Question 1025563
 1. {{{ x^2 - 6x + 5  = x - 5 }}} ==> {{{x^2-7x + 10 = 0}}} ==> (x-2)(x-5) = 0
==> the two graphs intersect at x = 2, 5.

==> Area is {{{int((x-5-(x^2 - 6x + 5)), dx, 2,5) = int((-x^2 + 7x - 10), dx, 2,5)}}}.
The antiderivative is {{{-x^3/3 + (7x^2)/2 -10x}}}.

The area is then {{{-5^3/3 + (7*5^2)/2 -10*5 - (-2^3/3 + (7*2^2)/2 -10*2) = 9/2}}}. 

2.  The graphs of {{{ y^2 = x }}} and {{{ x+ 2y = 3 }}} intersect at y = -3, 1.
==> Area =  {{{int((3-2y-y^2), dy, -3,1))}}}.

The antiderivative is {{{3y - y^2 - y^3/3}}}

==> area is {{{3*1 - 1^2 - 1^3/3 - (3*-3 - (-3)^2 - (-3)^3/3) = 32/3}}}