Question: Example $2.2$ Find the area of the region bounded by the line $y=x$ and the parabola $y=6-x^2$.
Graph: a blue line labeled $g(x)=x$ rises diagonally from bottom-left to top-right, and a red downward-opening parabola labeled $f(x)=6-x^2$ is centered near the top; the line and parabola intersect in the center area. position_hint = center
1. **State the problem.** We want the area of the region bounded by the line $y=x$ and the parabola $y=6-x^2$.
2. **Find the intersection points.** Set the two equations equal:
$$x=6-x^2$$
Move all terms to one side:
$$x^2+x-6=0$$
Factor the quadratic:
$$x^2+x-6=(x+3)(x-2)$$
Set each factor equal to $0$:
$$x+3=0 \quad \text{or} \quad x-2=0$$
So the intersection $x$-values are $x=-3$ and $x=2$.
Using $y=x$, the points are $(-3,-3)$ and $(2,2)$.
3. **Choose the top function and bottom function.** On the interval from $x=-3$ to $x=2$, the parabola $y=6-x^2$ is above the line $y=x$.
So the area formula is
$$A=\int_{-3}^{2} \big((6-x^2)-x\big)\,dx$$
4. **Simplify the integrand.**
$$A=\int_{-3}^{2} (6-x^2-x)\,dx$$
Now find an antiderivative:
$$\int (6-x^2-x)\,dx=6x-\frac{x^3}{3}-\frac{x^2}{2}$$
5. **Evaluate the definite integral.**
$$A=\left[6x-\frac{x^3}{3}-\frac{x^2}{2}\right]_{-3}^{2}$$
Plug in $x=2$:
$$6(2)-\frac{2^3}{3}-\frac{2^2}{2}=12-\frac{8}{3}-2=10-\frac{8}{3}=\frac{30}{3}-\frac{8}{3}=\frac{22}{3}$$
Plug in $x=-3$:
$$6(-3)-\frac{(-3)^3}{3}-\frac{(-3)^2}{2}=-18-\frac{-27}{3}-\frac{9}{2}$$
$$=-18+9-\frac{9}{2}=-9-\frac{9}{2}=\frac{-18}{2}-\frac{9}{2}=\frac{-27}{2}$$
Subtract:
$$A=\frac{22}{3}-\left(\frac{-27}{2}\right)=\frac{22}{3}+\frac{27}{2}$$
Find a common denominator:
$$\frac{22}{3}=\frac{44}{6}, \quad \frac{27}{2}=\frac{81}{6}$$
So
$$A=\frac{44}{6}+\frac{81}{6}=\frac{125}{6}$$
6. **Final answer.** The area of the region is $$\frac{125}{6}$$ square units.