Find the area bounded by y=x-2 and y=x^2-3

Hello!

The graph of the first function, `f(x)=x-2,` is a straight line, the graph of the second function, `g(x)=x^2-3,` is a parabola branches up. These graphs have two intersections, `(x_1,y_1)` and `(x_2,y_2),` we'll find them. The figure whose area we have to compute lies between these points of intersection, and the first graph is over the second between `x_1` and `x_2.`

Therefore the area `A` is equal to  `int_(x_1)^(x_2) (f(x)-g(x)) dx.`

To find `x_1` and `x_2` we have to solve the equation `f(x)=g(x),` i.e. `x-2=x^2-3,` or `x^2-x-1=0.` It is a quadratic equation and `x_1=(1-sqrt(5))/2,`  `x_2=(1+sqrt(5))/2.`

Thus `A=int_(x_1)^(x_2) (f(x)-g(x)) dx=int_(x_1)^(x_2) (-x^2+x+1) dx=`

`=(-1/3 x^3+1/2 x^2+x)|_(x_1)^(x_2)=-1/3(x_2^3-x_1^3)+1/2(x_2^2-x_1^2)+(x_2-x_1).`

We can simplify this slightly,

`A=(x_2-x_1)(-1/3(x_1^2+x_1x_2+x_2^2)+1/2(x_2+x_1)+1)=`

`=sqrt(5)*(-1/3*(3-1)+1/2+1)=sqrt(5)*(3/2-2/3)=sqrt(5)*5/6 approx 1.86.`

This is the answer.