From Förberedande kurs i matematik 2

Revision as of 10:05, 7 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_3_2a-1(2).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:2_3_2a-2(2).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (08:37, 29 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	Had the integral instead been
-	<center> [[Bild:2_3_2a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>\int e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}\,dx</math>}}
-	{{NAVCONTENT_START}}	+
-	<center> [[Bild:2_3_2a-2(2).gif]] </center>	+	it is quite obvious that we would substitute <math>u=\sqrt{x}</math>, but we are missing a factor <math>1/2\sqrt{x}</math> which would take account of the derivative of <math>u</math> which is needed when <math>dx</math> is replaced by
-	{{NAVCONTENT_STOP}}	+	<math>du</math>. In spite of this, we can try the substitution <math>u=\sqrt{x}</math> if we multiply top and bottom by what is missing,
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int e^{\sqrt{x}}\,dx
		+	&= \int e^{\sqrt{x}}\cdot 2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}\,dx\\[5pt]
		+	&= \left\{\begin{align}
		+	u &= \sqrt{x}\\[5pt]
		+	du &= \bigl(\sqrt{x}\,\bigr)'\,dx = \dfrac{1}{2\sqrt{x}}\,dx
		+	\end{align}\right\}\\[5pt]
		+	&= \int e^{u}\cdot 2u\,du\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (<math>2u</math> is the factor that we differentiate and <math>e^{u}</math> is the factor that we integrate),
		+
		+	{{Displayed math||<math>\begin{align}
		+	\int e^u\cdot 2u\,du
		+	&= e^u\cdot 2u - \int e^u\cdot 2\,du\\[5pt]
		+	&= 2ue^u - 2\int e^u\,du\\[5pt]
		+	&= 2ue^u - 2e^u + C\\[5pt]
		+	&= 2(u-1)e^u + C\,\textrm{.}
		+	\end{align}</math>}}
		+
		+	If we substitute back <math>u=\sqrt{x}</math>, we obtain the answer
		+
		+	{{Displayed math||<math>\int e^{\sqrt{x}}\,dx = 2(\sqrt{x}-1)e^{\sqrt{x}} + C\,\textrm{.}</math>}}
		+
		+	As can be seen, it is possible to mix different integration techniques and often we need to experiment with different approaches before we find the right one.

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Current revision

Had the integral instead been

\displaystyle \int e^{\sqrt{x}}\cdot\frac{1}{2\sqrt{x}}\,dx

it is quite obvious that we would substitute \displaystyle u=\sqrt{x}, but we are missing a factor \displaystyle 1/2\sqrt{x} which would take account of the derivative of \displaystyle u which is needed when \displaystyle dx is replaced by \displaystyle du. In spite of this, we can try the substitution \displaystyle u=\sqrt{x} if we multiply top and bottom by what is missing,

\displaystyle \begin{align} \int e^{\sqrt{x}}\,dx &= \int e^{\sqrt{x}}\cdot 2\sqrt{x}\cdot \frac{1}{2\sqrt{x}}\,dx\\[5pt] &= \left\{\begin{align} u &= \sqrt{x}\\[5pt] du &= \bigl(\sqrt{x}\,\bigr)'\,dx = \dfrac{1}{2\sqrt{x}}\,dx \end{align}\right\}\\[5pt] &= \int e^{u}\cdot 2u\,du\,\textrm{.} \end{align}

Now, we obtain instead another, not entirely simple, integral, but we can calculate the new integral by partial integration (\displaystyle 2u is the factor that we differentiate and \displaystyle e^{u} is the factor that we integrate),

\displaystyle \begin{align} \int e^u\cdot 2u\,du &= e^u\cdot 2u - \int e^u\cdot 2\,du\\[5pt] &= 2ue^u - 2\int e^u\,du\\[5pt] &= 2ue^u - 2e^u + C\\[5pt] &= 2(u-1)e^u + C\,\textrm{.} \end{align}

If we substitute back \displaystyle u=\sqrt{x}, we obtain the answer

\displaystyle \int e^{\sqrt{x}}\,dx = 2(\sqrt{x}-1)e^{\sqrt{x}} + C\,\textrm{.}

As can be seen, it is possible to mix different integration techniques and often we need to experiment with different approaches before we find the right one.