Consider the integrals `I_1 = int_0^1 e^(-y) cos^2 y dy`
`I_2 = int_0^1 e^(-y^2) cos^2 y dy , I_3 = int_0^1 e^(-y^2) dy`
`I_4 = int_0^1 e^(-1/2 y^2) dy`, where `0 lt y lt 1`. If I be the greatest amongest `I_1, I_2, I_3, ...I_4` then

Options

A. `I = I_1`
B. `3I = I_2`
C. `I = I_3`
D. `I = I_4`

Answer

To solve the problem, we need to compare the four integrals \( I_1, I_2, I_3, \) and \( I_4 \) and determine which one is the greatest.

### Step-by-Step Solution:

1. **Define the Integrals:**
\[ I_1 = \int_0^1 e^{-y} \cos^2 y \, dy \]
\[ I_2 = \int_0^1 e^{-y^2} \cos^2 y \, dy \]
\[ I_3 = \int_0^1 e^{-y^2} \, dy \]
\[ I_4 = \int_0^1 e^{-\frac{1}{2}y^2} \, dy \]

2. **Compare \( I_1 \) and \( I_2 \):**
Since \( y^2 < y \) for \( 0 < y < 1 \), we have:
\[ e^{-y^2} > e^{-y} \]
Therefore, multiplying both sides by \( \cos^2 y \) (which is always non-negative), we get:
\[ e^{-y^2} \cos^2 y > e^{-y} \cos^2 y \]
Thus:
\[ I_2 > I_1 \]

3. **Compare \( I_2 \) and \( I_3 \):**
Since \( \cos^2 y \) is always less than or equal to 1 for \( 0 < y < 1 \):
\[ e^{-y^2} \cos^2 y < e^{-y^2} \]
Therefore:
\[ I_2 < I_3 \]

4. **Compare \( I_3 \) and \( I_4 \):**
Since \( \frac{1}{2}y^2 < y^2 \) for \( 0 < y < 1 \), we have:
\[ e^{-\frac{1}{2}y^2} > e^{-y^2} \]
Thus:
\[ I_4 > I_3 \]

5. **Combine the Inequalities:**
From the comparisons, we have:
\[ I_4 > I_3 > I_2 > I_1 \]

6. **Conclusion:**
Therefore, the greatest integral among \( I_1, I_2, I_3, \) and \( I_4 \) is:
\[ I = I_4 \]

### Final Answer:
\[ I = I_4 \]