# Why Does an Integrand Equaling Zero at x=1 Not Determine the Integral's Value?

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• Hornbein
Hornbein
I'm trying to calculate the volume of a truncated hypersphere. As part of it I want this integral.

Clearly when x=1 the integrand is zero. But plugging this into the series give me a number greater than one. It is true that the series is not defined for x=1, but subtracting some tiny sum isn't going to make any difference.

I hope I'm missing something obvious.

I made a mistake in my previous post ( deleted it ).

\begin{align} \int (1-x^2)^n ~dx &= \int (-x^2+1)^n ~dx \nonumber\\ &=\int \sum_{k=0}^n \binom {n}{k} (-x^2)^k1^{n-k}~dx \nonumber\\ &=\sum_{k=0}^n (-1)^k \binom {n}{k} \int x^{2k}~dx \nonumber\\ &=\sum_{k=0}^n (-1)^k \frac { n!}{k!(n-k)!} \cdot \frac {x^{2k+1}}{2k+1} \nonumber\\ \end{align}

Last edited:
Hornbein said:
Clearly when x=1 the integrand is zero. But plugging this into the series give me a number greater than one. It is true that the series is not defined for x=1, but subtracting some tiny sum isn't going to make any difference.

I hope I'm missing something obvious.

Why would the integrand being zero at $x = 1$ tell you anything about the value of the integral?

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