A) \int_{0}^{1} x \cos x \, dx \ge 3/8
B) \int_{0}^{1} x \sin x \, dx \ge 3/10
C) \int_{0}^{1} x^2 \cos x \, dx \ge 1/2
D) \int_{0}^{1} x^2 \sin x \, dx \ge 2/9
Step-by-Step Solution
Key Concept: Use calculus to analyze the function $ f(x) = x - \ln(x) $, leveraging its monotonicity and critical points to establish the inequality.
We can establish these inequalities using standard Taylor series expansions for $\sin x$ and $\cos x$, which provide tight lower bounds on the interval $[0, 1]$.
1. **Evaluate Option A**:
On $[0, 1]$, we have the inequality $\cos x \ge 1 - \frac{x^2}{2}$.
Multiply by $x$ and integrate from 0 to 1:
$$\int_0^1 x \cos x \, dx \ge \int_0^1 \left( x - \frac{x^3}{2} \right) dx = \left[ \frac{x^2}{2} - \frac{x^4}{8} \right]_0^1 = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}$$
Thus, Option A is **TRUE**.
2. **Evaluate Option B**:
On $[0, 1]$, we have the inequality $\sin x \ge x - \frac{x^3}{6}$.
Multiply by $x$ and integrate from 0 to 1:
$$\int_0^1 x \sin x \, dx \ge \int_0^1 \left( x^2 - \frac{x^4}{6} \right) dx = \left[ \frac{x^3}{3} - \frac{x^5}{30} \right]_0^1 = \frac{1}{3} - \frac{1}{30} = \frac{9}{30} = \frac{3}{10}$$
Thus, Option B is **TRUE**.
3. **Evaluate Option C**:
Note that $\cos x \le 1$ on $[0, 1]$. Thus:
$$\int_0^1 x^2 \cos x \, dx \le \int_0^1 x^2 \, dx = \frac{1}{3} < \frac{1}{2}$$
Thus, Option C is **FALSE**.
4. **Evaluate Option D**:
On $[0, 1]$, using the bound $\sin x \ge x - \frac{x^3}{6}$:
Multiply by $x^2$ and integrate from 0 to 1:
$$\int_0^1 x^2 \sin x \, dx \ge \int_0^1 \left( x^3 - \frac{x^5}{6} \right) dx = \left[ \frac{x^4}{4} - \frac{x^6}{36} \right]_0^1 = \frac{1}{4} - \frac{1}{36} = \frac{8}{36} = \frac{2}{9}$$
Thus, Option D is **TRUE**.
5. **Conclusion**:
The correct options are A, B, and D.
Correct Answer: A, B, D