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Proposition 6 A function \( f : \rbrack a, b\lbrack \rightarrow \mathbb{R} \) that is differentiable on the open interval \( \rbrack a, b\left\lbrack \right. \) is convex (downward) on \( \rbrack a, b\lbrack \) if and only if its graph contains no points below any tangent drawn to it. In that case, a necessary and suff... | Proof Necessity. Let \( {x}_{0} \in \rbrack a, b\lbrack \) . The equation of the tangent line to the graph at \( \left( {{x}_{0}, f\left( {x}_{0}\right) }\right) \) has the form\n\n\[ y = f\left( {x}_{0}\right) + {f}^{\prime }\left( {x}_{0}\right) \left( {x - {x}_{0}}\right) ,\]\n\nso that\n\n\[ f\left( x\right) - y\le... | Yes |
The function \( f\left( x\right) = {\mathrm{e}}^{x} \) is strictly convex. | The straight line \( y = x + 1 \) is tangent to the graph of this function at \( \left( {0,1}\right) \), since \( f\left( 0\right) = {\mathrm{e}}^{0} = 1 \) and \( {f}^{\prime }\left( 0\right) = \) \( {\left. {\mathrm{e}}^{x}\right| }_{x = 0} = 1 \) . By Proposition 6 we conclude that for any \( x \in \mathbb{R} \)\n\n... | Yes |
When considering the function \( f\left( x\right) = \sin x \) in Example 12 we found the regions of convexity and concavity for its graph. We shall now show that the points of the graph with abscissas \( x = {\pi k}, k \in \mathbb{Z} \), are points of inflection. | Indeed, \( {f}^{\prime \prime }\left( x\right) = - \sin x \), so that \( {f}^{\prime \prime }\left( x\right) = 0 \) at \( x = {\pi k}, k \in \mathbb{Z} \) . Moreover, \( {f}^{\prime \prime }\left( x\right) \) changes sign as we pass through these points, which is a sufficient condition for a point of inflection (see Fi... | Yes |
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. | An example is easy to construct, by improving Example 5, which was given for just this purpose. Let\n\n\[ f\left( x\right) = \left\{ \begin{array}{ll} 2{x}^{3} + {x}^{3}\sin \frac{1}{{x}^{2}} & \text{ for }x \neq 0, \\ 0 & \text{ for }x = 0. \end{array}\right. \]\n\nThen \( {x}^{3} \leq f\left( x\right) \leq 3{x}^{3} \... | Yes |
Proposition 7 (Jensen’s inequality) \( {}^{19} \) If \( f : \rbrack a, b\lbrack \rightarrow \mathbb{R} \) is a convex function, \( \left. {{x}_{1},\ldots ,{x}_{n}\text{are points of}}\right\rbrack a, b\left\lbrack \right. \), and \( {\alpha }_{1},\ldots ,{\alpha }_{n} \) are nonnegative numbers such that\n\n\( {\alpha ... | Proof For \( n = 2 \), condition (5.95) is the same as the definition (5.92) of a convex function.\n\nWe shall now show that if (5.95) is valid for \( n = m - 1 \), it is also valid for \( n = m \) .\n\nFor 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) | \[ {\alpha }_{1}\ln {x}_{1} + \cdots + {\alpha }_{n}\ln {x}_{n} \leq \ln \left( {{\alpha }_{1}{x}_{1} + \cdots + {\alpha }_{n}{x}_{n}}\right) \] \[ {x}_{1}^{{\alpha }_{1}}\cdots {x}_{n}^{{\alpha }_{n}} \leq {\alpha }_{1}{x}_{1} + \cdots + {\alpha }_{n}{x}_{n} \] (5.97) for \( {x}_{i} \geq 0,{\alpha }_{i} \geq 0, i = 1,... | Yes |
Example 18 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}{\... | \[\mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}{b}_{i} \leq {\left( \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{i}^{p}\right) }^{1/p}{\left( \mathop{\sum }\limits_{{i = 1}}^{n}{b}_{i}^{q}\right) }^{1/q}\]\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 |
\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{{x}^{\alpha }} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\left( \frac{1}{x}\right) }{\alpha {x}^{\alpha - 1}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{1}{\alpha {x}^{\alpha }} = 0\;\text{ for }\alpha > 0. \] | \[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\ln x}{{x}^{\alpha }} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{\left( \frac{1}{x}\right) }{\alpha {x}^{\alpha - 1}} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{1}{\alpha {x}^{\alpha }} = 0\;\text{ for }\alpha > 0. \] | 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\[ y = \sin \left( {x}^{2}\right) \] | can be seen in Fig. 5.20.\n\nWe 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 for... | No |
Let us construct the graph of the function\n\n\\[ 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} \\) .\n\nWe now introduce a useful concept.\n\nDefinition 4 The line \\( {c}_{0} + {c}_{1}x \\) is called an asy... | Yes |
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}\\sin... | 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). | No |
Construct the graph of the function \( y = f\left( x\right) \) when\n\n\[ f\left( x\right) = \left| {x + 2}\right| {\mathrm{e}}^{-1/x}. \] | The function \( f\left( x\right) \) is defined for \( x \in \mathbb{R} \smallsetminus 0 \) . Since \( {\mathrm{e}}^{-1/x} \rightarrow 1 \) as \( x \rightarrow \infty \), it follows that\n\n\[ \left| {x + 2}\right| {\mathrm{e}}^{-1/x} \sim \left\{ \begin{array}{ll} - \left( {x + 2}\right) & \text{ as }x \rightarrow - \i... | Yes |
Example 28 Let \( \\left( {x, y}\\right) \) be Cartesian coordinates in the plane and suppose a moving point has coordinates\n\n\[ \nx = \\frac{t}{1 - {t}^{2}},\\;y = \\frac{t - 2{t}^{3}}{1 - {t}^{2}}\n\]\n\nat time \( t\\left( {t \\geq 0}\\right) \) . Describe the trajectory of the point. | We begin by sketching the graphs of each of the two coordinate functions \( x = \) \( x\\left( t\\right) \) and \( y = y\\left( t\\right) \) (Figs. 5.24a and 5.24b).\n\nThe second of these graphs is somewhat more interesting than the first, and so we shall describe how to construct it.\n\nWe can see the behavior of the... | 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 \( n \geq m > N \) . | It follows from the Cauchy criterion and the inequality\n\n\[ \left| {{z}_{m} + \cdots + {z}_{n}}\right| \leq \left| {z}_{m}\right| + \cdots + \left| {z}_{n}\right| \]\n\nthat if the series (5.110) converges absolutely, then it converges. | 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 rearranging \( {}^{24} \) 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 \( \mathop{\sum }\limits_{{n = N + 1}}^{\infty }\left| {z}_{n}\right| < \varepsilon \) .
We then find an index \( K \in \mathb... | 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 su... | 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 \[ \mathop{\sum }\limits_{{m + n = k}}\frac{1}{n!m!}{a}^{n}{b}^{m} = \frac{1}{k!}\mathop{\sum }\limits_{{n = 0}}^{k}\frac{k!}{n!\left( {k - n}\right) !}{a}^{n}{b}^{k - n} = \frac{1}{k!}{\left( a + b\right) }^{k}, \] and therefore we find that \[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{n!}{a}^{n} \cdot \ma... | Yes |
Example 11 See Fig. 5.27. | These correspondences follow from the equalities \( i = {\mathrm{e}}^{{i\pi }/2}, z = r{\mathrm{e}}^{i\varphi } \), and \( {iz} = \) \( r{\mathrm{e}}^{i\left( {r + \pi /2}\right) } \), that is, a rotation through angle \( \frac{\pi }{2} \) has occurred. | No |
Example 14 See Fig. 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 \( z = r{\mathrm{e}}^{i\varphi } \), then by (5.128), we have \( {z}^{n} = {r}^{n}{\mathrm{e}}^{in\varphi } \), 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 happens, at the ver... | 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 now ... | 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, the... | 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 |
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... | 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 |
[ \n\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 = \n] | [ \n= {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) = \n\n= {x}^{2}{\mathrm{e}}^{x} - {2x}{\mathrm{e}}^{x} + 2{\mathrm{e}}^{x} + c = \left( {{x}^{2} - {2x} + 2}\right) {\mathrm{e}}^{x} + c. \n] | Yes |
\[ \int \frac{t\mathrm{\;d}t}{1 + {t}^{2}} = \frac{1}{2}\int \frac{\mathrm{d}\left( {{t}^{2} + 1}\right) }{1 + {t}^{2}} = \frac{1}{2}\int \frac{\mathrm{d}x}{x} = \frac{1}{2}\ln \left| x\right| + c = \frac{1}{2}\ln \left( {{t}^{2} + 1}\right) + c. \] | \[ \int \frac{t\mathrm{\;d}t}{1 + {t}^{2}} = \frac{1}{2}\int \frac{\mathrm{d}\left( {{t}^{2} + 1}\right) }{1 + {t}^{2}} = \frac{1}{2}\int \frac{\mathrm{d}x}{x} = \frac{1}{2}\ln \left| x\right| + c = \frac{1}{2}\ln \left( {{t}^{2} + 1}\right) + c. \] | Yes |
\[ \int \frac{\mathrm{d}x}{\sin x} = \int \frac{\mathrm{d}x}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \int \frac{\mathrm{d}\left( \frac{x}{2}\right) }{\tan \frac{x}{2}{\cos }^{2}\frac{x}{2}} = \] | \[ = \int \frac{\mathrm{d}u}{\tan u{\cos }^{2}u} = \int \frac{\mathrm{d}\left( {\tan u}\right) }{\tan u} = \int \frac{\mathrm{d}v}{v} = \] \[ = \ln \left| v\right| + c = \ln \left| {\tan u}\right| + c = \ln \left| {\tan \frac{x}{2}}\right| + c. \] | 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 \si... | Yes |
\[\int \arcsin x\mathrm{\;d}x = \] | \[\int \arcsin x\mathrm{\;d}x = x\arcsin x - \int x\mathrm{\;d}\arcsin x = \]\n\n\[= x\arcsin x - \int \frac{x}{\sqrt{1 - {x}^{2}}}\mathrm{\;d}x = x\arcsin x + \frac{1}{2}\int \frac{\mathrm{d}\left( {1 - {x}^{2}}\right) }{\sqrt{1 - {x}^{2}}} = \]\n\n\[= x\arcsin x + \frac{1}{2}\int {u}^{-1/2}\mathrm{\;d}u = x\arcsin x ... | Yes |
\[ \int {\mathrm{e}}^{ax}\cos {bx}\mathrm{\;d}x = \frac{1}{a}\int \cos {bx}{\mathrm{{de}}}^{ax} = \] | \[ = \frac{1}{a}{\mathrm{e}}^{ax}\cos {bx} - \frac{1}{a}\int {\mathrm{e}}^{ax}\mathrm{\;d}\cos {bx} = \] \[ = \frac{1}{a}{\mathrm{e}}^{ax}\cos {bx} + \frac{b}{a}\int {\mathrm{e}}^{ax}\sin {bx}\mathrm{\;d}x = \] \[ = \frac{1}{a}{\mathrm{e}}^{ax}\cos {bx} + \frac{b}{{a}^{2}}\int \sin {bx}{\mathrm{{de}}}^{ax} = \] \[ = \f... | 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 |
\[ \int \frac{\mathrm{d}x}{2{\sin }^{2}{3x} - 3{\cos }^{2}{3x} + 1} = \int \frac{\mathrm{d}x}{{\cos }^{2}{3x}\left( {2{\tan }^{2}{3x} - 3 + \left( {1 + {\tan }^{2}{3x}}\right) }\right) } = \] | \[ = \frac{1}{3}\int \frac{\mathrm{d}\tan {3x}}{3{\tan }^{2}{3x} - 2} = \frac{1}{3}\int \frac{\mathrm{d}t}{3{t}^{2} - 2} = \] \[ = \frac{1}{3 \cdot 2}\sqrt{\frac{2}{3}}\int \frac{\mathrm{d}\sqrt{\frac{3}{2}}t}{\frac{3}{2}{t}^{2} - 1} = \] \[ = \frac{1}{3\sqrt{6}}\int \frac{\mathrm{d}u}{{u}^{2} - 1} = \frac{1}{6\sqrt{6}... | 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} - \f... | 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{2t}{1 - {t}^{3}} - 2\int \left( {\frac{1}{3\left( {1 - t}\rig... | Yes |
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 \( \left\lbrack {a, b}\right\rbrack \) . In short, \[ \left( {f \in \mathcal{R}\left\lbrack {a, b}\right\rbrac... | 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 point... | 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 \( \omega \left( {f;\left\lbrack {a, b}\right\rbrack }\right) = \mid f\left( b\right) - \) \( f\left( a\right) \mid \) . Suppose \( \varepsilon > 0 \) is given. We set \( \delta = \frac{\varepsilon }{\left| f\left( b\right)... | 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 ov... | Yes |
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) \rig... | 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 \mat... | 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; | 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\n\nb) This ass... | Yes |
The Dirichlet function\n\n\[ D\\left( x\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{ for }x \\in \\mathbb{Q}, \\\\ 0 & \\text{ for }x \\in \\mathbb{R} \\smallsetminus \\mathbb{Q}, \\end{array}\\right. \] on the interval \\( \\left\\lbrack {0,1}\\right\\rbrack \\) is not integrable on that interval, since for any ... | Then\n\n\[ \\sigma \\left( {f;P,{\\xi }^{\\prime }}\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{n}1 \\cdot \\Delta {x}_{i} = 1, \]\n\nwhile\n\n\[ \\sigma \\left( {f;P,{\\xi }^{\\prime \\prime }}\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{n}0 \\cdot \\Delta {x}_{i} = 0. \]\n\nThus the Riemann sums of the function \... | Yes |
Consider the Riemann function\n\n\[ \mathcal{R}\left( x\right) = \left\{ \begin{array}{ll} \frac{1}{n}, & \text{ if }x \in \mathbb{Q}\text{ and }x = \frac{m}{n}\text{ is in lowest terms,}n \in \mathbb{N}, \\ 0, & \text{ if }x \in \mathbb{R} \smallsetminus \mathbb{Q}. \end{array}\right. \] | We have already studied this function in Sect. 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 the L... | 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} + \beta... | 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 integr... | 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\[ \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\[ \n{\int }_{a}^{b}f\left( x\right) \mathrm{d}x + {\int }_{b}^{c}f\left( x\right) \mathrm{d}x - {\int }_{... | 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) \m... | 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\n\n\[{\int }_{a}^{b}{f}_{1}\left( x\right) \mathrm{d}x \leq {\int }_{a}^{b}{f}_{2}\left( x\right) \m... | Proof For \( a = b \) the assertion is trivial. If \( a < b \), it suffices to write the following inequality for the Riemann sums:\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{f}_{1}\left( {\xi }_{i}\right) \Delta {x}_{i} \leq \mathop{\sum }\limits_{{i = 1}}^{n}{f}_{2}\left( {\xi }_{i}\right) \Delta {x}_{i} \]\n\nwhich i... | 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 pa... | 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 \left... | 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 genera... | 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 \( i = \) \( 1,\ldots, n - 1 \),... | 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}_{1... | 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 \( f, g \in \mathcal{R}\left\lbrack {a, b \mid }\right. \) and \( g \) is a monotonic function on \( \left\lbrack {a, b}\right\rbrack \), then there exists a point \( \xi \in \lbrack a, b \mid \) such that\n\n\[{\int }_{a}^{b}\left( {f \cdot g}\right) \left( x\r... | 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, nondecreasing, and integrable on \( \left\lbrack {a, b}\right\rbrack \) . Applying formula (6.33), we find\n\n\[{\int }_{a}^{b}\left( {f \cdot G}\ri... | 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 (6.40) is differentiable at the point \( x \), and the following equality hol... | 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\l... | 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}\lef... | 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\right... | 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{\int }_{a}^{b}\left( {u \cdot {v}^{\prime }}\right) \left( x\right) \mathrm{d}x = {\left. \left( u \cdot v\right) \right| }_{a}^{b} - {\int }_{a}^{b}\... | 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 continuou... | 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 \( t \\rightarrow {f}^{\\left( n\\right) }\\left( t\\right) \) is continuous on that interval, the first mean-value theorem implies that there exists a point \( \\xi... | 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 \left... | Yes |
\[ {\int }_{-1}^{1}\sqrt{1 - {x}^{2}}\mathrm{\;d}x = {\int }_{-\pi /2}^{\pi /2}\sqrt{1 - {\sin }^{2}t}\cos t\mathrm{\;d}t = {\int }_{-\pi /2}^{\pi /2}{\cos }^{2}t\mathrm{\;d}t = \] | \[ = {\left. \frac{1}{2}{\int }_{-\pi /2}^{\pi /2}\left( 1 + \cos 2t\right) \mathrm{d}t = \frac{1}{2}\left( t + \frac{1}{2}\sin 2t\right) \right| }_{-\pi /2}^{\pi /2} = \frac{\pi }{2}\text{.} \] | Yes |
Example 2 Let us show that a) \( {\int }_{-\pi }^{\pi }\sin {mx}\cos {nx}\mathrm{\;d}x = 0,\; \) b) \( {\int }_{-\pi }^{\pi }{\sin }^{2}{mx}\mathrm{\;d}x = \pi \) , c) \( {\int }_{-\pi }^{\pi }{\cos }^{2}{nx}\mathrm{\;d}x = \pi \) for \( m, n \in \mathrm{N} \) . | a) \[ {\int }_{-\pi }^{\pi }\sin {mx}\cos {nx}\mathrm{\;d}x = \frac{1}{2}{\int }_{-\pi }^{\pi }\left( {\sin \left( {n + m}\right) x - \sin \left( {n - m}\right) x}\right) \mathrm{d}x = \] \[ = {\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\right) \right|... | Yes |
Example 3 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{array}{ll} 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{array}\... | 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 |
Example 4 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 \[ {\... | \[ {\int }_{a}^{a + T}f\left( x\right) \mathrm{d}x = {\int }_{a}^{0}f\left( x\right) \mathrm{d}x + {\int }_{0}^{T}f\left( x\right) \mathrm{d}x + {\int }_{T}^{a + T}f\left( x\right) \mathrm{d}x = \] \[ = {\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.\n\nWe take a different approach. When studying Taylor's formula in different... | 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\left... | 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 | \[ I\left( {\alpha ,\beta }\right) = {\int }_{\alpha }^{\beta }f\left( t\right) \mathrm{d}t \] We remark that in this case the function \( \mathcal{F}\left( x\right) \) is continuous on the closed interval \( \left\lbrack {a, b}\right\rbrack \) . | Yes |
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 1 as follows: the relation\n\n\... | 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 \le... | Yes |
Example 3 Let us test formula (6.52) on a familiar object. Suppose the point moves according to the law\n\n\[ \nx = R\cos {2\pi t} \]\n\n(6.53)\n\n\[ \ny = R\sin {2\pi t} \]\n\nOver the time interval \( \left\lbrack {0,1}\right\rbrack \) the point will traverse a circle of radius \( R \), that is, a path of length \( {... | 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\nDespite the encouraging agreement of the results, the reasoning just carried out ... | 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 equal... | 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{y... | 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 = {\\int }_{0}^{2\\pi }\\sqrt{{\\left( a\\cos \\psi \\right) }^{2} + {\\left( -b\\sin \\psi \\right) }^{2}}\\mathrm{\\;d}\\psi = {\\int }_{0}^{2\\pi }\\sqrt{{a}^{2} - \\left( {{a}^{2} - {b}^{2}}\\ri... | 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 (6.58): | \[ V = \pi {\int }_{-R}^{R}\left( {{R}^{2} - {x}^{2}}\right) \mathrm{d}x = \frac{4}{3}\pi {R}^{3}. \] | 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 |
For an arbitrary equation of the form \(\\ddot{s}\\left( t\\right) = f\\left( {s\\left( t\\right) }\\right)\), where \( f\\left( s\\right) \) is a given function, the sum \(\\frac{{\\dot{s}}^{2}}{2} + U\\left( s\\right) = E\) does not vary over time if \( {U}^{\\prime }\\left( s\\right) = - f\\left( s\\right) \) . | Indeed,\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( s\\r... | 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\[ \n{\int }_{1}^{b}\frac{\mathrm{d}x}{{x}^{\alpha }} = \left\{ \begin{array}{ll} {\left. \frac{1}{1 - \alpha }{x}^{1 - \alpha }\right| }_{1}^{b} & \text{ for }\alpha \neq 1, \\ {\left. \ln x\right| }_{1}^{b} & \text{ for }\alpha = 1, \end{array}\right.\n\]\n\nthe limit\n\n\[ \n\mathop{\lim }\limits_{{b \right... | 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\[ \n{\int }_{a}^{1}\frac{\mathrm{d}x}{{x}^{\alpha }} = \left\{ \begin{array}{ll} {\left. \frac{1}{1 - \alpha }{x}^{1 - \alpha }\right| }_{a}^{1}, & \text{ if }\alpha \neq 1, \\ {\left. \ln x\right| }_{a}^{1}, & \text{ if }\alpha = 1, \end{array}\right. \n\]\n\nit follows tha... | 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 \( x \rightarrow g\left( x\right) \) 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{\i... | 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 \( f \in \mathcal{R}\left\lbrack {a,\omega }\right\rbrack \) . | 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 \( {\int ... | 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 \( \left\lbrack {1, b}\right\rbrack \subset \lbrack 1, + \infty \lbrack \), then the s... | Proof It follows from the hypotheses that the inequalities\n\n\[ f\left( {n + 1}\right) \leq {\int }_{n}^{n + 1}f\left( x\right) \mathrm{d}x \leq f\left( n\right) \]\n\nhold for any \( n \in \mathbb{N} \) . After summing these inequalities, we obtain\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{k}f\left( {n + 1}\right) \leq ... | Yes |
Theorem 1 (Comparison theorem) Suppose the functions \( x \rightarrow f\\left( x\\right) \) and \( x \rightarrow g\\left( x\\right) \) are defined on the interval \( \\lbrack a,\\omega \\lbrack \) and integrable on any closed interval \( \\left\\lbrack {a, b}\\right\\rbrack \\subset \) \( \\lbrack a,\\omega \\lbrack \)... | 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 \\lbra... | Yes |
Example 4 The integral\n\n\[ \n{\int }^{+\infty }\frac{\sqrt{x}\mathrm{\;d}x}{\sqrt{1 + {x}^{4}}} \n\] | converges, 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 |
Example 6 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\[ \n{\int }^{+\infty }\frac{\mathrm{d}x}{\ln x} \n\] | diverges, since\n\n\[ \n\frac{1}{\ln x} > \frac{1}{x} \n\]\n\nfor sufficiently large values of \( x \) . | Yes |
Example 8 The Euler integral\n\n\\[ \n{\\int }_{0}^{\\pi /2}\\ln \\sin x\\mathrm{\\;d}x \n\\]\n\nconverges, | since\n\n\\[ \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\[ \n{\\int }_{0}^{1}\\frac{\\mathrm{d}x}{\\sqrt{\\left( {1 - {x}^{2}}\\right) \\left( {1 - {k}^{2}{x}^{2}}\\right) }} \n\]\n\nconverges for \( 0 \\leq {k}^{2} < 1 \) | since\n\n\[ \n\\sqrt{\\left( {1 - {x}^{2}}\\right) \\left( {1 - {k}^{2}{x}^{2}}\\right) } \\sim \\sqrt{2\\left( {1 - {k}^{2}}\\right) }{\\left( 1 - x\\right) }^{1/2} \n\]\n\nas \( x \\rightarrow 1 - 0 \) . | Yes |
The integral\n\n\[ \n{\int }_{0}^{\varphi }\frac{\mathrm{d}\theta }{\sqrt{\cos \theta - \cos \varphi }} \n\]\n\nconverges, since | \[ \n\sqrt{\cos \theta - \cos \varphi } = \sqrt{2\sin \frac{\varphi + \theta }{2}\sin \frac{\varphi - \theta }{2}} \sim \sqrt{\sin \varphi }{\left( \varphi - \theta \right) }^{1/2} \n\]\n\nas \( \theta \rightarrow \varphi - 0 \) . | Yes |
Example 11 The integral\n\n\[ T = 2\sqrt{\frac{L}{g}}{\int }_{0}^{{\varphi }_{0}}\frac{\mathrm{d}\psi }{\sqrt{{\sin }^{2}\frac{{\varphi }_{0}}{2}{\sin }^{2}\frac{\varphi }{2}}} \] | converges 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\[ \n{\int }_{\pi /2}^{+\infty }\frac{\sin x}{x}\mathrm{\;d}x \n\] | \[ \n{\int }_{\pi /2}^{+\infty }\frac{\sin x}{x}\mathrm{\;d}x = - {\left. \frac{\cos x}{x}\right| }_{\pi /2}^{+\infty } - {\int }_{\pi /2}^{+\infty }\frac{\cos x}{{x}^{2}}\mathrm{\;d}x = - {\int }_{\pi /2}^{+\infty }\frac{\cos x}{{x}^{2}}\mathrm{\;d}x \n\]\n\nprovided the last integral converges. But, as we saw in Exam... | Yes |
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