From Förberedande kurs i matematik 2

Revision as of 09:22, 7 April 2008 (edit) Oskar (Talk | contribs) (Ny sida: {{NAVCONTENT_START}} <center> Bild:2_2_1c).gif </center> {{NAVCONTENT_STOP}}) ← Previous diff		Current revision (12:48, 28 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	With the given variable substitution, <math>u=x^3</math>, we obtain
-	<center> [[Bild:2_2_1c).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{Displayed math||<math>du = \bigl(x^3\bigr)'\,dx = 3x^2\,dx</math>}}
		+
		+	and because the integral contains <math>x^2</math> as a factor, we can bundle it together with <math>dx</math> and replace the combination with <math>\tfrac{1}{3}\,du</math>,
		+
		+	{{Displayed math||<math>\int e^{x^3}x^2\,dx = \bigl\{\,u=x^3\,\bigr\} = \int e^u\tfrac{1}{3}\,du = \frac{1}{3}e^u + C\,\textrm{.}</math>}}
		+
		+	Thus, the answer is
		+
		+	{{Displayed math||<math>\int e^{x^3}x^2\,dx = \frac{1}{3}e^{x^3} + C\,,</math>}}
		+
		+	where <math>C</math> is an arbitrary constant.

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Current revision

With the given variable substitution, \displaystyle u=x^3, we obtain

\displaystyle du = \bigl(x^3\bigr)'\,dx = 3x^2\,dx

and because the integral contains \displaystyle x^2 as a factor, we can bundle it together with \displaystyle dx and replace the combination with \displaystyle \tfrac{1}{3}\,du,

\displaystyle \int e^{x^3}x^2\,dx = \bigl\{\,u=x^3\,\bigr\} = \int e^u\tfrac{1}{3}\,du = \frac{1}{3}e^u + C\,\textrm{.}

Thus, the answer is

\displaystyle \int e^{x^3}x^2\,dx = \frac{1}{3}e^{x^3} + C\,,

where \displaystyle C is an arbitrary constant.