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Lösung 2.1:2c

Aus Online Mathematik Brückenkurs 2

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Version vom 10:15, 11. Mär. 2009

If we recall that x=x12 , the integral can be written as

94x−1xdx=94x12−1x12dx=94x12−x−12dx.

This is a standard integral in which the integrand consists of two terms looking like xn, where n=12 and n=−12, respectively.

We obtain

94x12−x−12dx= x12+112+1−x−12+1−12+1 94= 32x1+12−12x12 94= 32xx−2x 94=3299−29−3244−24=3293−23−3242−22=18−6−316+4=16−316=3163−16=332.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:2c“