Area Between Curves — Question 2

Question 2 – Solution

1. Find the intersections.

$$x^2=x+1\quad\Longrightarrow\quad x^2-x-1=0.$$ The quadratic formula gives $x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-1)}}{2}=\dfrac{1\pm\sqrt5}{2}$. Write the two roots as $$a=\frac{1-\sqrt5}{2},\qquad b=\frac{1+\sqrt5}{2}.$$

2. Determine which curve is above the other.

Test $x=0$: the line has height $1$ and the parabola has height $0$. Between the roots, $(x+1)-x^2\ge0$, so the line is above the parabola.

3. Set up upper minus lower.

$$A=\int_a^b(-x^2+x+1)\,dx.$$

4. Complete the square to simplify the bounds.

Complete the square: $$-x^2+x+1=\frac54-\left(x-\frac12\right)^2.$$ Let $u=x-\tfrac12$. The new bounds are $-\sqrt5/2$ and $\sqrt5/2$. Here $du=dx$. The integrand is even, so the two halves have equal area.

5. Integrate and simplify.

$$\begin{align*} A&=2\int_0^{\sqrt5/2}\left(\frac54-u^2\right)du\\ &=2\left[\frac54u-\frac{u^3}{3}\right]_0^{\sqrt5/2}\\ &=2\left(\frac{5\sqrt5}{8}-\frac{5\sqrt5}{24}\right) =\boxed{\frac{5\sqrt5}{6}}. \end{align*}$$