`P rov et h a t4lt=int_1^3sqrt(3+x^2)lt=4sqrt(3)`

Explanation

To prove that \( 4 \leq \int_1^3 \sqrt{3 + x^2} \, dx \leq 4\sqrt{3} \), we will follow a systematic approach.

### Step 1: Establish the bounds for \( \sqrt{3 + x^2} \)

Given that \( x \) lies in the interval \( [1, 3] \):
- The minimum value of \( x^2 \) in this interval is \( 1^2 = 1 \).
- The maximum value of \( x^2 \) in this interval is \( 3^2 = 9 \).

Thus, we can establish:
\[ 3 + 1 \leq 3 + x^2 \leq 3 + 9 \]
which simplifies to:
\[ 4 \leq 3 + x^2 \leq 12 \]

### Step 2: Take the square root

Taking the square root of the inequalities:
\[ \sqrt{4} \leq \sqrt{3 + x^2} \leq \sqrt{12} \]
This simplifies to:
\[ 2 \leq \sqrt{3 + x^2} \leq 2\sqrt{3} \]

### Step 3: Integrate the bounds

Now, we will integrate the inequalities over the interval from 1 to 3:
\[ \int_1^3 2 \, dx \leq \int_1^3 \sqrt{3 + x^2} \, dx \leq \int_1^3 2\sqrt{3} \, dx \]

Calculating the left integral:
\[ \int_1^3 2 \, dx = 2[x]_1^3 = 2(3 - 1) = 2 \times 2 = 4 \]

Calculating the right integral:
\[ \int_1^3 2\sqrt{3} \, dx = 2\sqrt{3}[x]_1^3 = 2\sqrt{3}(3 - 1) = 2\sqrt{3} \times 2 = 4\sqrt{3} \]

### Step 4: Combine the results

From the integration results, we have:
\[ 4 \leq \int_1^3 \sqrt{3 + x^2} \, dx \leq 4\sqrt{3} \]

Thus, we have proved that:
\[ 4 \leq \int_1^3 \sqrt{3 + x^2} \, dx \leq 4\sqrt{3} \]

### Conclusion

Therefore, we conclude that:
\[ 4 \leq \int_1^3 \sqrt{3 + x^2} \, dx \leq 4\sqrt{3} \]
is indeed true.

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