Lösung 2.1:2b

Aus Online Mathematik Brückenkurs 2

Version vom 10:15, 11. Mär. 2009

There is no ready made standard formula for a primitive function to our integrand, but if we expand

2−1(x−2)(x+1)dx=2−1(x2+x−2x−2)dx=2−1(x2−x−2)dx

and write the last integral as

2−1(x2−x1−2x0)dx

we see that the integrand consists of three terms of the type xn and we can directly write down a primitive function,

2−1(x2−x1−2x0)dx= 3x3−2x2−2x1 2−1=323−222−212−3(−1)3−2(−1)2−21(−1)=38−24−4−−31−21+2=616−12−24+2+3−12=−627=−29.