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baz(z+v)ndz

Last modified by
on
Jul 24, 2020, 6:28:07 PM
Created by
on
Jan 16, 2014, 6:07:14 AM
baz(z+v)ndz=(b+v)n+1((n+1)b-v)(n+1)(n+2)-(a+v)n+1((n+1)a-v)(n+1)(n+2)
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e6cea2f0-da27-11e2-8e97-bc764e04d25f

This equation computes the definite integral for the expression F(z)=z(z+v)n.


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