2.2 Substitution

From Förberedande kurs i matematik 2

(Difference between revisions)
Jump to: navigation, search
m (Robot: Automated text replacement (-2.2 Variabelsubstitution +2.2 Substitution))
Current revision (14:43, 7 January 2009) (edit) (undo)
(Edited for UK idiom and terminology. Case for a section on substitutions of the form x = f(t)?)
 
(6 intermediate revisions not shown.)
Line 2: Line 2:
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" |  
| style="border-bottom:1px solid #797979" width="5px" |  
-
{{Vald flik|[[2.2 Substitution|Theory]]}}
+
{{Selected tab|[[2.2 Substitution|Theory]]}}
-
{{Ej vald flik|[[2.2 Övningar|Exercises]]}}
+
{{Not selected tab|[[2.2 Exercises|Exercises]]}}
| style="border-bottom:1px solid #797979" width="100%"|  
| style="border-bottom:1px solid #797979" width="100%"|  
|}
|}
Line 17: Line 17:
After this section, you will have learned to:
After this section, you will have learned to:
-
* Understand the derivation of the formula for variable substitution .
+
* Understand the derivation of the formula for substitution .
* Solve simple integration problems that require rewriting and / or substitution in one of the steps.
* Solve simple integration problems that require rewriting and / or substitution in one of the steps.
* Know how the limits of integration are to be changed after a variable substitution.
* Know how the limits of integration are to be changed after a variable substitution.
-
* Know when variable substitution is allowed.
+
* Know when substitution is allowed.
}}
}}
-
== Variable substitution ==
+
== Substitution, or change of variable ==
-
When you cannot directly determine a primitive function by using the usual rules of differentiation ”in the opposite direction” method, other methods or techniques are needed. One such is ''variable substitution'', which can be said to be based on the rule for the differentiation of composite functions — the so-called ''chain rule''.
+
When you cannot directly determine an indefinite integral by inspection (that is, by simple "differentiation in reverse"), other methods or techniques are needed. One such is ''substitution'' (sometimes called ''change of variable''), which can be said to be based on the rule for the differentiation of composite functions — the so-called ''chain rule''.
-
The chain rule <math>\ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \cdot u'(x)\ </math> can be written in integral form as
+
The chain rule <math>\ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \, u'(x)\ </math> can be written in integral form as
-
{{Fristående formel||<math>\int f^{\,\prime}(u(x)) \cdot u'(x) \, dx = f(u(x)) + C</math>}}
+
{{Displayed math||<math>\int f^{\,\prime}(u(x)) \, u'(x) \, dx = f(u(x)) + C</math>}}
or,
or,
<div class="regel">
<div class="regel">
-
{{Fristående formel||<math>\int f(u(x)) \cdot u'(x) \, dx = F (u(x)) + C\,\mbox{,}</math>}}
+
{{Displayed math||<math>\int f(u(x)) \, u'(x) \, dx = F (u(x)) + C\,\mbox{,}</math>}}
</div>
</div>
where ''F'' is a primitive function of ''f''. We compare this with the formula
where ''F'' is a primitive function of ''f''. We compare this with the formula
-
{{Fristående formel||<math>\int f(u) \, du = F(u) + C\,\mbox{.}</math>}}
+
{{Displayed math||<math>\int f(u) \, du = F(u) + C\,\mbox{.}</math>}}
-
We can see that we have replaced the term <math>u(x)</math> with variable <math>u</math> and the <math>u'(x)\, dx</math> with <math>du</math>. One thus can transform the more complicated integrand <math>f(u(x)) \cdot u'(x)</math> (with <math>x</math> as the variable) to the, let us hope, easier <math>f(u)</math> (with the <math>u</math> as the variable). The method is called variable substitution and can be used when the integrand can be written in the form <math>f(u(x)) \cdot u'(x)</math>.
+
We can see that we have replaced the term <math>u(x)</math> with variable <math>u</math> and the <math>u'(x)\, dx</math> with <math>du</math>. One thus can transform the more complicated integrand <math>f(u(x)) \, u'(x)</math> (with <math>x</math> as the variable) to the simpler (and possibly more tractable) <math>f(u)</math> (with the <math>u</math> as the variable). The method is called substitution, or change of variable, and can be used when the integrand can be written in the form <math>f(u(x)) \, u'(x)</math>.
Line 48: Line 48:
'' Note 2'' Replacing <math>u'(x) \, dx</math> with <math>du</math> also may be justified by studying the transition from the increment ratio to the derivative:
'' Note 2'' Replacing <math>u'(x) \, dx</math> with <math>du</math> also may be justified by studying the transition from the increment ratio to the derivative:
-
{{Fristående formel||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}}
+
{{Displayed math||<math>\lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}</math>}}
which, as <math>\Delta x</math> goes towards zero can be considered as a formal transition between variables
which, as <math>\Delta x</math> goes towards zero can be considered as a formal transition between variables
-
{{Fristående formel||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}}
+
{{Displayed math||<math>\Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}</math>}}
ie., a small change, <math>dx</math>, in the variable <math>x</math> gives rise to an approximate change <math>u'(x)\,dx</math> in the variable <math>u</math>.
ie., a small change, <math>dx</math>, in the variable <math>x</math> gives rise to an approximate change <math>u'(x)\,dx</math> in the variable <math>u</math>.
Line 65: Line 65:
If one puts <math>u(x)= x^2</math>, one gets <math>u'(x)= 2x</math>. The variable substitution replaces <math>e^{x^2}</math> with <math>e^u</math> and <math>u'(x)\,dx</math>, i.e. <math>2x\,dx</math>, with <math>du</math>
If one puts <math>u(x)= x^2</math>, one gets <math>u'(x)= 2x</math>. The variable substitution replaces <math>e^{x^2}</math> with <math>e^u</math> and <math>u'(x)\,dx</math>, i.e. <math>2x\,dx</math>, with <math>du</math>
-
{{Fristående formel||<math> \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \cdot 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}</math>}}
+
{{Displayed math||<math> \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \times 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}</math>}}
</div>
</div>
Line 72: Line 72:
''' Example 2'''
''' Example 2'''
-
Determine the integral <math>\ \int (x^3 + 1)^3 \cdot x^2 \, dx</math>.
+
Determine the integral <math>\ \int (x^3 + 1)^3 \, x^2 \, dx</math>.
<br>
<br>
<br>
<br>
Put <math>u=x^3 + 1</math>. This means <math>u'=3x^2</math>, or <math>du= 3x^2\, dx</math>, and
Put <math>u=x^3 + 1</math>. This means <math>u'=3x^2</math>, or <math>du= 3x^2\, dx</math>, and
-
{{Fristående formel||<math>\begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \cdot 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \times 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Line 89: Line 89:
After rewriting <math>\tan x</math> as <math>\sin x/\cos x</math> we substitute <math>u=\cos x</math>,
After rewriting <math>\tan x</math> as <math>\sin x/\cos x</math> we substitute <math>u=\cos x</math>,
-
{{Fristående formel||<math>\begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}</math>}}
</div>
</div>
Line 108: Line 108:
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\,dx</math>
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\,dx</math>
-
{{Fristående formel||<math>\begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}</math>}}
+
{{Displayed math||<math>\begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}</math>}}
Note that the limits of integration must be written in the form <math>x = 0</math> and <math>x = 2</math> when the variable of integration is not <math>x</math>. it is wrong to write
Note that the limits of integration must be written in the form <math>x = 0</math> and <math>x = 2</math> when the variable of integration is not <math>x</math>. it is wrong to write
-
{{Fristående formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ etc.}</math>}}
+
{{Displayed math||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ etc.}</math>}}
Line 119: Line 119:
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\, dx</math>. The limit of integration <math>x=0</math> is equivalent to <math>u=e^0 = 1</math> and <math>x=2</math> is equivalent to <math>u=e^2</math>
Put <math>u=e^x</math> which gives that <math>u'= e^x</math> and <math>du= e^x\, dx</math>. The limit of integration <math>x=0</math> is equivalent to <math>u=e^0 = 1</math> and <math>x=2</math> is equivalent to <math>u=e^2</math>
-
{{Fristående formel||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}</math>}}
+
{{Displayed math||<math>\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}</math>}}
</div>
</div>
Line 131: Line 131:
The substitution <math>u=\sin x</math> gives <math>du=\cos x\,dx</math> and the limits of integration become <math>u=\sin 0=0</math> and <math>u=\sin(\pi/2)=1</math>. The integral is
The substitution <math>u=\sin x</math> gives <math>du=\cos x\,dx</math> and the limits of integration become <math>u=\sin 0=0</math> and <math>u=\sin(\pi/2)=1</math>. The integral is
-
{{Fristående formel||<math>\int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}</math>}}
+
{{Displayed math||<math>\int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}</math>}}
-
<center>{{:2.2 - Figur - Area under y = sin³x cos x resp. y = u³}}</center>
+
<center>{{:2.2 - Figure - The area under y = sin³x cos x and y = u³, respectively}}</center>
{| width="80%" align="center"
{| width="80%" align="center"
||<small> The figure on the left shows the graph of the integrand sin³''x'' cos ''x'' and the figure on the right the graph of integrand ''u''³ which is obtained after the variable substitution. The change of variable modifies the integrand and the interval of the integration. The integrals value, the size of the area, is not changed however. </small>
||<small> The figure on the left shows the graph of the integrand sin³''x'' cos ''x'' and the figure on the right the graph of integrand ''u''³ which is obtained after the variable substitution. The change of variable modifies the integrand and the interval of the integration. The integrals value, the size of the area, is not changed however. </small>
Line 146: Line 146:
Examine the following calculation
Examine the following calculation
-
{{Fristående formel||<math>\int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}</math>}}
+
{{Displayed math||<math>\int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}</math>}}
{| width="100%"
{| width="100%"
Line 156: Line 156:
||
||
{|
{|
-
||{{:2.2 - Figur - Grafen till f(u) = 1/u²}}
+
||{{:2.2 - Figure - The graph of f(u) = 1/u²}}
|-
|-
||<small>Graph of ''f''(''u'') = 1/''u''²</small>
||<small>Graph of ''f''(''u'') = 1/''u''²</small>

Current revision

       Theory          Exercises      

Contents:

  • Integration by substitution

Learning outcomes:

After this section, you will have learned to:

  • Understand the derivation of the formula for substitution .
  • Solve simple integration problems that require rewriting and / or substitution in one of the steps.
  • Know how the limits of integration are to be changed after a variable substitution.
  • Know when substitution is allowed.

Substitution, or change of variable

When you cannot directly determine an indefinite integral by inspection (that is, by simple "differentiation in reverse"), other methods or techniques are needed. One such is substitution (sometimes called change of variable), which can be said to be based on the rule for the differentiation of composite functions — the so-called chain rule.

The chain rule \displaystyle \ \frac{d}{dx}f(u(x)) = f^{\,\prime} (u(x)) \, u'(x)\ can be written in integral form as

\displaystyle \int f^{\,\prime}(u(x)) \, u'(x) \, dx = f(u(x)) + C

or,

\displaystyle \int f(u(x)) \, u'(x) \, dx = F (u(x)) + C\,\mbox{,}

where F is a primitive function of f. We compare this with the formula

\displaystyle \int f(u) \, du = F(u) + C\,\mbox{.}

We can see that we have replaced the term \displaystyle u(x) with variable \displaystyle u and the \displaystyle u'(x)\, dx with \displaystyle du. One thus can transform the more complicated integrand \displaystyle f(u(x)) \, u'(x) (with \displaystyle x as the variable) to the simpler (and possibly more tractable) \displaystyle f(u) (with the \displaystyle u as the variable). The method is called substitution, or change of variable, and can be used when the integrand can be written in the form \displaystyle f(u(x)) \, u'(x).


Note 1 The method is based on the assumption that all the conditions for integration are satisfied; that is, \displaystyle u(x) is differentiable in the interval in question, and that \displaystyle f is continuous for all values of \displaystyle u in the range, that is, for all the values that \displaystyle u can take on in the interval.


Note 2 Replacing \displaystyle u'(x) \, dx with \displaystyle du also may be justified by studying the transition from the increment ratio to the derivative:

\displaystyle \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x} = \frac{du}{dx} = u'(x)\,\mbox{,}

which, as \displaystyle \Delta x goes towards zero can be considered as a formal transition between variables

\displaystyle \Delta u \approx u'(x) \Delta x \quad \to \quad du = u'(x) \, dx\,\mbox{,}

ie., a small change, \displaystyle dx, in the variable \displaystyle x gives rise to an approximate change \displaystyle u'(x)\,dx in the variable \displaystyle u.


Example 1

Determine the integral\displaystyle \ \int 2 x\, e^{x^2} \, dx.

If one puts \displaystyle u(x)= x^2, one gets \displaystyle u'(x)= 2x. The variable substitution replaces \displaystyle e^{x^2} with \displaystyle e^u and \displaystyle u'(x)\,dx, i.e. \displaystyle 2x\,dx, with \displaystyle du

\displaystyle \int 2 x\,e^{x^2} \, dx = \int e^{x^2} \times 2x \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\,\mbox{.}

Example 2

Determine the integral \displaystyle \ \int (x^3 + 1)^3 \, x^2 \, dx.

Put \displaystyle u=x^3 + 1. This means \displaystyle u'=3x^2, or \displaystyle du= 3x^2\, dx, and

\displaystyle \begin{align*}\int (x^3 + 1)^3 x^2 \, dx &= \int \frac{ (x^3 + 1)^3}{3} \times 3x^2\, dx = \int \frac{u^3}{3}\, du\\[4pt] &= \frac{u^4}{12} + C = \frac{1}{12} (x^3 + 1)^4 + C\,\mbox{.}\end{align*}

Example 3

Determine the integral \displaystyle \ \int \tan x \, dx\,\mbox{,}\ \ where \displaystyle -\pi/2 < x < \pi/2.

After rewriting \displaystyle \tan x as \displaystyle \sin x/\cos x we substitute \displaystyle u=\cos x,

\displaystyle \begin{align*}\int \tan x \, dx &= \int \frac{\sin x}{\cos x} \, dx = \left[\,\begin{align*} u &= \cos x\\ u' &= - \sin x\\ du &= - \sin x \, dx \end{align*}\,\right]\\[4pt] &= \int -\frac{1}{u}\, du = - \ln |u| +C = -\ln |\cos x| + C\,\mbox{.}\end{align*}


The limits of integration during variable substitution.

When calculating definite integrals, such as an area, one can go about using variable substitution in two ways. Either one can calculate the integral as usual and then switch back to the original variable and insert the original limits of integration. Alternatively one can change the limits of integration simultaneously with the variable substitution. The two methods are illustrated in the following example.

Example 4

Determine the integral \displaystyle \ \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx.


Method 1

Put \displaystyle u=e^x which gives that \displaystyle u'= e^x and \displaystyle du= e^x\,dx

\displaystyle \begin{align*}\int_{0}^{2} \frac{e^x}{1 + e^x} \, dx &= \int_{x=0}^{\,x=2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{x=0}^{x=2} = \Bigl[\,\ln (1+ e^x)\,\Bigr]_{0}^{2}\\[4pt] &= \ln (1+ e^2) - \ln 2 = \ln \frac{1+ e^2}{2}\,\mbox {.}\end{align*}

Note that the limits of integration must be written in the form \displaystyle x = 0 and \displaystyle x = 2 when the variable of integration is not \displaystyle x. it is wrong to write

\displaystyle \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{0}^{2} \frac{1}{1 + u} \, du \quad \text{ etc.}


Method 2

Put \displaystyle u=e^x which gives that \displaystyle u'= e^x and \displaystyle du= e^x\, dx. The limit of integration \displaystyle x=0 is equivalent to \displaystyle u=e^0 = 1 and \displaystyle x=2 is equivalent to \displaystyle u=e^2

\displaystyle \int_{0}^{2} \frac{e^x}{1 + e^x} \, dx = \int_{1}^{\,e^2} \frac{1}{1 + u} \, du = \Bigl[\,\ln |1+ u |\,\Bigr]_{1}^{e^2} = \ln (1+ e^2) - \ln 2 = \ln\frac{1+ e^2}{2}\,\mbox{.}

Example 5

Determine the integral \displaystyle \ \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx.

The substitution \displaystyle u=\sin x gives \displaystyle du=\cos x\,dx and the limits of integration become \displaystyle u=\sin 0=0 and \displaystyle u=\sin(\pi/2)=1. The integral is

\displaystyle \int_{0}^{\pi/2} \sin^3 x\,\cos x \, dx = \int_{0}^{1} u^3\,du = \Bigl[\,\tfrac{1}{4}u^4\,\Bigr]_{0}^{1} = \tfrac{1}{4} - 0 = \tfrac{1}{4}\,\mbox{.}


[Image]

The figure on the left shows the graph of the integrand sin³x cos x and the figure on the right the graph of integrand u³ which is obtained after the variable substitution. The change of variable modifies the integrand and the interval of the integration. The integrals value, the size of the area, is not changed however.

Example 6

Examine the following calculation

\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\cos x}{\sin^2 x}\, dx = \left[\,\begin{align*} &u = \sin x\\ &du = \cos x \, dx\\ &u(-\pi/2) = -1\\ &u (\pi/2) = 1\end{align*}\,\right ] = \int_{-1}^{1} \frac{1}{u^2} \, du = \Bigl[\, -\frac{1}{u}\, \Bigr]_{-1}^{1} = -1 - 1 = -2\,\mbox{.}

This calculation, however, is wrong, which is due to the fact that \displaystyle f(u)=1/u^2 is not continuous throughout the interval \displaystyle [-1,1].

A necessary condition in the theory is that \displaystyle f(u(x)) be defined and continuous for all values which \displaystyle u(x) can take in the interval in question. Otherwise one cannot be certain that the substitution \displaystyle u=u(x) will work.

[Image]

Graph of f(u) = 1/u²
Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-ImperialCollege/index.php/2.2_Substitution"