Let `I_(1)=int_(0)^(pi//4)e^(x^(2))dx, I_(2) = int_(0)^(pi//4) e^(x)dx, I_(3) = int_(0)^(pi//4)e^(x^(2)).sin x dx`, then :

Options

A. `I_(1)gt I_(2)gt I_(3)gt I_(4)`
B. `I_(2)gt I_(3)gt I_(4)gt I_(1)`
C. `I_(3)gt I_(4)gt I_(1)gt I_(2)`
D. `I_(2)gt I_(1)gt I_(3)gt I_(4)`

Answer

To solve the problem, we need to analyze the three integrals \( I_1 \), \( I_2 \), and \( I_3 \) defined as follows:

1. \( I_1 = \int_{0}^{\frac{\pi}{4}} e^{x^2} \, dx \)
2. \( I_2 = \int_{0}^{\frac{\pi}{4}} e^{x} \, dx \)
3. \( I_3 = \int_{0}^{\frac{\pi}{4}} e^{x^2} \sin x \, dx \)

We need to determine the relationship between these integrals.

### Step 1: Analyze \( I_1 \) and \( I_2 \)

First, we note that for \( x \) in the interval \( [0, \frac{\pi}{4}] \), the value of \( x^2 \) is less than \( x \) since \( \frac{\pi}{4} < 1 \). Therefore, we have:

\[ x^2 < x \quad \text{for } x \in [0, \frac{\pi}{4}] \]

This implies that:

\[ e^{x^2} < e^{x} \quad \text{for } x \in [0, \frac{\pi}{4}] \]

Thus, when we integrate both sides over the interval from \( 0 \) to \( \frac{\pi}{4} \):

\[ I_1 < I_2 \]

### Step 2: Analyze \( I_3 \)

Next, we analyze \( I_3 \). We know that \( \sin x \) is a positive function in the interval \( [0, \frac{\pi}{4}] \) and \( \sin x < 1 \) for \( x \in [0, \frac{\pi}{4}] \). Therefore, we can say:

\[ e^{x^2} \sin x < e^{x^2} \quad \text{for } x \in [0, \frac{\pi}{4}] \]

Integrating both sides gives:

\[ I_3 < I_1 \]

### Step 3: Combine the results

From our analysis, we have established the following inequalities:

\[ I_3 < I_1 < I_2 \]

### Conclusion

Thus, the relationship among the integrals is:

\[ I_3 < I_1 < I_2 \]

### Final Answer

The correct order is \( I_3 < I_1 < I_2 \).

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