Which of the following inequalities is/are TRUE ?

Options

A. `overset(1) underset(0)int x cos x dx ge (3)/(8)`
B. `overset(1) underset(0)int x sin x dx ge (3)/(10)`
C. `overset(1) underset(0)int x^(2) cos x dx ge (1)/(2)`
D. `overset(1) underset(0)int x^(2) sin x dx ge (2)/(9)`

Answer

To solve the inequalities given in the question, we will analyze each option step by step.

### Step 1: Analyze Option A
We need to check if:
\[ \int_0^1 x \cos x \, dx \geq \frac{3}{8} \]

**Solution:**
Using integration by parts, let:
- \( u = x \) → \( du = dx \)
- \( dv = \cos x \, dx \) → \( v = \sin x \)

Now, applying integration by parts:
\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx \]
\[ = x \sin x + \cos x \]

Evaluating from 0 to 1:
\[ \int_0^1 x \cos x \, dx = \left[ x \sin x + \cos x \right]_0^1 \]
\[ = (1 \cdot \sin 1 + \cos 1) - (0 + 1) \]
\[ = \sin 1 + \cos 1 - 1 \]

Using numerical values, \( \sin 1 \approx 0.8415 \) and \( \cos 1 \approx 0.5403 \):
\[ \sin 1 + \cos 1 - 1 \approx 0.8415 + 0.5403 - 1 \approx 0.3818 \]

Since \( 0.3818 \geq \frac{3}{8} \approx 0.375 \), option A is **TRUE**.

### Step 2: Analyze Option B
We need to check if:
\[ \int_0^1 x \sin x \, dx \geq \frac{3}{10} \]

**Solution:**
Using integration by parts again:
- \( u = x \) → \( du = dx \)
- \( dv = \sin x \, dx \) → \( v = -\cos x \)

Now, applying integration by parts:
\[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx \]
\[ = -x \cos x + \sin x \]

Evaluating from 0 to 1:
\[ \int_0^1 x \sin x \, dx = \left[ -x \cos x + \sin x \right]_0^1 \]
\[ = (-1 \cdot \cos 1 + \sin 1) - (0 + 0) \]
\[ = -\cos 1 + \sin 1 \]

Using numerical values:
\[ -\cos 1 + \sin 1 \approx -0.5403 + 0.8415 \approx 0.3012 \]

Since \( 0.3012 \geq \frac{3}{10} = 0.3 \), option B is **TRUE**.

### Step 3: Analyze Option C
We need to check if:
\[ \int_0^1 x^2 \cos x \, dx \geq \frac{1}{2} \]

**Solution:**
Using integration by parts:
- Let \( u = x^2 \) → \( du = 2x \, dx \)
- \( dv = \cos x \, dx \) → \( v = \sin x \)

Then:
\[ \int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx \]

Now we need to evaluate \( \int 2x \sin x \, dx \) using integration by parts again:
- Let \( u = 2x \) → \( du = 2 \, dx \)
- \( dv = \sin x \, dx \) → \( v = -\cos x \)

So:
\[ \int 2x \sin x \, dx = -2x \cos x + 2 \int \cos x \, dx \]
\[ = -2x \cos x + 2 \sin x \]

Now substituting back:
\[ \int_0^1 x^2 \cos x \, dx = \left[ x^2 \sin x \right]_0^1 - \left[ -2x \cos x + 2 \sin x \right]_0^1 \]
Calculating the limits:
\[ = (1 \cdot \sin 1) - (0) - \left[ -2 \cdot 1 \cdot \cos 1 + 2 \cdot \sin 1 \right] \]
\[ = \sin 1 - (-2 \cos 1 + 2 \sin 1) = \sin 1 + 2 \cos 1 - 2 \sin 1 \]
\[ = 2 \cos 1 - \sin 1 \]

Using numerical values:
\[ 2 \cdot 0.5403 - 0.8415 \approx 1.0806 - 0.8415 \approx 0.2391 \]

Since \( 0.2391 < \frac{1}{2} = 0.5 \), option C is **FALSE**.

### Step 4: Analyze Option D
We need to check if:
\[ \int_0^1 x^2 \sin x \, dx \geq \frac{2}{9} \]

**Solution:**
Using integration by parts:
- Let \( u = x^2 \) → \( du = 2x \, dx \)
- \( dv = \sin x \, dx \) → \( v = -\cos x \)

Then:
\[ \int x^2 \sin x \, dx = -x^2 \cos x + \int 2x \cos x \, dx \]

Now we need to evaluate \( \int 2x \cos x \, dx \) using integration by parts again:
- Let \( u = 2x \) → \( du = 2 \, dx \)
- \( dv = \cos x \, dx \) → \( v = \sin x \)

So:
\[ \int 2x \cos x \, dx = 2x \sin x - 2 \int \sin x \, dx \]
\[ = 2x \sin x + 2 \cos x \]

Now substituting back:
\[ \int_0^1 x^2 \sin x \, dx = \left[ -x^2 \cos x \right]_0^1 + \left[ 2x \sin x + 2 \cos x \right]_0^1 \]
Calculating the limits:
\[ = -1 \cdot \cos 1 + (2 \cdot 1 \cdot \sin 1 + 2 \cdot \cos 1) - (0 + 2) \]
\[ = -\cos 1 + (2 \sin 1 + 2 \cos 1) - 2 \]
\[ = 2 \sin 1 + \cos 1 - 2 \]

Using numerical values:
\[ = 2 \cdot 0.8415 + 0.5403 - 2 \approx 1.683 - 2 \approx -0.317 \]

Since \( -0.317 < \frac{2}{9} \approx 0.222 \), option D is **FALSE**.

### Final Conclusion
The true inequalities are:
- **Option A: TRUE**
- **Option B: TRUE**
- **Option C: FALSE**
- **Option D: FALSE**