Question:

Step 1: Use the property of inverse function integration. If $f$ is bijective, then: $$\int_0^{\pi} f^{-1}(x)\,dx = \pi \cdot f^{-1}(\pi) - 0 \cdot f^{-1}(0) - \int_0^{f^{-1}(\pi)} f(t)\,dt$$ Since $f(x) = x + \sin x$, $f(0) = 0$ and $f(\pi) = \pi + \sin\pi = \pi$, so $f^{-1}(\pi) = \pi$. Step 2: Apply the formula: $$I = \pi \cdot \pi - \int_0^{\pi}(t + \sin t)\,dt = \pi^2 - \left[\frac{t^2}{2} - \cos t\right]_0^{\pi}$$ $$= \pi^2 - \left(\frac{\pi^2}{2} - \cos\pi + \cos 0\right) = \pi^2 - \frac{\pi^2}{2} - 1 - 1 = \frac{\pi^2}{2} - 2$$ Step 3: Compute the numerical value. $$I = \frac{\pi^2}{2} - 2 \approx \frac{9.87}{2} - 2 \approx 4.935 - 2 = 2.935$$ So $2 < I < 3$. Option (c) is correct. Step 4: Check option (a). $\displaystyle\int_0^1 \frac{1}{1+x^3}\,dx < 1$ since the integrand $< 1$ on $(0,1)$. Since $I \approx 2.935 > 1$, option (a) is correct. Step 5: Check option (b). $\displaystyle\int_0^1 e^{x^2}\,dx < e^1 = e \approx 2.718 < I \approx 2.935$, so $I > \int_0^1 e^{x^2}\,dx$. Option (b) is incorrect. Step 6: Check option (d). $\frac{\pi}{4} \approx 0.785$ and $\frac{\pi}{2} \approx 1.571$, but $I \approx 2.935$, so option (d) is incorrect.