Area Between Curves — Question 1

Question 1 – Solution

1. Find the intersections.

$$x^2=2x\quad\Longrightarrow\quad x(x-2)=0 \quad\Longrightarrow\quad x=0,\;2.$$ Factoring gives $x=0$ or $x-2=0$. Substituting into $y=2x$ gives $y=0$ or $y=4$. The intersection points are $(0,0)$ and $(2,4)$.

2. Determine which curve is above the other.

On $0\le x\le2$, $2x-x^2=x(2-x)\ge0$, so the line is above the parabola.

3. Set up upper minus lower.

$$A=\int_0^2(\text{upper}-\text{lower})\,dx =\int_0^2(2x-x^2)\,dx.$$

4. Find an antiderivative.

By the power rule, $\int 2x\,dx=x^2$ and $\int x^2\,dx=x^3/3$.

5. Apply the bounds and simplify.

$$\begin{align*} A&=\left[x^2-\frac{x^3}{3}\right]_0^2\\ &=\left(4-\frac83\right)-0 =\boxed{\frac43}. \end{align*}$$