integration of limit 0 to 1 x(1-x)n
raghurajmuni:
is it (1-x) power n
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We can evaluate this integral using integration by substitution, or u-substitution. We pick some part of the integrand to set equal to some variable (such as u, but any variable is an option). Good places to look at first include under a radical or in the denominator. This is not always the case, but it is in this one.
We can set u=1−x
Therefore,
du=−1dx
−du=dx
We can substitute these values into our integral. We get:
−∫1√udu
Which we can rewrite as:
−∫u−12du
Integrating, we get:
−2u12
From here you have two options on evaluating for the given limits of integration. You can either choose now to substitute 1−x back in for uand evaluate from 0 to 1, or you can change the limits of integration and evaluate with u. I will demonstrate both options.
We can set u=1−x
Therefore,
du=−1dx
−du=dx
We can substitute these values into our integral. We get:
−∫1√udu
Which we can rewrite as:
−∫u−12du
Integrating, we get:
−2u12
From here you have two options on evaluating for the given limits of integration. You can either choose now to substitute 1−x back in for uand evaluate from 0 to 1, or you can change the limits of integration and evaluate with u. I will demonstrate both options.
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