Question

Here is an easy integration problem from
http://saab.org/calculus.cgi

Answer 1

what an odd problem

Answer 2

\[\int_{-1}^1 18t^5(1+t^6)^2 dt\]Let \(u=1+t^6\). By differentials, we get \(du=5t^5 dt\) and consenquently, \(18t^5 dt=\frac{18}{5}du\). If t = -1, then u = 2. If t = 1, then u = 2. Thus, integration by substitution gives\[=\int_{2}^2 u^2 \left(\frac{18}{5}du\right)\]Notice that, the limits of integration are equal, therefore this integral, and therefore the original integral, equals \(\boxed{0}\).

A shorter method for this is to realize that \(18t^5(1+t^6)^2\) is an odd expression, and that the limits of integration are additive inverses of each other. Thus, automatically, the definite integral equals zero.

Answer 3

Here is the solution generated by
http://saab.org/calculus.cgi

Answer 4

yep. :)