Question 831664
The curves intersect where {{{x^2+9x-4=x+3}}}
Solving:
{{{x^2+9x-4=x+3}}}
{{{x^2+9x-4-x-3=0}}}
{{{x^2+8x-7=0}}}
The solutions are calculated as
{{{x = (-8 +- sqrt(8^2-4*1*(-7) ))/(2*1) }}}
{{{x = (-8 +- sqrt(64+28))/2 }}}
{{{x = (-8 +- sqrt(92))/2 }}}
{{{x = (-8 +- sqrt(4*23))/2 }}}
{{{x = (-8 +- 2*sqrt(23))/2 }}}
{{{x =-4 +- sqrt(23)}}} (approximately -8.79 and 0.796)
{{{graph(300,300,-10,5,-25,5,x^2+9x-4,x+3)}}}
So the area between the curves can be calculated as
{{{int(((x^2+9x-4)-(x+3)), dx, -4-sqrt(23), -4+sqrt(23))=int((x^2+8x-7), dx, -4-sqrt(23), -4+sqrt(23))}}}
{{{f(x)=int((x^2+8x-7), dx)=x^3/3+4x^3-7x}}} so
{{{int(((x^2+9x-4)-(x+3)), dx, -4-sqrt(23), -4+sqrt(23))=int((x^2+8x-7), dx, -4-sqrt(23), -4+sqrt(23))=f(-4+sqrt(23))-f(-4-sqrt(23))=147.1}}} (rounded)