calculate the area of the region bounded by curves f(x)=x^2 +5x and g(x)=2-x^2

First we need to find the intersection points of the
curves.

==> f(x)=
g(x)

==> `x^2 +5x = 2-
x^2`

`==> 2x^2 +
5x -2 = 0`

`==>
x1= (-5 + sqrt(25+16))/4 =
(-5+sqrt41)/4`

class="AM">`==> x2=
(-5-sqrt(41))/4`

type="image/svg+xml"
src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="-7.5,7.5,-5,5,1,1,1,1,1,300,200,func,2-x^2,null,0,0,-7,7,black,1,none,func,x^2 +
5x,null,0,0,-7,7,red,1,none"/>

class="AM">`` Now we know that the area bounded by the curve
is:

`A = int g(x) dx - int f(x) dx =
int g(x)-f(x) dx`

`A =
int 2-x^2 - (x^2 +5x) dx = int -2x^2 - 5x +2
dx`

`A = -(2x^3) /3 -
(5x^2) /2 + 2x `

class="AM">``Now we will find the area bounded by (-5+sqrt41)/4  and
(-5-sqrt41)/4.

 `A1=-2/3
((-5+sqrt41)/4)^3 - 5/2 ((-5+sqrt41)/4)^2
+2(-5+sqrt41)/4`

`A2 =
-2/3 ((-5-sqrt41)/4)^3 -5/2 ((-5-sqrt41)/4)^2 +2
(-5-sqrt41)/4`

class="AM">``The area bounded is A = A1 -
A2

`A = -2/3 ((-5+sqrt41)/4)^3 +2/3
((-5+sqrt41)/3)^3 +(25sqrt41)/2 +
sqrt41`

class="AM">``

class="AM">``

class="AM">``