Question

How do you evaluate the integral `int 1/sqrt(1-x)dx` from 0 to 1?

Answer 1

Use integration by substitution (u-substitution).

Explanation:

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:

`-int1/sqrtudu`

Which we can rewrite as:

`-intu^(-1/2)du`

Integrating, we get:

`-2u^(1/2)`

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 `u` and evaluate from 0 to 1, or you can change the limits of integration and evaluate with u. I will demonstrate both options.

1. Substituting `1-x` back in for `u`,

`-2(1-x)^(1/2)`
`-2[(1-1)^(1/2)-(1-0)^(1/2)]`
`-2(-1)`

Final answer: 2

1. Changing limits of integration:

`u=1-x`

`u=1-(1)`
`u=0` (new upper limit)

`u=1-0`
`u=1` (new lower limit)

Evaluating, we have

`-2[(0)^(1/2)-(1)^(1/2)]`
`-2(-1)`

Final answer: 2

Hope this helps!