Question

If the area of region bounded by the curves $f(x) = |\sin x|$ and $g(x) = \frac{2}{11}\sin^{-1}(\sin x)$ lying between the lines $x=0$ and $x=2\pi$ is

Answer 1

4

Answer 2

To find the area of the region bounded by the curves $f(x) = |\sin x|$ and $g(x) = \frac{2}{11}\sin^{-1}(\sin x)$ between $x=0$ and $x=2\pi$, we first analyze the functions in the given interval.

1. Analyze $f(x) = |\sin x|$:

• For $x \in [0, \pi]$, $\sin x \ge 0$, so $f(x) = \sin x$.
• For $x \in (\pi, 2\pi]$, $\sin x < 0$, so $f(x) = -\sin x$.

The graph of $f(x)$ consists of two "humps" above the x-axis, each having an area of $\int_0^\pi \sin x dx = 2$. Thus, $\int_0^{2\pi} f(x) dx = \int_0^\pi \sin x dx + \int_\pi^{2\pi} (-\sin x) dx = [-\cos x]_0^\pi + [\cos x]_\pi^{2\pi} = (-(-1)+1) + (1-(-1)) = 2+2=4$.

2. Analyze $g(x) = \frac{2}{11}\sin^{-1}(\sin x)$:

The function $\sin^{-1}(\sin x)$ is defined piecewise in $[0, 2\pi]$:

• For $x \in [0, \frac{\pi}{2}]$, $\sin^{-1}(\sin x) = x$.
• For $x \in (\frac{\pi}{2}, \frac{3\pi}{2}]$, $\sin^{-1}(\sin x) = \pi - x$.
• For $x \in (\frac{3\pi}{2}, 2\pi]$, $\sin^{-1}(\sin x) = x - 2\pi$.

So, $g(x)$ is:

$g(x) = \begin{cases} \frac{2}{11}x & \text{if } 0 \le x \le \frac{\pi}{2} \\ \frac{2}{11}(\pi - x) & \text{if } \frac{\pi}{2} < x \le \frac{3\pi}{2} \\ \frac{2}{11}(x - 2\pi) & \text{if } \frac{3\pi}{2} < x \le 2\pi \end{cases}$

Let's evaluate $g(x)$ at key points: $g(0)=0$, $g(\frac{\pi}{2}) = \frac{2}{11}\frac{\pi}{2} = \frac{\pi}{11}$, $g(\pi) = \frac{2}{11}(\pi-\pi) = 0$, $g(\frac{3\pi}{2}) = \frac{2}{11}(\pi-\frac{3\pi}{2}) = -\frac{\pi}{11}$, $g(2\pi) = \frac{2}{11}(2\pi-2\pi) = 0$.

3. Compare $f(x)$ and $g(x)$:

• For $x \in [0, \pi]$: $f(x) = \sin x \ge 0$. $g(x)$ is positive or zero (from $0$ to $\pi/11$ and back to $0$). The maximum value of $g(x)$ is $\frac{\pi}{11} \approx 0.285$. The maximum value of $f(x)$ is $1$. Consider $k(x) = f(x) - g(x)$. For $x \in [0, \pi/2]$, $k(x) = \sin x - \frac{2}{11}x$. $k(0)=0$. $k(\pi/2) = 1 - \frac{\pi}{11} > 0$. $k'(x) = \cos x - \frac{2}{11}$. $k'(x)=0$ implies $\cos x = 2/11$. At this point, $k(x)$ has a local maximum. Since $k(0)=0$ and $k(\pi/2)>0$, and it's differentiable, $k(x) \ge 0$ for $x \in [0, \pi/2]$. For $x \in [\pi/2, \pi]$, $k(x) = \sin x - \frac{2}{11}(\pi-x)$. $k(\pi/2) = 1 - \frac{\pi}{11} > 0$. $k(\pi)=0$. $k'(x) = \cos x + \frac{2}{11}$. Similar analysis shows $k(x) \ge 0$ for $x \in [\pi/2, \pi]$. Therefore, $f(x) \ge g(x)$ for $x \in [0, \pi]$.

• For $x \in [\pi, 2\pi]$: $f(x) = -\sin x \ge 0$. $g(x)$ is negative or zero (from $0$ to $-\pi/11$ and back to $0$). Since $f(x) \ge 0$ and $g(x) \le 0$ in this interval, it is clear that $f(x) \ge g(x)$ for $x \in [\pi, 2\pi]$.

Since $f(x) \ge g(x)$ for all $x \in [0, 2\pi]$, the area bounded by the curves is given by $\int_0^{2\pi} (f(x) - g(x)) dx$.

4. Calculate the area:

Area $A = \int_0^{2\pi} (f(x) - g(x)) dx = \int_0^{2\pi} f(x) dx - \int_0^{2\pi} g(x) dx$.

We already calculated $\int_0^{2\pi} f(x) dx = 4$.

Now, calculate $\int_0^{2\pi} g(x) dx$: $\int_0^{2\pi} g(x) dx = \int_0^{\pi/2} \frac{2}{11}x dx + \int_{\pi/2}^{3\pi/2} \frac{2}{11}(\pi-x) dx + \int_{3\pi/2}^{2\pi} \frac{2}{11}(x-2\pi) dx$

First integral: $\int_0^{\pi/2} \frac{2}{11}x dx = \frac{2}{11} \left[\frac{x^2}{2}\right]_0^{\pi/2} = \frac{1}{11} \left(\frac{\pi^2}{4} - 0\right) = \frac{\pi^2}{44}$.

Second integral: $\int_{\pi/2}^{3\pi/2} \frac{2}{11}(\pi-x) dx = \frac{2}{11} \left[\pi x - \frac{x^2}{2}\right]_{\pi/2}^{3\pi/2}$ $= \frac{2}{11} \left[ \left(\pi\frac{3\pi}{2} - \frac{(3\pi/2)^2}{2}\right) - \left(\pi\frac{\pi}{2} - \frac{(\pi/2)^2}{2}\right) \right]$ $= \frac{2}{11} \left[ \left(\frac{3\pi^2}{2} - \frac{9\pi^2}{8}\right) - \left(\frac{\pi^2}{2} - \frac{\pi^2}{8}\right) \right]$ $= \frac{2}{11} \left[ \left(\frac{12\pi^2 - 9\pi^2}{8}\right) - \left(\frac{4\pi^2 - \pi^2}{8}\right) \right]$ $= \frac{2}{11} \left[ \frac{3\pi^2}{8} - \frac{3\pi^2}{8} \right] = 0$.

Third integral: $\int_{3\pi/2}^{2\pi} \frac{2}{11}(x-2\pi) dx = \frac{2}{11} \left[\frac{x^2}{2} - 2\pi x\right]_{3\pi/2}^{2\pi}$ $= \frac{2}{11} \left[ \left(\frac{(2\pi)^2}{2} - 2\pi(2\pi)\right) - \left(\frac{(3\pi/2)^2}{2} - 2\pi\frac{3\pi}{2}\right) \right]$ $= \frac{2}{11} \left[ \left(2\pi^2 - 4\pi^2\right) - \left(\frac{9\pi^2}{8} - 3\pi^2\right) \right]$ $= \frac{2}{11} \left[ -2\pi^2 - \left(\frac{9\pi^2 - 24\pi^2}{8}\right) \right]$ $= \frac{2}{11} \left[ -2\pi^2 - \left(-\frac{15\pi^2}{8}\right) \right]$ $= \frac{2}{11} \left[ -2\pi^2 + \frac{15\pi^2}{8} \right] = \frac{2}{11} \left[ \frac{-16\pi^2 + 15\pi^2}{8} \right] = \frac{2}{11} \left(-\frac{\pi^2}{8}\right) = -\frac{\pi^2}{44}$.

Summing the integrals for $g(x)$: $\int_0^{2\pi} g(x) dx = \frac{\pi^2}{44} + 0 - \frac{\pi^2}{44} = 0$.

Finally, the total area is $A = 4 - 0 = 4$.