ldots ,{x}_{n} \) are points of \( \rbrack a, b\lbrack \), and \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are nonnegative numbers such that \( {\alpha }_{1} + \cdots + {\alpha }_{... | Proof. For \( n = 2 \), condition (5.95) is the same as the definition (5.92) of a convex function. We shall now show that if (5.95) is valid for \( n = m - 1 \), it is also valid for \( n = m \) . For the sake of definiteness, assume that \( {\alpha }_{n} \neq 0 \) in the set \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) ... | Yes |
The function \( f\left( x\right) = \ln x \) is strictly convex upward on the set of positive numbers, and so by (5.96)\n\n\[{\alpha }_{1}\ln {x}_{1} + \cdots + {\alpha }_{n}\ln {x}_{n} \leq \ln \left( {{\alpha }_{1}{x}_{1} + \cdots + {\alpha }_{n}{x}_{n}}\right)\] | or,\n\n\[{x}_{1}^{{\alpha }_{1}}\cdots {x}_{n}^{{\alpha }_{n}} \leq {\alpha }_{1}{x}_{1} + \cdots + {\alpha }_{n}{x}_{n}\]\n\n(5.97)\n\nfor \( {x}_{i} \geq 0,{\alpha }_{i} \geq 0, i = 1,\ldots, n \), and \( \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i} = 1 \) .\n\nIn particular, if \( {\alpha }_{1} = \cdots = {\alph... | Yes |
Let \( f\left( x\right) = {x}^{p}, x \geq 0, p > 1 \) . Since such a function is convex, we have\n\n\[{\left( \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{x}_{i}\right) }^{p} \leq \mathop{\sum }\limits_{{i = 1}}^{n}{\alpha }_{i}{x}_{i}^{p}\]\n\nSetting \( q\; = \;\frac{p}{p - 1},\;{\alpha }_{i}\; = \;{b}_{i}^{q}{\l... | \[ \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{b}_{i} \leq {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{p}\right) }^{1/p}{\left( \mathop{\sum }\limits_{{i = 1}}^{n}{b}_{i}^{q}\right) }^{1/q} \]\n\nwhere \( \frac{1}{p} + \frac{1}{q} = 1 \) and \( p > 1 \) . | Yes |
Example 19. \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x}{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\cos x}{1} = 1 \) . | This example should not be looked on as a new, independent proof of the relation \( \frac{\sin x}{x} \rightarrow 1 \) as \( x \rightarrow 0 \) . The fact is that in deriving the relation \( {\sin }^{\prime }x = \cos x \) we already made use of the limit just calculated. | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{\alpha }}{{a}^{x}} = 0 \] | \[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{\alpha }}{{a}^{x}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\alpha {x}^{\alpha - 1}}{{a}^{x}\ln a} = \cdots = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\alpha \left( {\alpha - 1}\right) \cdots \left( {\alpha - n + 1}\right) {x}^... | Yes |
Let us construct a sketch of the graph of the function\n\n\[ h = {\log }_{{x}^{2} - {3x} - 2}2. \] | Taking account of the relation\n\n\[ y = {\log }_{{x}^{2} - {3x} + 2}2 = \frac{1}{{\log }_{2}\left( {{x}^{2} - {3x} + 2}\right) } = \frac{1}{{\log }_{2}\left( {x - 1}\right) \left( {x - 2}\right) }, \]\n\nwe construct successively the graph of the quadratic trinomial \( {y}_{1} = {x}^{2} - {3x} + 2 \) , then \( {y}_{2}... | No |
The construction of a sketch of the graph of the function\n\n\[ y = \sin \left( {x}^{2}\right) \] | We have constructed this graph using certain characteristic points for this function, the points where \( \sin \left( {x}^{2}\right) = - 1,\sin \left( {x}^{2}\right) = 0 \), or \( \sin \left( {x}^{2}\right) = 1 \) . Between two adjacent points of this type the function is monotonic. The form of the graph near the point... | Yes |
Let us construct the graph of the function \[ y = x + \arctan \left( {{x}^{3} - 1}\right) \] | As \( x \rightarrow - \infty \) the graph is well approximated by the line \( y = x - \frac{\pi }{2} \) , while for \( x \rightarrow + \infty \) it is approximated by \( y = x + \frac{\pi }{2} \). | No |
Example 26. Let \( \\left( {\\rho ,\\varphi }\\right) \) be polar coordinates in the plane and suppose a point is moving in the plane in such a way that\n\n\[ \n\\rho = \\rho \\left( t\\right) = 1 - {\\mathrm{e}}^{-t}\\cos \\frac{\\pi }{2}t \n\]\n\n\[ \n\\varphi = \\varphi \\left( t\\right) = 1 - {\\mathrm{e}}^{-t}\\si... | In order to do this, we first draw the graphs of \( \\rho \\left( t\\right) \) and \( \\varphi \\left( t\\right) \) (Figs. 5.22a and 5.22b).\n\nThen, looking simultaneously at both of the graphs just constructed, we can describe the general form of the trajectory of the point (Fig. 5.22c).\n\n![63ec862d-f82b-43c3-b65c-... | Yes |
Proposition 2. The series (5.110) converges if and only if for every \( \varepsilon > 0 \) there exists \( N \in \mathbb{N} \) such that\n\n\[ \left| {{z}_{m} + \cdots + {z}_{n}}\right| < \varepsilon \]\n\n(5.111)\n\nfor any natural numbers \( n \geq m > N \) . | From this one can see that a necessary condition for convergence of the series (5.110) is that \( {z}_{n} \rightarrow 0 \) as \( n \rightarrow \infty \) . (This, however, is also clear from the very definition of convergence.)\n\nAs in the real case, the series (5.110) is absolutely convergent if the series\n\n\[ \left... | No |
Proposition 4. If a series \( {z}_{1} + {z}_{2} + \cdots + {z}_{n} + \cdots \) of complex numbers converges absolutely, then a series \( {z}_{{n}_{1}} + {z}_{{n}_{2}} + \cdots + {z}_{{n}_{k}} + \cdots \) obtained by rear- \( {\text{ranging}}^{24} \) its terms also converges absolutely and has the same sum. | Proof. Using the convergence of the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {z}_{n}\right| \), given a number \( \varepsilon > 0 \), we choose \( N \in \mathbb{N} \) such that \( \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\left| {z}_{n}\right| < \varepsilon \) .\n\nWe then find an index \( K \in \ma... | Yes |
Proposition 5. The product of absolutely convergent series is an absolutely convergent series whose sum equals the product of the sums of the factor series. | Proof. We begin by remarking that whatever finite sum \( \sum {a}_{i}{b}_{j} \) of terms of the form \( {a}_{i}{b}_{j} \) we take, we can always find \( N \) such that the product of the sums \( {A}_{N} = {a}_{1} + \cdots + {a}_{N} \) and \( {B}_{N} = {b}_{1} + \cdots + {b}_{N} \) contains all the terms in that sum. Th... | Yes |
The series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}{a}^{n} \) and \( \mathop{\sum }\limits_{{m = 0}}^{\infty }\frac{1}{m!}{b}^{m} \) converge absolutely. In the product of these series let us group together all monomials of the form \( {a}^{n}{b}^{m} \) having the same total degree \( n + m = k \) . We ... | But\n\n\[ \mathop{\sum }\limits_{{m + n = k}}\frac{1}{n!m!}{a}^{n}{b}^{m} = \frac{1}{k!}\mathop{\sum }\limits_{{n = 0}}^{k}\frac{k!}{n!\left( {k - n}\right) !}{a}^{n}{b}^{k - n} = \frac{1}{k!}{\left( a + b\right) }^{k}, \]\n\nand therefore we find that\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}{a}^{n} ... | Yes |
Example 14. \n\nFig. 5.30. | It is clear from Examples 12 and 13 that under this function the unit disk maps into itself, but is covered twice. | No |
If \( z = r{\mathrm{e}}^{\mathrm{i}\varphi } \), then by (5.128), we have \( {z}^{n} = {r}^{n}{\mathrm{e}}^{\mathrm{i}{n\varphi }} \), so that in this case the image of the disk of radius \( r \) is the disk of radius \( {r}^{n} \), each point of which is the image of \( n \) points in the original disk (located, as it... | The only exception is the point \( w = 0 \), whose pre-image is the point \( z = 0 \) . However, as \( z \rightarrow 0 \), the function \( {z}^{n} \) is an infinitesimal of order \( n \), and so we say that at \( z = 0 \) the function has a zero of order \( n \) . Taking account of this kind of multiplicity, one can no... | Yes |
Every polynomial \( P\left( z\right) = {c}_{0} + \cdots + {c}_{n}{z}^{n} \) of degree \( n \geq 1 \) with complex coefficients admits a representation in the form\n\n\[ P\left( z\right) = {c}_{n}\left( {z - {z}_{1}}\right) \cdots \left( {z - {z}_{n}}\right) \]\n\nwhere \( {z}_{1},\ldots ,{z}_{n} \in \mathbb{C} \) (and ... | Proof. From the long division algorithm for dividing one polynomial \( P\left( z\right) \) by another polynomial \( Q\left( z\right) \) of lower degree, we find that \( P\left( z\right) = q\left( z\right) Q\left( z\right) + r\left( z\right) \) , where \( q\left( z\right) \) and \( r\left( z\right) \) are polynomials, t... | Yes |
Every polynomial \( P\left( z\right) = {a}_{0} + \cdots + {a}_{n}{z}^{n} \) with real coefficients can be expanded as a product of linear and quadratic polynomials with real coefficients. | This follows from Corollary 1 and Remark 2, by virtue of which for any root \( {z}_{k} \) of \( P\left( z\right) \) the number \( {\bar{z}}_{k} \) is also a root. Then, carrying out the multiplication \( \left( {z - {z}_{k}}\right) \left( {z - {\bar{z}}_{k}}\right) \) in the product (5.132), we obtain the quadratic pol... | Yes |
Corollary 3. Every root \( {z}_{j} \) of multiplicity \( {k}_{j} > 1 \) of a polynomial \( P\left( z\right) \) is a root of multiplicity \( {k}_{j} - 1 \) of the derivative \( {P}^{\prime }\left( z\right) \) . | Indeed, by the Euclidean algorithm, we first find the greatest common divisor \( q\left( z\right) \) of \( P\left( z\right) \) and \( {P}^{\prime }\left( z\right) \) . By Corollary 3, the expansion (5.133), and Theorem 2, the polynomial \( q\left( z\right) \) is equal, apart from a constant factor, to \( {\left( z - {z... | Yes |
If \( Q\left( z\right) = {\left( z - {z}_{1}\right) }^{{k}_{1}}\cdots {\left( z - {z}_{p}\right) }^{{k}_{p}} \) and \( \frac{P\left( z\right) }{Q\left( z\right) } \) is a proper fraction, there exists a unique representation of the fraction \( \frac{P\left( z\right) }{Q\left( z\right) } \) in the form | \[ \frac{P\left( z\right) }{Q\left( z\right) } = \mathop{\sum }\limits_{{j = 1}}^{p}\left( {\mathop{\sum }\limits_{{k = 1}}^{{k}_{j}}\frac{{a}_{jk}}{{\left( z - {z}_{j}\right) }^{k}}}\right) . \] | Yes |
Find the partial-fraction expansion (5.135) of the fraction \( \frac{P\left( x\right) }{Q\left( x\right) } \). | First of all, the problem is complicated by the fact that we do not know the factors of the polynomial \( Q\left( x\right) \). Let us try to simplify the situation by eliminating any multiple roots there may be of \( Q\left( x\right) \). We find\n\n\[ {Q}^{\prime }\left( x\right) = 7{x}^{6} + {18}{x}^{5} + {25}{x}^{4} ... | Yes |
The function \( F\left( x\right) = \arctan x \) is a primitive of \( f\left( x\right) = \frac{1}{1 + {x}^{2}} \) on the entire real line. | since \( {\arctan }^{\prime }x = \frac{1}{1 + {x}^{2}} \) | Yes |
The function \( F\left( x\right) = \operatorname{arccot}\frac{1}{x} \) is a primitive of \( f\left( x\right) = \frac{1}{1 + {x}^{2}} \) on the set of positive real numbers and on the set of negative real numbers, since for \( x \neq 0 \) | \[ {F}^{\prime }\left( x\right) = - \frac{1}{1 + {\left( \frac{1}{x}\right) }^{2}} \cdot \left( {-\frac{1}{{x}^{2}}}\right) = \frac{1}{1 + {x}^{2}} = f\left( x\right) . \] | Yes |
\[\int {\left( x + \frac{1}{\sqrt{x}}\right) }^{2}\mathrm{\;d}x = \int \left( {{x}^{2} + 2\sqrt{x} + \frac{1}{x}}\right) \mathrm{d}x =\] | \[\int {x}^{2}\mathrm{\;d}x + 2\int {x}^{1/2}\mathrm{\;d}x + \int \frac{1}{x}\mathrm{\;d}x = \frac{1}{3}{x}^{3} + \frac{4}{3}{x}^{3/2} + \ln \left| x\right| + c.\] | Yes |
\[\int {\cos }^{2}\frac{x}{2}\;\mathrm{d}x\] | \[\int {\cos }^{2}\frac{x}{2}\;\mathrm{d}x = \int \frac{1}{2}\left( {1 + \cos x}\right) \;\mathrm{d}x = \frac{1}{2}\int \left( {1 + \cos x}\right) \;\mathrm{d}x = \frac{1}{2}\int 1\mathrm{\;d}x + \frac{1}{2}\int \cos x\mathrm{\;d}x = \frac{1}{2}x + \frac{1}{2}\sin x + c.\] | Yes |
\[\int \ln x\mathrm{\;d}x = x\ln x - \int x\mathrm{\;d}\ln x = x\ln x - \int x \cdot \frac{1}{x}\mathrm{\;d}x =\] | \[\int \ln x\mathrm{\;d}x = x\ln x - \int x\mathrm{\;d}\ln x = x\ln x - \int x \cdot \frac{1}{x}\mathrm{\;d}x = x\ln x - \int 1\mathrm{\;d}x = x\ln x - x + c.\] | Yes |
\[ \int {x}^{2}{\mathrm{e}}^{x}\mathrm{\;d}x = \int {x}^{2}{\mathrm{{de}}}^{x} = {x}^{2}{\mathrm{e}}^{x} - \int {\mathrm{e}}^{x}\mathrm{\;d}{x}^{2} = {x}^{2}{\mathrm{e}}^{x} - 2\int x{\mathrm{e}}^{x}\mathrm{\;d}x = \] | \[ = {x}^{2}{\mathrm{e}}^{x} - 2\int x\;{\mathrm{{de}}}^{x} = {x}^{2}{\mathrm{e}}^{x} - 2\left( {x{\mathrm{e}}^{x}-\int {\mathrm{e}}^{x}\;\mathrm{d}x}\right) = \] \[ = {x}^{2}{\mathrm{e}}^{x} - {2x}{\mathrm{e}}^{x} + 2{\mathrm{e}}^{x} + c = \left( {{x}^{2} - {2x} + 2}\right) {\mathrm{e}}^{x} + c. \] | Yes |
\[\int \frac{\mathrm{d}x}{\sin x}\] | \[\int \frac{\mathrm{d}x}{\sin x} = \int \frac{\mathrm{d}x}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \int \frac{\mathrm{d}\left( \frac{x}{2}\right) }{\tan \frac{x}{2}{\cos }^{2}\frac{x}{2}} = \int \frac{\mathrm{d}u}{\tan u{\cos }^{2}u} = \int \frac{\mathrm{d}\left( {\tan u}\right) }{\tan u} = \int \frac{dv}{v} = \ln \left|... | Yes |
\[\int \sin {2x}\cos {3x}\mathrm{\;d}x\] | \[\int \sin {2x}\cos {3x}\mathrm{\;d}x = \frac{1}{2}\int \left( {\sin {5x} - \sin x}\right) \mathrm{d}x =\] \[= \frac{1}{2}\left( {\;\int \sin {5x}\;\mathrm{d}x-\int \sin x\;\mathrm{d}x}\right) = \frac{1}{2}\left( {\frac{1}{5}\int \sin {5x}\;\mathrm{d}\left( {5x}\right) + \cos x}\right) =\] \[= \frac{1}{10}\int \sin u\... | Yes |
\[ \int {\mathrm{e}}^{ax}\cos {bx}\mathrm{\;d}x = \frac{1}{a}\int \cos {bx}{\mathrm{{de}}}^{ax} = \] | \[ = \frac{1}{a}{\mathrm{e}}^{ax}\cos {bx} - \frac{1}{a}\int {\mathrm{e}}^{ax}\mathrm{\;d}\cos {bx} = \frac{1}{a}{\mathrm{e}}^{ax}\cos {bx} + \frac{b}{a}\int {\mathrm{e}}^{ax}\sin {bx}\mathrm{\;d}x = \] \[ = \frac{1}{a}{\mathrm{e}}^{ax}\cos {bx} + \frac{b}{{a}^{2}}\int \sin {bx}{\mathrm{{de}}}^{ax} = \frac{1}{a}{\mathr... | Yes |
Let us calculate \( \int \frac{2{x}^{2} + {5x} + 5}{\left( {{x}^{2} - 1}\right) \left( {x + 2}\right) }\mathrm{d}x \) . | Since the integrand is a proper fraction, and the factorization of the denominator into the product \( \left( {x - 1}\right) \left( {x + 1}\right) \left( {x + 2}\right) \) is also known, we immediately seek a partial fraction expansion\n\n\[ \frac{2{x}^{2} + {5x} + 5}{\left( {x - 1}\right) \left( {x + 1}\right) \left( ... | Yes |
\[ \int \frac{\mathrm{d}x}{3 + \sin x} = \int \frac{1}{3 + \frac{2t}{1 + {t}^{2}}} \cdot \frac{2\mathrm{\;d}t}{1 + {t}^{2}} = \] | \[ = 2\int \frac{\mathrm{d}t}{3{t}^{2} + {2t} + 3} = \frac{2}{3}\int \frac{\mathrm{d}\left( {t + \frac{1}{3}}\right) }{{\left( t + \frac{1}{3}\right) }^{2} + \frac{8}{9}} = \frac{2}{3}\int \frac{\mathrm{d}u}{{u}^{2} + {\left( \frac{2\sqrt{2}}{3}\right) }^{2}} = \] \[ = \frac{1}{\sqrt{2}}\arctan \frac{3u}{2\sqrt{2}} + c... | Yes |
\[\int \frac{\mathrm{d}x}{{\left( \sin x + \cos x\right) }^{2}}\] | \[\int \frac{\mathrm{d}x}{{\left( \sin x + \cos x\right) }^{2}} = \int \frac{\mathrm{d}x}{{\cos }^{2}x{\left( \tan x + 1\right) }^{2}} = \int \frac{\mathrm{d}\tan x}{{\left( \tan x + 1\right) }^{2}} = \int \frac{\mathrm{d}t}{{\left( t + 1\right) }^{2}} = - \frac{1}{t + 1} + c = c - \frac{1}{1 + \tan x}.\] | Yes |
\[ \int \frac{\mathrm{d}x}{2{\sin }^{2}{3x} - 3{\cos }^{2}{3x} + 1} \] | \[ \int \frac{\mathrm{d}x}{2{\sin }^{2}{3x} - 3{\cos }^{2}{3x} + 1} = \int \frac{\mathrm{d}x}{{\cos }^{2}{3x}\left( {2{\tan }^{2}{3x} - 3 + \left( {1 + {\tan }^{2}{3x}}\right) }\right) } = \] \[ = \frac{1}{3}\int \frac{\mathrm{d}\tan {3x}}{3{\tan }^{2}{3x} - 2} = \frac{1}{3}\int \frac{\mathrm{d}t}{3{t}^{2} - 2} = \frac... | Yes |
\[ \int \frac{{\cos }^{3}x}{{\sin }^{7}x}\mathrm{\;d}x \] | \[ \int \frac{{\cos }^{3}x}{{\sin }^{7}x}\mathrm{\;d}x = \int \frac{{\cos }^{2}x\mathrm{\;d}\sin x}{{\sin }^{7}x} = \int \frac{\left( {1 - {t}^{2}}\right) \mathrm{d}t}{{t}^{7}} = \int \left( {{t}^{-7} - {t}^{-5}}\right) \mathrm{d}t = - \frac{1}{6}{t}^{-6} + \frac{1}{4}{t}^{-4} + c = \frac{1}{4{\sin }^{4}x} - \frac{1}{6... | Yes |
\[ \int \sqrt[3]{\frac{x - 1}{x + 1}}\mathrm{\;d}x = \int t\mathrm{\;d}\left( \frac{{t}^{3} + 1}{1 - {t}^{3}}\right) = t \cdot \frac{{t}^{3} + 1}{1 - {t}^{3}}\mathrm{\;d}t - \int \frac{{t}^{3} + 1}{1 - {t}^{3}}\mathrm{\;d}t = \] | \[ = t \cdot \frac{{t}^{3} + 1}{1 - {t}^{3}} - \int \left( {\frac{2}{1 - {t}^{3}} - 1}\right) \mathrm{d}t = \] \[ = t \cdot \frac{{t}^{3} + 1}{1 - {t}^{3}} + t - 2\int \frac{\mathrm{d}t}{\left( {1 - t}\right) \left( {1 + t + {t}^{2}}\right) } = \] \[ = \frac{t}{1 - {t}^{3}} - 2\int \left( {\frac{1}{3\left( {1 - t}\righ... | Yes |
\[ \int \frac{\mathrm{d}x}{x + \sqrt{{x}^{2} + {2x} + 2}} = \int \frac{\mathrm{d}x}{x + \sqrt{{\left( x + 1\right) }^{2} + 1}} = \int \frac{\mathrm{d}t}{t - 1 + \sqrt{{t}^{2} + 1}}. \] | Setting \( \sqrt{{t}^{2} + 1} = u - t \), we have \( 1 = {u}^{2} - {2tu} \), from which it follows that \( t = \frac{{u}^{2} - 1}{2u} \) . Therefore\n\n\[ \int \frac{\mathrm{d}t}{t - 1 + \sqrt{{t}^{2} + 1}} = \frac{1}{2}\int \frac{1}{u - 1}\left( {1 + \frac{1}{{u}^{2}}}\right) \mathrm{d}u = \frac{1}{2}\int \frac{1}{u -... | Yes |
Proposition 1. A necessary condition for a function \( f \) defined on a closed interval \( \left\lbrack {a, b}\right\rbrack \) to be Riemann integrable on \( \left\lbrack {a, b}\right\rbrack \) is that \( f \) be bounded on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. If \( f \) is not bounded on \( \left\lbrack {a, b}\right\rbrack \), then for any partition \( P \) of \( \left\lbrack {a, b}\right\rbrack \) the function \( f \) is unbounded on at least one of the intervals \( \left\lbrack {{x}_{i - 1},{x}_{i}}\right\rbrack \) of \( P \) . This means that, by choosing the poin... | Yes |
Corollary 1. \( \left( {f \in C\left\lbrack {a, b}\right\rbrack }\right) \Rightarrow \left( {f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack }\right) \), that is, every continuous function on a closed interval is integrable on that closed interval. | Proof. If a function is continuous on a closed interval, it is bounded there, so that the necessary condition for integrability is satisfied in this case. But a continuous function on a closed interval is uniformly continuous on that interval. Therefore, for every \( \varepsilon > 0 \) there exists \( \delta > 0 \) suc... | Yes |
Corollary 3. A monotonic function on a closed interval is integrable on that interval. | Proof. It follows from the monotonicity of \( f \) on \( \left\lbrack {a, b}\right\rbrack \) that \( \omega \left( {f;\left\lbrack {a, b}\right\rbrack }\right) = \) \( \left| {f\left( b\right) - f\left( a\right) }\right| \) . Suppose \( \varepsilon > 0 \) is given. We set \( \delta = \frac{\varepsilon }{\left| f\left( ... | Yes |
\[ s\left( {f;P}\right) = \mathop{\inf }\limits_{\xi }\sigma \left( {f;P,\xi }\right) \] \[ S\left( {f;P}\right) = \mathop{\sup }\limits_{\xi }\sigma \left( {f;P,\xi }\right) \] | Proof. Let us verify, for example, that the upper Darboux sum corresponding to a partition \( P \) of the closed interval \( \left\lbrack {a, b}\right\rbrack \) is the least upper bound of the Riemann sums corresponding to the partitions with distinguished points \( \left( {P,\xi }\right) \), the supremum being taken o... | No |
Proposition 3. A bounded real-valued function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is Riemann-integrable on \( \left\lbrack {a, b}\right\rbrack \) if and only if the following limits exist and are equal to each other:\n\n\[ \underline{I} = \mathop{\lim }\limits_{{\lambda \left( P\right) \ri... | Proof. Indeed, if the limits (6.9) exist and are equal, we conclude by the properties of limits and by (6.7) that the Riemann sums have a limit and that\n\n\[ \underline{I} = \mathop{\lim }\limits_{{\lambda \left( P\right) \rightarrow 0}}\sigma \left( {f;P,\xi }\right) = \bar{I} \]\n\nOn the other hand, if \( f \in \ma... | Yes |
Proposition 4. If \( f, g \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), then\n\na) \( \left( {f + g}\right) \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \)\n\nb) \( \left( {\alpha f}\right) \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), where \( \alpha \) is a numerical coefficient;\n\nc) \( \left| f\righ... | Proof. a) This assertion is obvious since\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\left( {f + g}\right) \left( {\xi }_{i}\right) \Delta {x}_{i} = \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {\xi }_{i}\right) \Delta {x}_{i} + \mathop{\sum }\limits_{{i = 1}}^{n}g\left( {\xi }_{i}\right) \Delta {x}_{i}. \]\n\nb) This asse... | Yes |
Lemma 2. a) A single point and a finite number of points are sets of measure zero. | Proof. a) A point can be covered by one interval of length less than any preassigned number \( \varepsilon > 0 \) ; therefore a point is a set of measure zero. The rest of a) then follows from b). | No |
The Dirichlet function \[ \mathcal{D}\left( x\right) = \left\{ \begin{array}{l} 1\text{ for }x \in \mathbb{Q}, \\ 0\text{ for }x \in \mathbb{R} \smallsetminus \mathbb{Q}, \end{array}\right. \] on the interval \( \left\lbrack {0,1}\right\rbrack \) is not integrable on that interval, since for any partition \( P \) of \(... | Then \[ \sigma \left( {f;P,{\xi }^{\prime }}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}1 \cdot \Delta {x}_{i} = 1 \] while \[ \sigma \left( {f;P,{\xi }^{\prime \prime }}\right) = \mathop{\sum }\limits_{{i = 1}}^{n}0 \cdot \Delta {x}_{i} = 0. \] Thus the Riemann sums of the function \( \mathcal{D}\left( x\right) \) ca... | Yes |
Consider the Riemann function\n\n\[ \mathcal{R}\left( x\right) = \left\{ \begin{array}{l} \frac{1}{n},\text{ if }x \in \mathbb{Q}\text{ and }x = \frac{m}{n}\text{ is in lowest terms }, \\ 0,\text{ if }x \in \mathbb{R} \smallsetminus \mathbb{Q}. \end{array}\right. \] | We have already studied this function in Subsect. 4.1.2, and we know that \( \mathcal{R}\left( x\right) \) is continuous at all irrational points and discontinuous at all rational points except 0 . Thus the set of points of discontinuity of \( \mathcal{R}\left( x\right) \) is countable and hence has measure zero. By th... | Yes |
Theorem 1. If \( f \) and \( g \) are integrable functions on the closed interval \( \left\lbrack {a, b}\right\rbrack \) , a linear combination of them \( {\alpha f} + {\beta g} \) is also integrable on \( \left\lbrack {a, b}\right\rbrack \), and\n\n\[{\int }_{a}^{b}\left( {{\alpha f} + {\beta g}}\right) \left( x\right... | Proof. Consider a Riemann sum for the integral on the left-hand side of (6.11), and transform it as follows:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\left( {{\alpha f} + {\beta g}}\right) \left( {\xi }_{i}\right) \Delta {x}_{i} = \alpha \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {\xi }_{i}\right) \Delta {x}_{i} + \bet... | Yes |
Lemma 1. If \( a < b < c \) and \( f \in \mathcal{R}\left\lbrack {a, c}\right\rbrack \), then \( {\left. f\right| }_{\left\lbrack a, b\right\rbrack } \in \mathcal{R}\left\lbrack {a, b}\right\rbrack ,{\left. f\right| }_{\left\lbrack b, c\right\rbrack } \in \mathcal{R}\left\lbrack {b, c}\right\rbrack \), and the followin... | Proof. We first note that the integrability of the restrictions of \( f \) to the closed intervals \( \left\lbrack {a, b}\right\rbrack \) and \( \left\lbrack {b, c}\right\rbrack \) is guaranteed by Proposition 4 of Sect. 6.1.\n\nNext, since \( f \in \mathcal{R}\left\lbrack {a, c}\right\rbrack \), in computing the integ... | Yes |
Theorem 2. Let \( a, b, c \in \mathbb{R} \) and let \( f \) be a function integrable over the largest closed interval having two of these points as endpoints. Then the restriction of \( f \) to each of the other closed intervals is also integrable over those intervals and the following equality holds:\n\n\[{\int }_{a}^... | Proof. By the symmetry of Eq. (6.17) in \( a, b \), and \( c \), we may assume without loss of generality that \( a = \min \{ a, b, c\} \).\n\nIf \( \max \{ a, b, c\} = c \) and \( a < b < c \), then by Lemma 1\n\n\[{\int }_{a}^{b}f\left( x\right) \mathrm{d}x + {\int }_{b}^{c}f\left( x\right) \mathrm{d}x - {\int }_{a}^... | Yes |
Theorem 3. If \( a \leq b \) and \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), then \( \left| f\right| \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) and the following inequality holds:\n\n\[ \left| {{\int }_{a}^{b}f\left( x\right) \mathrm{d}x}\right| \leq {\int }_{a}^{b}\left| f\right| \left( x\right) \... | Proof. For \( a = b \) the assertion is trivial, and so we shall assume that \( a < b \) .\n\nTo prove the theorem it now suffices to recall that \( \left| f\right| \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) (see Proposition 4 of Sect. 6.1), and write the following estimate for the Riemann sum \( \sigma \left( ... | Yes |
Theorem 4. If \( a \leq b,{f}_{1},{f}_{2} \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), and \( {f}_{1}\left( x\right) \leq {f}_{2}\left( x\right) \) at each point \( x \in \left\lbrack {a, b}\right\rbrack \), then \[ {\int }_{a}^{b}{f}_{1}\left( x\right) \mathrm{d}x \leq {\int }_{a}^{b}{f}_{2}\left( x\right) \mat... | Proof. For \( a = b \) the assertion is trivial. If \( a < b \), it suffices to write the following inequality for the Riemann sums: \[ \mathop{\sum }\limits_{{i = 1}}^{n}{f}_{1}\left( {\xi }_{i}\right) \Delta {x}_{i} \leq \mathop{\sum }\limits_{{i = 1}}^{n}{f}_{2}\left( {\xi }_{i}\right) \Delta {x}_{i} \] which is val... | Yes |
Corollary 1. If \( a \leq b, f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), and \( m \leq f\left( x\right) \leq M \) at each \( x \in \left\lbrack {a, b}\right\rbrack \) , then\n\n\[ m \cdot \left( {b - a}\right) \leq {\int }_{a}^{b}f\left( x\right) \mathrm{d}x \leq M \cdot \left( {b - a}\right) ,\]\n\nand, in p... | Proof. Relation (6.22) is obtained by integrating each term in the inequality \( m \leq f\left( x\right) \leq M \) and using Theorem 4. | Yes |
Corollary 2. If \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack, m = \mathop{\inf }\limits_{{x \in \left\lbrack {a, b}\right\rbrack }}f\left( x\right) \), and \( M = \mathop{\sup }\limits_{{x \in \left\lbrack {a, b}\right\rbrack }}f\left( x\right) \), then there exists a number \( \mu \in \left\lbrack {m, M}\right... | Proof. If \( a = b \), the assertion is trivial. If \( a \neq b \), we set \( \mu = \frac{1}{b - a}{\int }_{a}^{b}f\left( x\right) \mathrm{d}x \) . It then follows from (6.22) that \( m \leq \mu \leq M \) if \( a < b \) . But both sides of (6.23) reverse sign if \( a \) and \( b \) are interchanged, and therefore (6.23... | Yes |
Corollary 3. If \( f \in C\left\lbrack {a, b}\right\rbrack \), there is a point \( \xi \in \left\lbrack {a, b}\right\rbrack \) such that\n\n\[{\int }_{a}^{b}f\left( x\right) \mathrm{d}x = f\left( \xi \right) \left( {b - a}\right) .\] | Proof. By the intermediate-value theorem for a continuous function, there is a point \( \xi \) on \( \left\lbrack {a, b}\right\rbrack \) at which \( f\left( \xi \right) = \mu \) if\n\n\[m = \mathop{\min }\limits_{{x \in \left\lbrack {a, b}\right\rbrack }}f\left( x\right) \leq \mu \leq \mathop{\max }\limits_{{x \in \lef... | Yes |
Theorem 5. (First mean-value theorem for the integral). Let \( f, g \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) , \( m = \mathop{\inf }\limits_{{x \in \left\lbrack {a, b}\right\rbrack }}f\left( x\right) \), and \( M = \mathop{\sup }\limits_{{x \in \left\lbrack {a, b}\right\rbrack }}f\left( x\right) \) . If \( g ... | Proof. Since interchanging the limits of integration leads to a simultaneous sign reversal on both sides of Eq. (6.25), it suffices to verify this equality for the case \( a < b \) . Reversing the sign of \( g\left( x\right) \) also reverses the signs of both sides of (6.25), so that we may assume without loss of gener... | Yes |
Lemma 2. If the numbers \( {A}_{k} = \mathop{\sum }\limits_{{i = 1}}^{k}{a}_{i}\left( {k = 1,\ldots, n}\right) \) satisfy the inequalities \( m \leq {A}_{k} \leq M \) and the numbers \( {b}_{i}\left( {i = 1,\ldots, n}\right) \) are nonnegative and \( {b}_{i} \geq {b}_{i + 1} \) for \( i = 1,\ldots, n - 1 \), then\n\n\[... | Proof. Using the fact that \( {b}_{n} \geq 0 \) and \( {b}_{i} - {b}_{i + 1} \geq 0 \) for \( i = 1,\ldots, n - 1 \), we obtain from (6.29),\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{b}_{i} \leq M{b}_{n} + \mathop{\sum }\limits_{{i = 1}}^{{n - 1}}M\left( {{b}_{i} - {b}_{i + 1}}\right) = M{b}_{n} + M\left( {{b}_{... | Yes |
Lemma 3. If \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), then for any \( x \in \left\lbrack {a, b}\right\rbrack \) the function\n\n\[ F\left( x\right) = {\int }_{a}^{x}f\left( t\right) \mathrm{d}t \]\n\n(6.31)\n\nis defined and \( F\left( x\right) \in C\left\lbrack {a, b}\right\rbrack \) . | Proof. The existence of the integral in (6.31) for any \( x \in \left\lbrack {a, b}\right\rbrack \) is already known from Proposition 4 of Sect. 6.1; therefore it remains only for us to verify that the function \( F\left( x\right) \) is continuous. Since \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), we have ... | Yes |
Theorem 6. (Second mean-value theorem for the integral). If \( f, g \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) and \( g \) is a monotonic function on \( \left\lbrack {a, b}\right\rbrack \), then there exists a point \( \xi \in \left\lbrack {a, b}\right\rbrack \) such that\n\n\[{\int }_{a}^{b}\left( {f \cdot g}\... | Proof. Let \( g \) be a nondecreasing function on \( \left\lbrack {a, b}\right\rbrack \) . Then \( G\left( x\right) = g\left( b\right) - g\left( x\right) \) is nonnegative, nonincreasing, and integrable on \( \left\lbrack {a, b}\right\rbrack \) . Applying formula (6.33), we find\n\n\[{\int }_{a}^{b}\left( {f \cdot G}\r... | Yes |
Lemma 1. If \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) and the function \( f \) is continuous at a point \( x \in \) \( \left\lbrack {a, b}\right\rbrack \), then the function \( F \) defined on \( \left\lbrack {a, b}\right\rbrack \) by \( \left( {6.40}\right) \) is differentiable at the point \( x \), and ... | Proof. Let \( x, x + h \in \left\lbrack {a, b}\right\rbrack \) . Let us estimate the difference \( F\left( {x + h}\right) - F\left( x\right) \) . It follows from the continuity of \( f \) at \( x \) that \( f\left( t\right) = f\left( x\right) + \Delta \left( t\right) \), where \( \Delta \left( t\right) \rightarrow 0 \)... | Yes |
Theorem 1. Every continuous function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) on the closed interval \( \left\lbrack {a, b}\right\rbrack \) has a primitive, and every primitive of \( f \) on \( \left\lbrack {a, b}\right\rbrack \) has the form\n\n\[ \mathcal{F}\left( x\right) = {\int }_{a}^{x}f\... | Proof. We have the implication \( \left( {f \in C\left\lbrack {a, b}\right\rbrack }\right) \Rightarrow \left( {f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack }\right) \), so that by Lemma 1 the function (6.40) is a primitive for \( f \) on \( \left\lbrack {a, b}\right\rbrack \) . But two primitives \( \mathcal{F}\le... | Yes |
Theorem 2. If \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is a bounded function with a finite number of points of discontinuity, then \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) and\n\n\[{\int }_{a}^{b}f\left( x\right) \mathrm{d}x = \mathcal{F}\left( b\right) - \mathcal{F}\left( a\righ... | Proof. We already know that a bounded function on a closed interval having only a finite number of discontinuities is integrable (see Corollary 2 after Proposition 2 in Sect. 6.1). The existence of a generalized primitive \( \mathcal{F}\left( x\right) \) of the function \( f \) on \( \left\lbrack {a, b}\right\rbrack \)... | Yes |
If the functions \( u\left( x\right) \) and \( v\left( x\right) \) are continuously differentiable on a closed interval with endpoints \( a \) and \( b \), then\n\n\[ \n{\left. {\int }_{a}^{b}\left( u \cdot {v}^{\prime }\right) \left( x\right) \mathrm{d}x = \left( u \cdot v\right) \right| }_{a}^{b} - {\int }_{a}^{b}\le... | Proof. By the rule for differentiating a product of functions, we have\n\n\[ \n{\left( u \cdot v\right) }^{\prime }\left( x\right) = \left( {{u}^{\prime } \cdot v}\right) \left( x\right) + \left( {u \cdot {v}^{\prime }}\right) \left( x\right) . \n\]\n\nBy hypothesis, all the functions in this last equality are continuo... | Yes |
Proposition 2. If the function \( t \mapsto f\\left( t\\right) \) has continuous derivatives up to order \( n \) inclusive on the closed interval with endpoints a and \( x \), then Taylor’s formula holds:\n\n\[ f\\left( x\\right) = f\\left( a\\right) + \\frac{1}{1!}{f}^{\\prime }\\left( a\\right) \\left( {x - a}\\right... | We note that the function \( {\\left( x - t\\right) }^{n - 1} \) does not change sign on the closed interval with endpoints \( a \) and \( x \), and since \( t \mapsto {f}^{\\left( n\\right) }\\left( t\\right) \) is continuous on that interval, the first mean-value theorem implies that there exists a point \( \\xi \) s... | Yes |
Proposition 3. If \( \varphi : \left\lbrack {\alpha ,\beta }\right\rbrack \rightarrow \left\lbrack {a, b}\right\rbrack \) is a continuously differentiable mapping of the closed interval \( \alpha \leq t \leq \beta \) into the closed interval \( a \leq x \leq b \) such that \( \varphi \left( \alpha \right) = a \) and \(... | Proof. Let \( \mathcal{F}\left( x\right) \) be a primitive of \( f\left( x\right) \) on \( \left\lbrack {a, b}\right\rbrack \) . Then, by the theorem on differentiation of a composite function, the function \( \mathcal{F}\left( {\varphi \left( t\right) }\right) \) is a primitive of the function \( f\left( {\varphi \lef... | Yes |
Theorem 3. Let \( \varphi : \left\lbrack {\alpha ,\beta }\right\rbrack \rightarrow \left\lbrack {a, b}\right\rbrack \) be a continuously differentiable strictly monotonic mapping of the closed interval \( \alpha \leq t \leq \beta \) into the closed interval \( a \leq x \leq b \) with the correspondence \( \varphi \left... | Proof. Since \( \varphi \) is a strictly monotonic mapping of \( \left\lbrack {\alpha ,\beta }\right\rbrack \) onto \( \left\lbrack {a, b}\right\rbrack \) with endpoints corresponding to endpoints, every partition \( {P}_{t}\left( {\alpha = {t}_{0} < \cdots < {t}_{n} = \beta }\right) \) of the closed interval \( \left\... | Yes |
\[ {\int }_{-1}^{1}\sqrt{1 - {x}^{2}}\mathrm{\;d}x = {\int }_{-\pi /2}^{\pi /2}\sqrt{1 - {\sin }^{2}t}\cos t\mathrm{\;d}t = {\int }_{-\pi /2}^{\pi /2}{\cos }^{2}t\mathrm{\;d}t = \] | \[ = \frac{1}{2}{\int }_{-\pi /2}^{\pi /2}\left( {1 + \cos {2t}}\right) \mathrm{d}t = {\left. \frac{1}{2}\left( t + \frac{1}{2}\sin 2t\right) \right| }_{-\pi /2}^{\pi /2} = \frac{\pi }{2}. \] | Yes |
Let us show that\n\n\[ \text{a)}{\int }_{-\pi }^{\pi }\sin {mx}\cos {nx}\mathrm{\;d}x = 0\text{, b)}{\int }_{-\pi }^{\pi }{\sin }^{2}{mx}\mathrm{\;d}x = \pi \text{, c)}{\int }_{-\pi }^{\pi }{\cos }^{2}{nx}\mathrm{\;d}x = \pi \]\n\nfor \( m, n \in \mathbb{N} \) . | \n\n\[ \text{a)}{\int }_{-\pi }^{\pi }\sin {mx}\cos {nx}\mathrm{\;d}x = \frac{1}{2}{\int }_{-\pi }^{\pi }\left( {\sin \left( {n + m}\right) x - \sin \left( {n - m}\right) x}\right) \mathrm{d}x = \]\n\n\[ = {\left. \frac{1}{2}\left( -\frac{1}{n + m}\cos \left( n + m\right) x + \frac{1}{n - m}\cos \left( n - m\right) x\r... | Yes |
Let \( f \in \mathcal{R}\left\lbrack {-a, a}\right\rbrack \) . We shall show that\n\n\[{\int }_{-a}^{a}f\left( x\right) \mathrm{d}x = \left\{ \begin{matrix} 2{\int }_{0}^{a}f\left( x\right) \mathrm{d}x, & \text{ if }f\text{ is an even function }, \\ 0, & \text{ if }f\text{ is an odd function }. \end{matrix}\right.\] | If \( f\left( {-x}\right) = f\left( x\right) \), then\n\n\[{\int }_{-a}^{a}f\left( x\right) \mathrm{d}x = {\int }_{-a}^{0}f\left( x\right) \mathrm{d}x + {\int }_{0}^{a}f\left( x\right) \mathrm{d}x = {\int }_{a}^{0}f\left( {-t}\right) \left( {-1}\right) \mathrm{d}t + {\int }_{0}^{a}f\left( x\right) \mathrm{d}x =\n\]\n\[... | Yes |
Let \( f \) be a function defined on the entire real line \( \mathbb{R} \) and having period \( T \), that is \( f\left( {x + T}\right) = f\left( x\right) \) for all \( x \in \mathbb{R} \). If \( f \) is integrable on each finite closed interval, then for any \( a \in \mathbb{R} \) we have the equality \[ {\int }_{a}^{... | \[ {\int }_{a}^{a + T}f\left( x\right) \mathrm{d}x = {\int }_{a}^{0}f\left( x\right) \mathrm{d}x + {\int }_{0}^{T}f\left( x\right) \mathrm{d}x + {\int }_{T}^{a + T}f\left( x\right) \mathrm{d}x = \] \[ = {\int }_{0}^{T}f\left( x\right) \mathrm{d}x + {\int }_{a}^{0}f\left( x\right) \mathrm{d}x + {\int }_{0}^{a}f\left( {t... | Yes |
Suppose we need to compute the integral \( {\int }_{0}^{1}\sin \left( {x}^{2}\right) \mathrm{d}x \), for example within \( {10}^{-2} \) . | We know that the primitive \( \int \sin \left( {x}^{2}\right) \mathrm{d}x \) (the Fresnel integral) cannot be expressed in terms of elementary functions, so that it is impossible to use the Newton-Leibniz formula here in the traditional sense. We take a different approach. When studying Taylor's formula in differential... | Yes |
We shall show that \( {F}_{\delta }\left( x\right) \) (called the average of \( f \) ) is, compared to \( f \) , more regular. More precisely, if \( f \) is integrable on any interval \( \left\lbrack {a, b}\right\rbrack \), then \( {F}_{\delta }\left( x\right) \) is continuous on \( \mathbb{R} \), and if \( f \in C\lef... | We verify first that \( {F}_{\delta }\left( x\right) \) is continuous:\n\n\[ \left| {{F}_{\delta }\left( {x + h}\right) - {F}_{\delta }\left( x\right) }\right| = \frac{1}{2\delta }\left| {{\int }_{x + \delta }^{x + \delta + h}f\left( t\right) \mathrm{d}t + {\int }_{x - \delta + h}^{x - \delta }f\left( t\right) \mathrm{... | Yes |
If \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \), the function \( \mathcal{F}\left( x\right) = {\int }_{a}^{x}f\left( t\right) \mathrm{d}t \) generates via formula (6.50) the additive function\n\n\[ I\left( {\alpha ,\beta }\right) = {\int }_{\alpha }^{\beta }f\left( t\right) \mathrm{d}t \] | We remark that in this case the function \( \mathcal{F}\left( x\right) \) is continuous on the closed interval \( \left\lbrack {a, b}\right\rbrack \) . | No |
Suppose the interval \( \left\lbrack {0,1}\right\rbrack \) is a weightless string with a bead of unit mass attached to the string at the point \( x = 1/2 \). Let \( \mathcal{F}\left( x\right) \) be the amount of mass located in the closed interval \( \left\lbrack {0, x}\right\rbrack \) of the string. Then by hypothesis... | Since the function \( \mathcal{F} \) is discontinuous, the additive function \( I\left( {\alpha ,\beta }\right) \) in this case cannot be represented as the Riemann integral of a function - a mass density. (This density, that is, the limit of the ratio of the mass in an interval to the length of the interval, would hav... | Yes |
Proposition 1. Suppose the additive function \( I\left( {\alpha ,\beta }\right) \) defined for points \( \alpha ,\beta \) of a closed interval \( \left\lbrack {a, b}\right\rbrack \) is such that there exists a function \( f \in \mathcal{R}\left\lbrack {a, b}\right\rbrack \) connected with \( I \) as follows: the relati... | Proof. Let \( P \) be an arbitrary partition \( a = {x}_{0} < \cdots < {x}_{n} = b \) of the closed interval \( \left\lbrack {a, b}\right\rbrack \), let \( {m}_{i} = \mathop{\inf }\limits_{{x \in \left\lbrack {{x}_{i - 1},{x}_{i}}\right\rbrack }}f\left( x\right) \), and let \( {M}_{i} = \mathop{\sup }\limits_{{x \in \l... | Yes |
Example 3. Let us test formula (6.52) on a familiar object. Suppose the point moves according to the law\n\n\\[ \n x = R\\cos {2\\pi t} \n\\]\n\n(6.53)\n\n\\[ \n y = R\\sin {2\\pi t} \n\\]\n\nOver the time interval \\( \\left\\lbrack {0,1}\\right\\rbrack \\) the point will traverse a circle of radius \\( R \\) , that... | Let us carry out the computation according to formula (6.52):\n\n\\[ \nl\\left\\lbrack {0,1}\\right\\rbrack = {\\int }_{0}^{1}\\sqrt{{\\left( -2\\pi R\\sin 2\\pi t\\right) }^{2} + {\\left( 2\\pi R\\cos 2\\pi t\\right) }^{2}}\\mathrm{\\;d}t = {2\\pi R}.\n\\] | Yes |
Proposition 2. If a smooth path \( \widetilde{\Gamma } : \left\lbrack {\alpha ,\beta }\right\rbrack \rightarrow {\mathbb{R}}^{3} \) is obtained from a smooth path \( \Gamma : \left\lbrack {a, b}\right\rbrack \rightarrow {\mathbb{R}}^{3} \) by an admissible change of parameter, then the lengths of the two paths are equa... | Proof. Let \( \widetilde{\Gamma } : \left\lbrack {\alpha ,\beta }\right\rbrack \rightarrow {\mathbb{R}}^{3} \) and \( \Gamma : \left\lbrack {a, b}\right\rbrack \rightarrow {\mathbb{R}}^{3} \) be defined respectively by the triples of smooth functions \( \tau \mapsto \left( {\widetilde{x}\left( \tau \right) ,\widetilde{... | Yes |
Let us find the length of the ellipse defined by the canonical equation\n\n\\[ \n\\frac{{x}^{2}}{{a}^{2}} + \\frac{{y}^{2}}{{b}^{2}} = 1\\;\\left( {a \\geq b > 0}\\right) .\n\\]\n\n(6.56) | Taking the parametrization \\( x = a\\sin \\psi, y = b\\cos \\psi ,0 \\leq \\psi \\leq {2\\pi } \\), we obtain\n\n\\[ \nl = \\mathop{\\int }\\limits_{0}^{{2\\pi }}\\sqrt{{\\left( a\\cos \\psi \\right) }^{2} + {\\left( -b\\sin \\psi \\right) }^{2}}\\mathrm{\\;d}\\psi = \\mathop{\\int }\\limits_{0}^{{2\\pi }}\\sqrt{{a}^{... | Yes |
Let us use formula (6.57) to compute the area of the ellipse given by the canonical equation (6.56). | By the symmetry of the figure and the assumed additivity of areas, it suffices to find the area of just the part of the ellipse in the first quadrant, then quadruple the result. Here are the computations:\n\n\[ S = 4{\int }_{0}^{a}\sqrt{{b}^{2}\left( {1 - \frac{{x}^{2}}{{a}^{2}}}\right) }\mathrm{d}x = {4b}{\int }_{0}^{... | Yes |
By revolving about the \( x \) -axis the semicircle bounded by the closed interval \( \left\lbrack {-R, R}\right\rbrack \) of the axis and the arc of the circle \( y = \sqrt{{R}^{2} - {x}^{2}} \) , \( - R \leq x \leq R \), one can obtain a three-dimensional ball of radius \( R \) whose volume is easily computed from (6... | \[ V = \pi {\int }_{-R}^{R}\left( {{R}^{2} - {x}^{2}}\right) \mathrm{d}x = \frac{4}{3}\pi {R}^{3}. \] | Yes |
The work that must be performed against the force of gravity to lift a body of mass \( m \) vertically from height \( {h}_{1} \) above the surface of the Earth to height \( {h}_{2} \) is, by the definition just given, \( {mg}\left( {{h}_{2} - {h}_{1}}\right) \) . | It is assumed that the entire operation occurs near the surface of the Earth, so that the variation of the gravitational force \( {mg} \) can be neglected. | Yes |
Suppose we have a perfectly elastic spring, one end of which is attached at the point 0 of the real line, while the other is at the point \( x \) . It is known that the force necessary to hold this end of the spring is \( {kx} \), where \( k \) is the modulus of the spring.\n\nLet us compute the work that must be done ... | Regarding the work \( A\left( {\alpha ,\beta }\right) \) as an additive function of the interval \( \left\lbrack {\alpha ,\beta }\right\rbrack \) and assuming valid the estimates\n\n\[ \n\mathop{\inf }\limits_{{x \in \left\lbrack {\alpha ,\beta }\right\rbrack }}\left( {kx}\right) \left( {\beta - \alpha }\right) \leq A\... | Yes |
We begin by remarking that by analogy with the function (6.60), which was written for a particular mechanical system satisfying Eq. (6.59), one can verify that for an arbitrary equation of the form\n\n\\[ \n\\ddot{s}\\left( t\\right) = f\\left( {s\\left( t\\right) }\\right) \n\\]\n\n(6.61)\n\nwhere \\( f\\left( s\\righ... | Indeed,\n\n\\[ \n\\frac{\\mathrm{d}E}{\\mathrm{\\;d}t} = \\frac{1}{2}\\frac{\\mathrm{d}{\\dot{s}}^{2}}{\\mathrm{\\;d}t} + \\frac{\\mathrm{d}U\\left( s\\right) }{\\mathrm{d}t} = \\dot{s}\\ddot{s} + \\frac{\\mathrm{d}U}{\\mathrm{\\;d}s} \\cdot \\frac{\\mathrm{d}s}{\\mathrm{\\;d}t} = \\dot{s}\\left( {\\ddot{s} - f\\left( ... | Yes |
Let us investigate the values of the parameter \( \alpha \) for which the improper integral\n\n\[ \n{\int }_{1}^{+\infty }\frac{\mathrm{d}x}{{x}^{\alpha }} \n\]\n\n(6.69)\n\nconverges, or what is the same, is defined. | Since\n\[ \n{\int }_{1}^{b}\frac{\mathrm{d}x}{{x}^{\alpha }} = \left\{ \begin{matrix} {\left. \frac{1}{1 - \alpha }{x}^{1 - \alpha }\right| }_{1}^{b}\text{ for }\alpha \neq 1, \\ {\left. \ln x\right| }_{1}^{b}\;\text{ for }\alpha = 1, \end{matrix}\right. \n\]\n\nthe limit\n\n\[ \n\mathop{\lim }\limits_{{b \rightarrow +... | Yes |
Let us investigate the values of the parameter \( \alpha \) for which the integral\n\n\[ \n{\int }_{0}^{1}\frac{\mathrm{d}x}{{x}^{\alpha }}\n\]\n\nconverges. | Since for \( a \in \rbrack 0,1\rbrack \)\n\n\[ \n{\int }_{a}^{1}\frac{\mathrm{d}x}{{x}^{\alpha }} = \left\{ \begin{matrix} {\left. \frac{1}{1 - \alpha }{x}^{1 - \alpha }\right| }_{a}^{1},\text{ if }\alpha \neq 1, \\ {\left. \ln x\right| }_{a}^{1},\;\text{ if }\alpha = 1, \end{matrix}\right.\n\]\n\nit follows that the l... | Yes |
\[ {\int }_{-\infty }^{0}{\mathrm{e}}^{x}\mathrm{\;d}x = \mathop{\lim }\limits_{{a \rightarrow - \infty }}{\int }_{a}^{0}{\mathrm{e}}^{x}\mathrm{\;d}x \] | \[ {\int }_{-\infty }^{0}{\mathrm{e}}^{x}\mathrm{\;d}x = \mathop{\lim }\limits_{{a \rightarrow - \infty }}{\int }_{a}^{0}{\mathrm{e}}^{x}\mathrm{\;d}x = \mathop{\lim }\limits_{{a \rightarrow - \infty }}\left( {\left. {\mathrm{e}}^{x}\right| }_{a}^{0}\right) = \mathop{\lim }\limits_{{a \rightarrow - \infty }}\left( {1 -... | Yes |
Proposition 1. Suppose \( x \mapsto f\left( x\right) \) and \( x \mapsto g\left( x\right) \) are functions defined on an interval \( \lbrack a,\omega \lbrack \) and integrable on every closed interval \( \left\lbrack {a, b}\right\rbrack \subset \lbrack a,\omega \lbrack \). Suppose the improper integrals\n\n\[ \n{\int }... | Proof. Part a) follows from the continuity of the function\n\n\[ \n\mathcal{F}\left( b\right) = {\int }_{a}^{b}f\left( x\right) \mathrm{d}x \n\]\n\non the closed interval \( \left\lbrack {a,\omega }\right\rbrack \) on which \( f \in \mathcal{R}\left\lbrack {a,\omega }\right\rbrack \) .\n\nPart b) follows from the fact ... | Yes |
Proposition 2. (Cauchy criterion for convergence of an improper integral). If the function \( x \mapsto f\left( x\right) \) is defined on the interval \( \lbrack a,\omega \lbrack \) and integrable on every closed interval \( \left\lbrack {a, b}\right\rbrack \subset \lbrack a,\omega \lbrack \), then the integral \( {\in... | Proof. As a matter of fact, we have\n\n\[ {\int }_{{b}_{1}}^{{b}_{2}}f\left( x\right) \mathrm{d}x = {\int }_{a}^{{b}_{2}}f\left( x\right) \mathrm{d}x - {\int }_{a}^{{b}_{1}}f\left( x\right) \mathrm{d}x = \mathcal{F}\left( {b}_{2}\right) - \mathcal{F}\left( {b}_{1}\right) ,\]\n\nand therefore the condition is simply the... | Yes |
Proposition 3. If a function \( f \) satisfies the hypotheses of Definition 3 and \( f\left( x\right) \geq 0 \) on \( \lbrack a,\omega \lbrack \), then the improper integral (6.71) exists if and only if the function (6.74) is bounded on \( \lbrack a,\omega \lbrack \) . | Proof. Indeed, if \( f\left( x\right) \geq 0 \) on \( \lbrack a,\omega \lbrack \), then the function (6.74) is nondecreasing on \( \lbrack a,\omega \lbrack \), and therefore it has a limit as \( b \rightarrow \omega, b \in \lbrack a,\omega \lbrack \), if and only if it is bounded. | Yes |
Corollary 1. (Integral test for convergence of a series). If the function \( x \mapsto f\\left( x\\right) \) is defined on the interval \( \\lbrack 1, + \\infty \\lbrack \), nonnegative, nonincreasing, and integrable on each closed interval \( \\left\\lbrack {1, b}\\right\\rbrack \\subset \\lbrack 1, + \\infty \\lbrack... | Proof. It follows from the hypotheses that the inequalities\n\n\[ \nf\\left( {n + 1}\\right) \\leq {\\int }_{n}^{n + 1}f\\left( x\\right) \\mathrm{d}x \\leq f\\left( n\\right) \n\]\n\nhold for any \( n \\in \\mathbb{N} \) . After summing these inequalities, we obtain\n\n\[ \n\\mathop{\\sum }\\limits_{{n = 1}}^{k}f\\lef... | Yes |
Theorem 1. (Comparison theorem). Suppose the functions \( x \mapsto f\left( x\right) \) and \( x \mapsto g\left( x\right) \) are defined on the interval \( \lbrack a,\omega \lbrack \) and integrable on any closed interval \( \left\lbrack {a, b}\right\rbrack \subset \lbrack a,\omega \lbrack \) . If\n\n\[ 0 \leq f\left( ... | Proof. From the hypotheses of the theorem and the inequalities for proper Riemann integrals we have\n\n\[ \mathcal{F}\left( b\right) = {\int }_{a}^{b}f\left( x\right) \mathrm{d}x \leq {\int }_{a}^{b}g\left( x\right) \mathrm{d}x = \mathcal{G}\left( b\right) \]\n\nfor any \( b \in \lbrack a,\omega \lbrack \) . Since both... | Yes |
The integral\n\n\[ \n{\int }^{+\infty }\frac{\sqrt{x}\mathrm{\;d}x}{\sqrt{1 + {x}^{4}}} \n\]\n\nconverges, since | \n\n\[ \n\frac{\sqrt{x}}{\sqrt{1 + {x}^{4}}} \sim \frac{1}{{x}^{3/2}} \n\]\n\nas \( x \rightarrow + \infty \) . | Yes |
The integral\n\n\[ \n{\int }_{1}^{+\infty }\frac{\cos x}{{x}^{2}}\mathrm{\;d}x \n\]\n\nconverges absolutely, since | \[ \n\left| \frac{\cos x}{{x}^{2}}\right| \leq \frac{1}{{x}^{2}} \n\]\n\nfor \( x \geq 1 \) . Consequently,\n\n\[ \n\left| {{\int }_{1}^{+\infty }\frac{\cos x}{{x}^{2}}\mathrm{\;d}x}\right| \leq {\int }_{1}^{+\infty }\left| \frac{\cos x}{{x}^{2}}\right| \mathrm{d}x \leq {\int }_{1}^{+\infty }\frac{1}{{x}^{2}}\mathrm{\;... | Yes |
The integral\n\n\[ \n{\int }_{1}^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x \n\]\n\nconverges, since \( {\mathrm{e}}^{-{x}^{2}} < {\mathrm{e}}^{-x} \) for \( x > 1 \) and | \[ \n{\int }_{1}^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x < {\int }_{1}^{+\infty }{\mathrm{e}}^{-x}\mathrm{\;d}x = \frac{1}{\mathrm{e}}. \n\] | Yes |
Example 7. The integral\n\n\[ \n{\int }^{+\infty }\frac{\mathrm{d}x}{\ln x} \n\]\n\ndiverges, since | \[ \n\frac{1}{\ln x} > \frac{1}{x} \n\]\n\nfor sufficiently large values of \( x \) . | No |
Example 8. The Euler integral\n\n\[ \n{\int }_{0}^{\pi /2}\ln \sin x\mathrm{\;d}x \n\]\n\nconverges, since | \[ \n\left| {\ln \sin x}\right| \sim \left| {\ln x}\right| < \frac{1}{\sqrt{x}} \n\]\n\nas \( x \rightarrow + 0 \) . | Yes |
The elliptic integral\n\n\[ \mathop{\int }\limits_{0}^{1}\frac{\mathrm{d}x}{\sqrt{\left( {1 - {x}^{2}}\right) \left( {1 - {k}^{2}{x}^{2}}\right) }} \]\n\nconverges for \( 0 \leq {k}^{2} < 1 \) | since\n\n\[ \sqrt{\left( {1 - {x}^{2}}\right) \left( {1 - {k}^{2}{x}^{2}}\right) } \sim \sqrt{2\left( {1 - {k}^{2}}\right) }{\left( 1 - x\right) }^{1/2} \]\n\nas \( x \rightarrow 1 - 0 \) . | Yes |
Example 10. The integral\n\n\[ \n{\int }_{0}^{\varphi }\frac{\mathrm{d}\theta }{\sqrt{\cos \theta - \cos \varphi }} \n\] | converges, since\n\n\[ \n\sqrt{\cos \theta - \cos \varphi } = \sqrt{2\sin \frac{\varphi + \theta }{2}\sin \frac{\varphi - \theta }{2}} \sim \sqrt{\sin \varphi }{\left( \varphi - \theta \right) }^{1/2} \n\]\n\nas \( \theta \rightarrow \varphi - 0 \) . | Yes |
The integral\n\n\[ T = 2\sqrt{\frac{L}{g}}{\int }_{0}^{{\varphi }_{0}}\frac{\mathrm{d}\psi }{\sqrt{{\sin }^{2}\frac{{\varphi }_{0}}{2} - {\sin }^{2}\frac{\psi }{2}}} \]\n\nconverges for \( 0 < {\varphi }_{0} < \pi \) | since as \( \psi \rightarrow {\varphi }_{0} - 0 \) we have\n\n\[ \sqrt{{\sin }^{2}\frac{{\varphi }_{0}}{2} - {\sin }^{2}\frac{\psi }{2}} \sim \sqrt{\sin {\varphi }_{0}}{\left( {\varphi }_{0} - \psi \right) }^{1/2}. \] | Yes |
Using Remark 1, by the formula for integration by parts in an improper integral, we find that\n\n\[ \n{\int }_{\pi /2}^{+\infty }\frac{\sin x}{x}\mathrm{\;d}x \n\] | \[ \n{\int }_{\pi /2}^{+\infty }\frac{\sin x}{x}\mathrm{\;d}x = - {\left. \frac{\cos x}{x}\right| }_{\pi /2}^{+\infty } - {\int }_{\pi /2}^{+\infty }\frac{\cos x}{{x}^{2}}\mathrm{\;d}x = - {\int }_{\pi /2}^{+\infty }\frac{\cos x}{{x}^{2}}\mathrm{\;d}x \n\]\nprovided the last integral converges. But, as we saw in Exampl... | Yes |
Proposition 4. (Abel-Dirichlet test for convergence of an integral). Let \( x \mapsto \) \( f\\left( x\\right) \) and \( x \mapsto g\\left( x\\right) \) be functions defined on an interval \( \\lbrack a,\\omega \\lbrack \) and integrable on every closed interval \( \\left\\lbrack {a, b}\\right\\rbrack \\subset \\lbrack... | Proof. For any \( {b}_{1} \) and \( {b}_{2} \) in \( \\lbrack a,\\omega \\lbrack \) we have, by the second mean-value theorem,\n\n\[ \n{\\int }_{{b}_{1}}^{{b}_{2}}\\left( {f \\cdot g}\\right) \\left( x\\right) \\mathrm{d}x = g\\left( {b}_{1}\\right) {\\int }_{{b}_{1}}^{\\xi }f\\left( x\\right) \\mathrm{d}x + g\\left( {... | Yes |
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