# Integrating $\int [n (T - x) ^{n - 1} - 1] dx$ for constants $T$ and $n$

It's been way too long, and I'm having trouble integrating a function (with a practical application) that should be easy to do with high school calculus. It seems very simple compared to the questions I see here. But I would like some help integrating it.

Here's what I've got, for constants $T$ and $n$

$$\int n (T - x) ^{n - 1} - {1} dx$$

Here's where I landed:

$${n (T - x) ^ n \over -n} - x + C$$

$$-(T - x)^{n} - x + C$$

But for $T=6$ and $n=2$ and $x=0, \ldots, 5$, I expect to get $11, 20, 27, 32, 35, 36$.

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## migrated from meta.math.stackexchange.comApr 5 '13 at 7:43

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I don't understand your question... What do you want to do ? What do you mean with "for T=6 and n=2 and x=0,…,5, I expect to get 11,20,27,32,35,36" ? – mwoua Apr 5 '13 at 9:49
Don't forget to consider the case $n=0$. The result of your integral would then be a $\ln$ function. – mwoua Apr 5 '13 at 9:52
@mwoua: You will find that a certain factor in the integrand will make the logarithm disappear, when $n=0$. – Jyrki Lahtonen Apr 5 '13 at 11:06
I think I made a hash of this; I'll take another look at the original function. I think I have a -1 out of place. Thank you for the help! – Rob Apr 5 '13 at 16:30
I cleared some comments made obsolete by the migration and subsequent edits. – Willie Wong Apr 8 '13 at 8:45