Lösung 2.2:2d

Aus Online Mathematik Brückenkurs 2

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=u=1−xdu=1−xdx=−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 we change sign at the same time, i.e.

−013udu=+103udu 

All that is now left is routine calculations:

103udu=10u31du=u31+131+110=43u3410=43134−034=43