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| {{Calculus |Integral}}
| | If your computer is running slow, you have possibly gone through the different stages of rage plus frustration. Having such a wonderful tool like a computer may appear like a curse along with a blessing at the same time whenever this arises. It is good when it is actually running rapidly plus smooth, however then when it starts acting weird plus slows means down, frustration sets inside. How could anything as great as a computer create a individual so mad?<br><br>Firstly, you should use the Antivirus or security tool plus run a scan on the computer. It can be done that the computer is afflicted with virus or malware which slows down your computer. If there is nothing found in the scanning report, it will be your RAM which cause the problem.<br><br>The 'registry' is a central database which shops information, settings plus choices for the computer. It's actually the most common reason why XP runs slow and in the event you fix this issue, we could make a computer run a lot quicker. The issue is the fact that the 'registry' shops a lot of settings plus details regarding the PC... plus because Windows must employ a lot of of these settings, any corrupted or damaged ones may straight affect the speed of your system.<br><br>The problem with most of the individuals is that they never like to spend income. In the damaged version 1 does not have to pay anything and will download it from internet really easily. It is easy to install also. But, the problem comes when it really is not able to detect all possible viruses, spyware and malware in the program. This is considering it happens to be obsolete inside nature plus refuses to get any standard changes within the site downloaded. Thus, a program is accessible to issues like hacking.<br><br>To fix the issue that is caused by registry error, we need to use a [http://bestregistrycleanerfix.com/regzooka regzooka]. That is the safest plus easiest means for average PC consumers. However there are thousands of registry cleaners available available. You should find a wise 1 which can definitely solve a problem. If you use a terrible 1, you can expect more difficulties.<br><br>Why this problem arises frequently? What are the causes of it? In fact, there are 3 major causes that can lead to the PC freezing problem. To solve the problem, you should take 3 procedures in the following paragraphs.<br><br>To accelerate a computer, you simply should be capable to receive rid of all these junk files, allowing your computer to find exactly what it wants, when it wants. Luckily, there's a tool which enables you to do this easily plus swiftly. It's a tool called a 'registry cleaner'.<br><br>Registry cleaners could assist your computer run inside a better mode. Registry cleaners ought to be part of the standard scheduled repair system for a computer. You don't have to wait forever for the computer or the programs to load plus run. A little repair may bring back the speed we lost. |
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| In [[calculus]], '''integration by substitution''', also known as '''u-substitution''', is a method for finding [[integral]]s. Using the [[fundamental theorem of calculus]] often requires finding an antiderivative. For this and other reasons, integration by substitution is an important tool for mathematicians. It is the counterpart to the [[chain rule]] of [[derivative|differentiation]].
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| == Substitution for single variable ==
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| === Relation to the fundamental theorem of calculus ===
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| Let {{math|''I'' ⊆ ℝ}} be an interval and {{math|''{{Unicode|ϕ}}'' : [''a'',''b''] → ''I''}} be a [[smooth function|continuously differentiable]] function. Suppose that {{math|''{{Unicode|ƒ}}'' : ''I'' → ℝ}} is a [[continuous function]]. Then
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| :<math>
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| \int_{\phi(a)}^{\phi(b)} f(x)\,dx = \int_a^b f(\phi(t))\phi'(t)\, dt.
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| </math>
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| Using [[Leibniz notation]]: the substitution {{math|''x'' {{=}} ''{{Unicode|ϕ}}''(''t'')}} yields {{math|''{{sfrac|dx|dt}}'' {{=}} {{Unicode|''ϕ''′}}(''t'')}} and thus, formally, {{math|''dx'' {{=}} {{unicode|''ϕ''′}}(''t'') ''dt''}}, which is the required substitution for {{math|''dx''}}. (One could view the method of integration by substitution as a major justification of Leibniz's notation for integrals and derivatives.)
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| The formula is used to transform one integral into another integral that is easier to compute. Thus, the formula can be used from left to right or from right to left in order to simplify a given integral. When used in the latter manner, it is sometimes known as '''u-substitution''' or '''w-substitution'''.
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| Integration by substitution can be derived from the [[fundamental theorem of calculus]] as follows. Let {{math|''ƒ''}} and {{math|''{{unicode|ϕ}}''}} be two functions satisfying the above hypothesis that {{math|''ƒ''}} is continuous on {{math|''I''}} and {{math|{{unicode|''ϕ''′}}}} is continuous on the closed interval {{math|[''a'',''b'']}}. Then the function {{math|''ƒ''(''{{unicode|ϕ}}''(''t'')){{unicode|''ϕ''′}}(''t'')}} is also continuous on {{math|[''a'',''b'']}}. Hence the integrals
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| :<math>
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| \int_{\phi(a)}^{\phi(b)} f(x)\,dx
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| </math> | |
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| and
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| :<math>
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| \int_a^b f(\phi(t))\phi'(t)\,dt
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| </math>
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| in fact exist, and it remains to show that they are equal.
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| Since {{math|''ƒ''}} is continuous, it possesses an [[antiderivative]] {{math|''F''}}. The [[function composition|composite function]] {{math|''F''∘{{unicode|''ϕ''}}}} is then defined. Since {{math|''F''}} and {{math|{{unicode|''ϕ''}}}} are differentiable, the [[chain rule]] gives
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| :<math>
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| (F \circ \phi)'(t) = F'(\phi(t))\phi'(t) = f(\phi(t))\phi'(t).
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| </math>
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| Applying the [[fundamental theorem of calculus]] twice gives
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| : <math>
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| \begin{align}
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| \int_a^b f(\phi(t))\phi'(t)\,dt & {} = (F \circ \phi)(b) - (F \circ \phi)(a) \\
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| & {} = F(\phi(b)) - F(\phi(a)) \\
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| & {} = \int_{\phi(a)}^{\phi(b)} f(x)\,dx,
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| \end{align}
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| </math>
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| which is the substitution rule.
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| === Examples ===
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| Consider the integral
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| :<math>
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| \int_{0}^2 x \cos(x^2+1) \,dx
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| </math>
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| If we make the substitution ''u'' = ''x''<sup>2</sup> + 1, we obtain ''du'' = 2''x'' ''dx'' and thence ''x'' ''dx'' = ½''du''
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| '''(1) Definite integral'''<br>
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| : <math>
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| \begin{align}
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| \int_{x=0}^{x=2} x \cos(x^2+1) \,dx & {} = \frac{1}{2} \int_{u=1}^{u=5}\cos(u)\,du \\
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| & {} = \frac{1}{2}(\sin(5)-\sin(1)).
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| \end{align}
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| </math>
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| It is important to note that since the lower limit ''x'' = 0 was replaced with ''u'' = 0<sup>2</sup> + 1 = 1, and the upper limit ''x'' = 2 replaced with ''u'' = 2<sup>2</sup> + 1 = 5, a transformation back into terms of ''x'' was unnecessary.
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| For the integral
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| :<math>
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| \int_0^1 \sqrt{1-x^2}\; dx
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| </math>
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| the formula needs to be used from right to left: | |
| the substitution ''x'' = sin(''u''), ''dx'' = cos(''u'') ''du'' is useful, because <math>\sqrt{(1-\sin^2(u))} = \cos(u)</math>:
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| :<math>
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| \int_0^1 \sqrt{1-x^2}\; dx = \int_0^\frac{\pi}{2} \sqrt{1-\sin^2(u)} \cos(u)\;du = \int_0^\frac{\pi}{2} \cos^2(u)\;du=\frac{\pi}{4}
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| </math>
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| The resulting integral can be computed using [[integration by parts]] or a [[List of trigonometric identities#Double-.2C_triple-.2C_and_half-angle_formulae|double angle formula]] followed by one more substitution. One can also note that the function being integrated is the upper right quarter of a circle with a radius of one, and hence integrating the upper right quarter from zero to one is the geometric equivalent to the area of one quarter of the unit circle, or π/4.
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| '''(2) Antiderivatives''' <br>
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| Substitution can be used to determine [[antiderivative]]s. One chooses a relation between ''x'' and ''u'', determines the corresponding relation between ''dx'' and ''du'' by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between ''u'' and ''x'' is then undone.
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| Similar to our first example above, we can determine the following antiderivative with this method:
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| :<math>
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| \begin{align}
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| & {} \quad \int x \cos(x^2+1) \,dx = \frac{1}{2} \int 2x \cos(x^2+1) \,dx \\
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| & {} = \frac{1}{2} \int\cos u\,du = \frac{1}{2}\sin u + C = \frac{1}{2}\sin(x^2+1) + C
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| \end{align}
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| </math>
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| where ''C'' is an arbitrary [[constant of integration]].
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| Note that there were no integral boundaries to transform, but in the last step we had to revert the original substitution ''u'' = ''x''<sup>2</sup> + 1.
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| == Substitution for multiple variables ==
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| One may also use substitution when integrating functions of several variables.
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| Here the substitution function (''v''<sub>1</sub>,...,''v''<sub>''n''</sub>) = ''φ''(''u''<sub>1</sub>, ..., ''u''<sub>''n''</sub> ) needs to be [[injective]] and continuously differentiable, and the differentials transform as
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| :<math>dv_1 \cdots dv_n = |\det(\operatorname{D}\phi)(u_1, \ldots, u_n)| \, du_1 \cdots du_n</math> | |
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| where det(D''φ'')(''u''<sub>1</sub>, ..., ''u''<sub>''n''</sub> ) denotes the [[determinant]] of the [[Jacobian matrix]] containing the [[partial derivative]]s of ''φ''. This formula expresses the fact that the [[absolute value]] of the determinant of a matrix equals the volume of the [[parallelotope]] spanned by its columns or rows.
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| More precisely, the ''[[change of variables]]'' formula is stated in the next theorem:
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| '''Theorem'''. Let ''U'' be an open set in '''R'''<sup>''n''</sup> and ''φ'' : ''U'' → '''R'''<sup>''n''</sup> an [[Injective function|injective]] differentiable function with continuous partial derivatives, the Jacobian of which is nonzero for every ''x'' in ''U''. Then for any real-valued, compactly supported, continuous function ''f'', with support contained in ''φ''(''U''),
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| :<math> \int_{\phi(U)} f(\mathbf{v})\, d \mathbf{v} = \int_U f(\phi(\mathbf{u})) \left|\det(\operatorname{D}\phi)(\mathbf{u})\right| \,d \mathbf{u}.</math>
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| The conditions on the theorem can be weakened in various ways. First, the requirement that φ be continuously differentiable can be replaced by the weaker assumption that φ be merely differentiable and have a continuous inverse {{harv|Rudin|1987|loc=Theorem 7.26}}. This is guaranteed to hold if φ is continuously differentiable by the [[inverse function theorem]]. Alternatively, the requirement that Det(Dφ)≠0 can be eliminated by applying [[Sard's theorem]] {{harv|Spivak|1965}}.
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| For Lebesgue measurable functions, the theorem can be stated in the following form {{harv|Fremlin|2010|loc=Theorem 263D}}:
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| '''Theorem'''. Let ''U'' be a measurable subset of '''R'''<sup>''n''</sup> and ''φ'' : ''U'' → '''R'''<sup>''n''</sup> an [[Injective function|injective]] function, and suppose for every ''x'' in ''U'' there exists ''<span>φ'</span>''(''x'') in '''R'''<sup>''n'',''n''</sup> such that ''φ''(''y'') = ''φ''(''x'') + ''<span>φ'</span>''(''x'') (''y'' − ''x'') + [[Landau_symbol#Related_asymptotic_notations|''o'']](||''y'' − ''x''||) as ''y'' → ''x''. Then ''φ''(''U'') is measurable, and for any real-valued function ''f'' defined on ''φ''(''U''),
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| :<math> \int_{\phi(U)} f(v)\, dv \;=\; \int_U f(\phi(u)) \; \left|\det \phi'(u)\right| \,du</math>
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| in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.
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| Another very general version in [[measure theory]] is the following {{harv|Hewitt|Stromberg|1965|loc=Theorem 20.3}}:
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| '''Theorem'''. Let ''X'' be a [[locally compact]] [[Hausdorff space]] equipped with a finite [[Radon measure]] μ, and let ''Y'' be a [[Σ-compact space|σ-compact]] Hausdorff space with a [[sigma finite measure|σ-finite]] Radon measure ρ. Let ''φ'' : ''X'' → ''Y'' be a [[continuous function|continuous]] and [[absolutely continuous]] function (where the latter means that ρ(''φ''(''E'')) = 0 whenever μ(''E'') = 0). Then there exists a real-valued [[Borel algebra|Borel measurable function]] ''w'' on ''X'' such that for every [[Lebesgue integral|Lebesgue integrable]] function ''f'' : ''Y'' → '''R''', the function (''f'' <math> \circ </math> ''φ'')''w'' is Lebesgue integrable on ''X'', and
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| :<math>\int_Y f(y)\,d\rho(y) = \int_X f\circ\phi(x)w(x)\,d\mu(x).</math>
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| Furthermore, it is possible to write
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| :<math>w(x) = g\circ\phi(x)</math>
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| for some Borel measurable function ''g'' on ''Y''.
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| In [[geometric measure theory]], integration by substitution is used with [[Lipschitz function]]s. A bi-Lipschitz function is a Lipschitz function ''φ'' : ''U'' → '''R'''<sup>n</sup> which is one-to-one, and such that its inverse function ''φ''<sup>−1</sup> : ''φ''(''U'') → ''U'' is also Lipschitz. By [[Rademacher's theorem]] a bi-Lipschitz mapping is differentiable [[almost everywhere]]. In particular, the Jacobian determinant of a bi-Lipschitz mapping det D'''φ'' is well-defined almost everywhere. The following result then holds:
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| '''Theorem.''' Let ''U'' be an open subset of '''R'''<sup>n</sup> and ''φ'' : ''U'' → '''R'''<sup>n</sup> be a bi-Lipschitz mapping. Let ''f'' : ''φ''(''U'') → '''R''' be measurable. Then | |
| :<math>\int_U (f\circ \phi)|\det D\phi| = \int_{\phi(U)}f</math>
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| in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.
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| The above theorem was first proposed by [[Euler]] when he developed the notion of [[double integrals]] in 1769. Although generalized to triple integrals by [[Lagrange]] in 1773, and used by [[Adrien-Marie Legendre|Legendre]], [[Laplace]], [[Gauss]], and first generalized to ''n'' variables by [[Mikhail Ostrogradski]] in 1836, it resisted a fully rigorous formal proof for a surprisingly long time, and was first satisfactorily resolved 125 years later, by [[Élie Cartan]] in a series of papers beginning in the mid-1890s ({{harnvb|Katz|1982}}; {{harvnb|Ferzola|1994}}).
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| ==Application in probability==
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| Substitution can be used to answer the following important question in probability: given a random variable <math>X</math> with probability density <math>p_x</math> and another random variable <math>Y</math> related to <math>X</math> by the equation <math>y=\phi(x)</math>, what is the probability density for <math>Y</math>?
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| It is easiest to answer this question by first answering a slightly different question: what is the probability that <math>Y</math> takes a value in some particular subset <math>S</math>? Denote this probability <math>P(Y \in S)</math>. Of course, if <math>Y</math> has probability density <math>p_y</math> then the answer is
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| :<math>P(Y \in S) = \int_S p_y(y)\,dy, </math>
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| but this isn't really useful because we don't know ''p''<sub>''y''</sub>; it's what we're trying to find in the first place. We can make progress by considering the problem in the variable <math>X</math>. <math>Y</math> takes a value in ''S'' whenever ''X'' takes a value in <math>\phi^{-1}(S)</math>, so
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| :<math> P(Y \in S) = \int_{\phi^{-1}(S)} p_x(x)\,dx. </math>
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| Changing from variable ''x'' to ''y'' gives
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| :<math>
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| P(Y \in S) = \int_{\phi^{-1}(S)} p_x(x)~dx = \int_S p_x(\phi^{-1}(y)) ~ \left|\frac{d\phi^{-1}}{dy}\right|~dy. </math>
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| Combining this with our first equation gives
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| :<math>
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| \int_S p_y(y)~dy = \int_S p_x(\phi^{-1}(y)) ~ \left|\frac{d\phi^{-1}}{dy}\right|~dy </math>
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| so
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| :<math>
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| p_y(y) = p_x(\phi^{-1}(y)) ~ \left|\frac{d\phi^{-1}}{dy}\right|. </math>
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| In the case where <math>X</math> and <math>Y</math> depend on several uncorrelated variables, i.e. <math>p_x=p_x(x_1\ldots x_n)</math>, and <math>y=\phi(x)</math>, <math>p_y</math> can be found by substitution in several variables discussed above. The result is
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| :<math>
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| p_y(y) = p_x(\phi^{-1}(y)) ~ \left|\det \left[ D\phi ^{-1}(y) \right] \right|. </math>
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| ==See also==
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| {{Wikiversity|Integration by Substitution}}
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| {{Wikibooks|Calculus|Integration#The_Substitution_Rule|The Substitution Rule}}
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| *[[Substitution of variables]]
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| *[[Probability density function]]
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| *[[Tangent half-angle substitution]]
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| ==References==
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| * {{citation|first1=Edwin|last1=Hewitt|first2=Karl|last2=Stromberg|title=Real and abstract analysis|publisher=Springer-Verlag|year=1965|isbn=978-0-387-04559-7}}.
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| *{{citation|first=Anthony P.|last=Ferzola|url=http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=2688|title=Euler and differentials|journal=The College Mathematics Journal|volume=25|issue=2|year=1994|pages=102–111|doi=10.2307/2687130}}
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| * {{citation|first=D.H.|last=Fremlin|title=Measure Theory, Volume 2|publisher=Torres Fremlin|year=2010|isbn=978-0-9538129-7-4}}.
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| * {{citation|first=V.|last=Katz|title=Change of variables in multiple integrals: Euler to Cartan|journal=Mathematics Magazine|volume=55|year=1982|pages=3–11|doi=10.2307/2689856|issue=1}}
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| * {{citation|first=Walter|last=Rudin|authorlink=Walter Rudin|title=Real and complex analysis|publisher=McGraw-Hill|year=1987|isbn=978-0-07-054234-1}}.
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| * {{citation|first=Michael|last=Spivak|authorlink=Michael Spivak|title=Calculus on manifolds|publisher=Westview Press|year=1965|isbn=978-0-8053-9021-6}}.
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| ==Extternal links==
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| * [http://www.encyclopediaofmath.org/index.php/Integration_by_substitution Integration by substitution] at [http://www.encyclopediaofmath.org/ Encyclopedia of Mathematics]
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| * [http://www.encyclopediaofmath.org/index.php/Area_formula Area formula] at [http://www.encyclopediaofmath.org/ Encyclopedia of Mathematics]
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| [[Category:Articles containing proofs]]
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| [[Category:Integral calculus]]
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| [[es:Métodos de integración#Método de integración por sustitución]]
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If your computer is running slow, you have possibly gone through the different stages of rage plus frustration. Having such a wonderful tool like a computer may appear like a curse along with a blessing at the same time whenever this arises. It is good when it is actually running rapidly plus smooth, however then when it starts acting weird plus slows means down, frustration sets inside. How could anything as great as a computer create a individual so mad?
Firstly, you should use the Antivirus or security tool plus run a scan on the computer. It can be done that the computer is afflicted with virus or malware which slows down your computer. If there is nothing found in the scanning report, it will be your RAM which cause the problem.
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The problem with most of the individuals is that they never like to spend income. In the damaged version 1 does not have to pay anything and will download it from internet really easily. It is easy to install also. But, the problem comes when it really is not able to detect all possible viruses, spyware and malware in the program. This is considering it happens to be obsolete inside nature plus refuses to get any standard changes within the site downloaded. Thus, a program is accessible to issues like hacking.
To fix the issue that is caused by registry error, we need to use a regzooka. That is the safest plus easiest means for average PC consumers. However there are thousands of registry cleaners available available. You should find a wise 1 which can definitely solve a problem. If you use a terrible 1, you can expect more difficulties.
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To accelerate a computer, you simply should be capable to receive rid of all these junk files, allowing your computer to find exactly what it wants, when it wants. Luckily, there's a tool which enables you to do this easily plus swiftly. It's a tool called a 'registry cleaner'.
Registry cleaners could assist your computer run inside a better mode. Registry cleaners ought to be part of the standard scheduled repair system for a computer. You don't have to wait forever for the computer or the programs to load plus run. A little repair may bring back the speed we lost.