Processing Math: Done

Aus Online Mathematik Brückenkurs 2

Version vom 13:00, 10. Mär. 2009 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) ← Zum vorherigen Versionsunterschied

Version vom 10:20, 11. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Solution 2.2:1a moved to Lösung 2.2:1a: Robot: moved page) Zum nächsten Versionsunterschied →

---

Version vom 10:20, 11. Mär. 2009

A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with.

When we carry out a substitution of variables u=u(x), there are three things which are affected in the integral:

1. the integral must be rewritten in terms of the new variable u;
2. the element of integration, dx, is replaced by du, according to the formula du=u(x)dx;
3. the limits of integration are for x and must be changed to limits of integration for the variable u.

In this case, we will perform the change of variables u=3x−1, mainly because the integrand 1(3x−1)4 will then be replaced by 1u4.

The relation between dx and du reads

du=u(x)dx=(3x−1)dx=3dx

which means that dx is replaced by 31du.

Furthermore, when x=1 in the lower limit of integration, the corresponding u-value becomes u=31−1=2, and when x=2, we obtain the u-value u=32−1=5.

One usually writes the whole substitution of variables as

21dx(3x−1)4=udu=3x−1=3dx=52u431du.

Sometimes, we are more brief and hide the details,

21dx(3x−1)4=u=3x−1=52u431du.

After the substitution of variables, we have a standard integral which is easy to compute.

In summary, the whole calculation is,

21dx(3x−1)4=udu=3x−1=3dx=52u431du=3152u−4du=31 u−4+1−4+1 52=−91 1u3 52=−91153−123=−91235323−53=117322353=3213322353=132353=131000.