Lösung 2.2:2d

Aus Online Mathematik Brückenkurs 2

Version vom 10:22, 11. Mär. 2009

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

1031−xdx=udu=1−x=(1−x)dx=−dx=−013udu.

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.

−013udu=+103udu.

All that is now left is routine calculations,

103udu=10u13du= 31+1u13+1 10=43 u43 10=43143−043=43.