If `I=int_(0)^(1) (1+e^(-x^2)) dx` then, s

Options

A. `I= in (1,2)`
B. `I in (0,1)`
C. `I in (1+(1)/(e),2]`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{1} (1 + e^{-x^2}) \, dx \), we can break it down into two separate integrals:

\[ I = \int_{0}^{1} 1 \, dx + \int_{0}^{1} e^{-x^2} \, dx \]

### Step 1: Calculate the first integral

The first integral is straightforward:

\[ \int_{0}^{1} 1 \, dx = [x]_{0}^{1} = 1 - 0 = 1 \]

### Step 2: Analyze the second integral

Now, we need to evaluate the second integral \( \int_{0}^{1} e^{-x^2} \, dx \). We do not have an elementary function for this integral, but we can estimate its value.

### Step 3: Estimate \( \int_{0}^{1} e^{-x^2} \, dx \)

We know that \( e^{-x^2} \) is a decreasing function on the interval \( [0, 1] \). Therefore, we can find bounds for this integral.

- At \( x = 0 \), \( e^{-0^2} = e^0 = 1 \).
- At \( x = 1 \), \( e^{-1^2} = e^{-1} \).

Thus, we can say:

\[ e^{-1} < e^{-x^2} < 1 \quad \text{for } x \in [0, 1] \]

### Step 4: Set up the inequalities for the integral

Integrating the inequalities over the interval from 0 to 1 gives:

\[ \int_{0}^{1} e^{-1} \, dx < \int_{0}^{1} e^{-x^2} \, dx < \int_{0}^{1} 1 \, dx \]

This simplifies to:

\[ e^{-1} < \int_{0}^{1} e^{-x^2} \, dx < 1 \]

### Step 5: Combine results

Adding 1 to all parts of the inequality gives:

\[ 1 + e^{-1} < I < 2 \]

### Step 6: Conclusion

Thus, we have established that:

\[ I \in \left( 1 + \frac{1}{e}, 2 \right) \]

This means that the correct option is \( I \) belongs to \( 1 + \frac{1}{e} \) to \( 2 \).

### Final Answer

The correct option is \( I \in \left( 1 + \frac{1}{e}, 2 \right) \).

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