Area Between Curves — Question 3

Question 3

Find the exact area of the region between the curves $$y=\sin x \quad\text{and}\quad y=1-\cos x,$$ between $x=0$ and $x=\pi$.

Question 3 – Solution

1. Find where the curves cross on $[0,\pi]$.

Using half-angle identities, $$\begin{align*} \sin x=1-\cos x &\iff 2\sin(x/2)\bigl(\cos(x/2)-\sin(x/2)\bigr)=0\\ &\iff x=0\quad\text{or}\quad x=\frac\pi2. \end{align*}$$

2. Determine the upper curve on each interval.

On $(0,\pi/2)$, $\sin x>1-\cos x$; on $(\pi/2,\pi)$, the order reverses. Test $x=\pi/4$: $\sin x=\sqrt2/2>1-\sqrt2/2$. At $x=3\pi/4$, $1-\cos x=1+\sqrt2/2>\sqrt2/2$. The line $x=\pi$ closes the region on the right.

3. Set up the sum of the two areas.

$$A=\int_0^{\pi/2}(\sin x-1+\cos x)\,dx +\int_{\pi/2}^{\pi}(1-\cos x-\sin x)\,dx.$$

4. Integrate each piece and apply its bounds.

$$\begin{align*} A_1&=\left[-\cos x-x+\sin x\right]_0^{\pi/2} =2-\frac\pi2,\\[6pt] A_2&=\left[x-\sin x+\cos x\right]_{\pi/2}^{\pi} =(\pi-1)-\left(\frac\pi2-1\right)=\frac\pi2. \end{align*}$$

5. Add the positive areas.

$$A=A_1+A_2=\left(2-\frac\pi2\right)+\frac\pi2=\boxed{2}.$$