integrate integral from 1 to 0 (r^3)/(square root of 4+(r^2))dr.
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Originally Posted by twilightstr integrate integral from 1 to 0 (r^3)/(square root of 4+(r^2))dr. If you make the substitution $\displaystyle u = r^2 + 4$ and convert to a new integral involving u, things should work out.
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