Statement I- The area of the curve `y=sin^2 x "from" 0 "to" pi` will be more than that of the curve `y=sin x "from" 0 "to" pi`.
Statement II -`x^(2)gtx, if xgt1`.

Options

A. Statement I is true, Statement II is also true, Statement II is the correct explanation of Statement I.
B. Statement I is true, Statement II is also true, Statement II is not the correct explanation of Statement I.
C. Statement I is true , Statement II is false
D. Statement I is false , Statement II is true

Answer

To solve the problem, we need to analyze both statements one by one.

### Step 1: Analyze Statement I
We need to compare the areas under the curves \( y = \sin^2 x \) and \( y = \sin x \) from \( x = 0 \) to \( x = \pi \).

1. **Area under \( y = \sin^2 x \)**:
\[ A_1 = \int_0^\pi \sin^2 x \, dx \]
We can use the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \):
\[ A_1 = \int_0^\pi \frac{1 - \cos(2x)}{2} \, dx = \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_0^\pi \]
Evaluating this:
\[ A_1 = \frac{1}{2} \left[ \pi - 0 \right] = \frac{\pi}{2} \]

2. **Area under \( y = \sin x \)**:
\[ A_2 = \int_0^\pi \sin x \, dx \]
Evaluating this integral:
\[ A_2 = \left[ -\cos x \right]_0^\pi = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]

3. **Comparison**:
Now we compare \( A_1 \) and \( A_2 \):
\[ A_1 = \frac{\pi}{2} \quad \text{and} \quad A_2 = 2 \]
Since \( \frac{\pi}{2} \approx 1.57 < 2 \), we conclude:
\[ A_1 < A_2 \]
Thus, Statement I is **False**.

### Step 2: Analyze Statement II
We need to check if \( x^2 > x \) for \( x > 1 \).

1. **Rearranging the inequality**:
\[ x^2 - x > 0 \implies x(x - 1) > 0 \]
The product \( x(x - 1) \) is positive when both factors are positive or both are negative.

2. **Finding the intervals**:
- For \( x > 1 \): Both \( x > 0 \) and \( x - 1 > 0 \) are true, hence \( x(x - 1) > 0 \).

Thus, Statement II is **True**.

### Conclusion
- Statement I is **False**.
- Statement II is **True**.

The correct answer is that Statement I is False and Statement II is True.