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`

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

`` Now we know that the area bounded by the curve is:

`A...

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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`

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

`` 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 `

``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`

``The area bounded is A = A1 - A2

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

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