From Förberedande kurs i matematik 2

Revision as of 09:19, 7 April 2008 (edit) Oskar (Talk | contribs) ← Previous diff		Current revision (12:31, 28 October 2008) (edit) (undo) Tek (Talk | contribs) m
(3 intermediate revisions not shown.)
Line 1:		Line 1:
-	{{NAVCONTENT_START}}	+	A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with.
-	<center> [[Bild:2_2_1a-1(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	When we carry out a substitution of variables <math>u=u(x)</math>, there are three things which are affected in the integral:
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:2_2_1a-2(3).gif]] </center>	+	# the integral must be rewritten in terms of the new variable <math>u</math>;
-	{{NAVCONTENT_STOP}}	+	# the element of integration, <math>dx</math>, is replaced by <math>du</math>, according to the formula <math>du=u'(x)\,dx</math>;
-	{{NAVCONTENT_START}}	+	# the limits of integration are for <math>x</math> and must be changed to limits of integration for the variable <math>u</math>.
-	<center> [[Bild:2_2_1a-3(3).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	In this case, we will perform the change of variables <math>u=3x-1</math>, mainly because the integrand <math>1/(3x-1)^4</math> will then be replaced by <math>1/u^4</math>.
		+
		+	The relation between <math>dx</math> and <math>du</math> reads
		+
		+	{{Displayed math||<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}
		+
		+	which means that <math>dx</math> is replaced by <math>\tfrac{1}{3}\,du</math>.
		+
		+	Furthermore, when <math>x=1</math> in the lower limit of integration, the corresponding ''u''-value becomes <math>u=3\cdot 1-1=2</math>, and when
		+	<math>x=2</math>, we obtain the ''u''-value <math>u=3\cdot 2-1=5\,</math>.
		+
		+	One usually writes the whole substitution of variables as
		+
		+	{{Displayed math||<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}
		+	u &= 3x-1\\[5pt]
		+	du &= 3\,dx
		+	\end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}
		+
		+	Sometimes, we are more brief and hide the details,
		+
		+	{{Displayed math||<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}
		+
		+	After the substitution of variables, we have a standard integral which is easy to compute.
		+
		+	In summary, the whole calculation is,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int\limits_1^2 \frac{dx}{(3x-1)^4}
		+	&= \left\{\begin{align}
		+	u &= 3x-1\\[5pt]
		+	du &= 3\,dx
		+	\end{align}\right\}
		+	= \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}
		+	= \frac{1}{3}\int\limits_2^5 u^{-4}\,du\\[5pt]
		+	&= \frac{1}{3}\Bigl[\ \frac{u^{-4+1}}{-4+1}\ \Bigr]_2^5
		+	= -\frac{1}{9}\Bigl[\ \frac{1}{u^3}\ \Bigr]_2^5
		+	= -\frac{1}{9}\Bigl(\frac{1}{5^3} - \frac{1}{2^3} \Bigr)
		+	= -\frac{1}{9}\cdot\frac{2^3-5^3}{2^3\cdot 5^3}\\[5pt]
		+	&= \frac{117}{3^2\cdot 2^3\cdot 5^3}
		+	= \frac{3^2\cdot 13}{3^2\cdot 2^3\cdot 5^3}
		+	= \frac{13}{2^3\cdot 5^3}
		+	= \frac{13}{1000}\,\textrm{.}
		+	\end{align}</math>}}

---

Current revision

A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with.

When we carry out a substitution of variables \displaystyle u=u(x), there are three things which are affected in the integral:

1. the integral must be rewritten in terms of the new variable \displaystyle u;
2. the element of integration, \displaystyle dx, is replaced by \displaystyle du, according to the formula \displaystyle du=u'(x)\,dx;
3. the limits of integration are for \displaystyle x and must be changed to limits of integration for the variable \displaystyle u.

In this case, we will perform the change of variables \displaystyle u=3x-1, mainly because the integrand \displaystyle 1/(3x-1)^4 will then be replaced by \displaystyle 1/u^4.

The relation between \displaystyle dx and \displaystyle du reads

\displaystyle du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,

which means that \displaystyle dx is replaced by \displaystyle \tfrac{1}{3}\,du.

Furthermore, when \displaystyle x=1 in the lower limit of integration, the corresponding u-value becomes \displaystyle u=3\cdot 1-1=2, and when \displaystyle x=2, we obtain the u-value \displaystyle u=3\cdot 2-1=5\,.

One usually writes the whole substitution of variables as

\displaystyle \int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align} u &= 3x-1\\[5pt] du &= 3\,dx \end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}

Sometimes, we are more brief and hide the details,

\displaystyle \int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}

After the substitution of variables, we have a standard integral which is easy to compute.

In summary, the whole calculation is,

\displaystyle \begin{align} \int\limits_1^2 \frac{dx}{(3x-1)^4} &= \left\{\begin{align} u &= 3x-1\\[5pt] du &= 3\,dx \end{align}\right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4} = \frac{1}{3}\int\limits_2^5 u^{-4}\,du\\[5pt] &= \frac{1}{3}\Bigl[\ \frac{u^{-4+1}}{-4+1}\ \Bigr]_2^5 = -\frac{1}{9}\Bigl[\ \frac{1}{u^3}\ \Bigr]_2^5 = -\frac{1}{9}\Bigl(\frac{1}{5^3} - \frac{1}{2^3} \Bigr) = -\frac{1}{9}\cdot\frac{2^3-5^3}{2^3\cdot 5^3}\\[5pt] &= \frac{117}{3^2\cdot 2^3\cdot 5^3} = \frac{3^2\cdot 13}{3^2\cdot 2^3\cdot 5^3} = \frac{13}{2^3\cdot 5^3} = \frac{13}{1000}\,\textrm{.} \end{align}