Question

If $f(x) = x + \sin x$ and I denotes the value of integral $\int_{\pi}^{2\pi} [f^{-1}(x) + \sin x]dx$ then the value of $[\frac{2I}{3}]$ (where [.] denotes greatest integer function)

• A. 2
• B. 9
• C. 8
• D. 10

Answer 1

Answer: (B)

9

Answer 2

Let $f(x) = x + \sin x$. We need to evaluate $I = \int_{\pi}^{2\pi} [f^{-1}(x) + \sin x]dx$. The derivative of $f(x)$ is $f'(x) = 1 + \cos x$. For $x \in [\pi, 2\pi]$, $f'(x) \ge 0$, and $f'(x) = 0$ only at $x = \pi$. Thus, $f(x)$ is strictly increasing in this interval and its inverse function $f^{-1}(x)$ exists. Evaluating $f(x)$ at the limits: $f(\pi) = \pi + \sin \pi = \pi$. $f(2\pi) = 2\pi + \sin 2\pi = 2\pi$. So, for $x \in [\pi, 2\pi]$, the range of $f(x)$ is $[\pi, 2\pi]$. This implies $f^{-1}(x)$ maps $[\pi, 2\pi]$ to $[\pi, 2\pi]$.

We can split the integral $I$ into two parts: $I = \int_{\pi}^{2\pi} f^{-1}(x)dx + \int_{\pi}^{2\pi} \sin x dx$.

The second part of the integral is: $\int_{\pi}^{2\pi} \sin x dx = [-\cos x]_{\pi}^{2\pi} = -\cos(2\pi) - (-\cos(\pi)) = -1 - (-(-1)) = -1 - 1 = -2$.

For the first part, $\int_{\pi}^{2\pi} f^{-1}(x)dx$, we use the property: $\int_{a}^{b} f(x)dx + \int_{f(a)}^{f(b)} f^{-1}(y)dy = b f(b) - a f(a)$. Here, $a = \pi$, $b = 2\pi$, $f(a) = \pi$, and $f(b) = 2\pi$. First, calculate $\int_{\pi}^{2\pi} f(x)dx$: $\int_{\pi}^{2\pi} f(x)dx = \int_{\pi}^{2\pi} (x + \sin x)dx = \left[\frac{x^2}{2} - \cos x\right]_{\pi}^{2\pi}$ $= \left(\frac{(2\pi)^2}{2} - \cos(2\pi)\right) - \left(\frac{\pi^2}{2} - \cos(\pi)\right)$ $= \left(2\pi^2 - 1\right) - \left(\frac{\pi^2}{2} - (-1)\right) = 2\pi^2 - 1 - \frac{\pi^2}{2} - 1 = \frac{3\pi^2}{2} - 2$.

Using the property: $\int_{\pi}^{2\pi} f(x)dx + \int_{\pi}^{2\pi} f^{-1}(x)dx = (2\pi) f(2\pi) - \pi f(\pi)$ $= (2\pi)(2\pi) - \pi(\pi) = 4\pi^2 - \pi^2 = 3\pi^2$.

Therefore, $\int_{\pi}^{2\pi} f^{-1}(x)dx = 3\pi^2 - \int_{\pi}^{2\pi} f(x)dx$ $= 3\pi^2 - \left(\frac{3\pi^2}{2} - 2\right) = 3\pi^2 - \frac{3\pi^2}{2} + 2 = \frac{3\pi^2}{2} + 2$.

Now, we find the value of $I$: $I = \int_{\pi}^{2\pi} f^{-1}(x)dx + \int_{\pi}^{2\pi} \sin x dx = \left(\frac{3\pi^2}{2} + 2\right) + (-2) = \frac{3\pi^2}{2}$.

The problem asks for the value of $[\frac{2I}{3}]$. $\frac{2I}{3} = \frac{2}{3} \left(\frac{3\pi^2}{2}\right) = \pi^2$.

We need to find the greatest integer of $\pi^2$. Since $\pi \approx 3.14159$, $\pi^2 \approx (3.14159)^2 \approx 9.8696$. The greatest integer of $\pi^2$ is $9$.

$[\frac{2I}{3}] = [\pi^2] = 9$.