From Förberedande kurs i matematik 2

Revision as of 09:33, 7 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_2d.gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (13:56, 28 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	What makes the integral not entirely simple is the expression <math>1-x</math> under the root sign, so we try the substitution <math>u=1-x</math>,
-	<center> [[Bild:2_2_2d.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
		+	u &= 1-x\\[5pt]
		+	du &= (1-x)'\,dx = -\,dx
		+	\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}</math>}}
		+
		+	Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.
		+
		+	{{Displayed math||<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}
		+
		+	All that is now left is routine calculations,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int\limits_0^1 \sqrt[3]{u}\,du
		+	&= \int\limits_0^1 u^{1/3}\,du
		+	= \biggl[\ \frac{u^{1/3+1}}{\tfrac{1}{3}+1}\ \biggr]_0^1\\[5pt]
		+	&= \frac{3}{4}\Bigl[\ u^{4/3}\ \Bigr]_0^1
		+	= \frac{3}{4}\bigl( 1^{4/3}-0^{4/3} \bigr) = \frac{3}{4}\,\textrm{.}
		+	\end{align}</math>}}

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Current revision

What makes the integral not entirely simple is the expression \displaystyle 1-x under the root sign, so we try the substitution \displaystyle u=1-x,

\displaystyle \int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align} u &= 1-x\\[5pt] du &= (1-x)'\,dx = -\,dx \end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}

Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.

\displaystyle -\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}

All that is now left is routine calculations,

\displaystyle \begin{align} \int\limits_0^1 \sqrt[3]{u}\,du &= \int\limits_0^1 u^{1/3}\,du = \biggl[\ \frac{u^{1/3+1}}{\tfrac{1}{3}+1}\ \biggr]_0^1\\[5pt] &= \frac{3}{4}\Bigl[\ u^{4/3}\ \Bigr]_0^1 = \frac{3}{4}\bigl( 1^{4/3}-0^{4/3} \bigr) = \frac{3}{4}\,\textrm{.} \end{align}