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Evaluate \[ \int_{-1}^1 18 t^5 ( 1+ t^6) dt \] There are two for doing this one short way that does not need any computation and one that a bit longer. Either way will be fine.
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you can multiply out easily to get \(18t^5+18t^{11}\) you can also use a u-sub but that is unnecessary
think there is a typo here is it \[\int_{-1}^1 18t^5(1+t^6)dt\]?
if so compute nothing you have an odd function so integral will be zero
From -1 to 1
then it is zero for sure
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if \(f\) is odd then \(\int_{-a}^a f(x)dx=0\)
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