```Question 344815
<pre>
I will put {{{x=""}}} on the limits to show what variable the
limits are on, namely x, since we will be changing the limits:

{{{ int( (1-x)^2010, dx, x=0, x=1 ) }}}

Let {{{u=1-x}}}
Then {{{(du)/(dx)=-1}}}
or {{{du=-dx}}}

and {{{dx=-du}}}

So we substitute {{{u}}} for {{{(1-x)}}} and {{{-du}}} for {{{dx}}}

{{{ int( u^2010, (-du), x=0, x=1 ) }}}

And we can take out the negative sign in front of the integral:

{{{ -int( u^2010, du, x=0, x=1 ) }}}

But the limits are on x.  We want to change the limits
so that they will be on u rather than x.

To do that we substitute the limits for x in the equation
for u which is {{{u=1-x}}}

when {{{x=0}}}, {{{u=1-0=1}}} and
when {{{x=1}}}, {{{u=1-1=0}}},
so we change the limits to:

{{{ -int( u^2010, du, u=1, u=0 ) }}}

and we use the power formula {{{int( u^n, du)=u^(n+1)/(n+1)+C}}}

{{{drawing(400,100,-.5,5,-.5,.5, locate(.7,.3, -u^2011/2011),

line(1.6,-.3,1.6,.3),locate(1.7,.35,u=0), locate(1.7,-.22,u=1),

locate(2,.3,""=(-0^2011/2011)-(-1^2011/2011)=1/2011)

)}}}

Edwin</pre>
```