The value of the integral `int_(0)^(3) sqrt(3+x^(3))dx`lies in the interval

Options

A. (1,3)
B. (2,30)
C. `(4,2sqrt(30))`
D. none of these

Answer

To find the value of the integral \( I = \int_{0}^{3} \sqrt{3 + x^3} \, dx \) and determine the interval in which it lies, we can follow these steps:

### Step 1: Establish the bounds of the integrand
We first analyze the function \( \sqrt{3 + x^3} \) over the interval \( [0, 3] \).

- When \( x = 0 \):
\[ \sqrt{3 + 0^3} = \sqrt{3} \]

- When \( x = 3 \):
\[ \sqrt{3 + 3^3} = \sqrt{3 + 27} = \sqrt{30} \]

Thus, as \( x \) varies from 0 to 3, \( \sqrt{3 + x^3} \) varies from \( \sqrt{3} \) to \( \sqrt{30} \).

### Step 2: Set up inequalities for the integral
Since \( \sqrt{3 + x^3} \) is increasing in the interval \( [0, 3] \), we can establish the following inequalities:
\[ \sqrt{3} \leq \sqrt{3 + x^3} \leq \sqrt{30} \]

### Step 3: Integrate the inequalities
Now, we integrate the inequalities over the interval from 0 to 3:
\[ \int_{0}^{3} \sqrt{3} \, dx \leq \int_{0}^{3} \sqrt{3 + x^3} \, dx \leq \int_{0}^{3} \sqrt{30} \, dx \]

Calculating the left-hand side:
\[ \int_{0}^{3} \sqrt{3} \, dx = \sqrt{3} \cdot \left[ x \right]_{0}^{3} = \sqrt{3} \cdot (3 - 0) = 3\sqrt{3} \]

Calculating the right-hand side:
\[ \int_{0}^{3} \sqrt{30} \, dx = \sqrt{30} \cdot \left[ x \right]_{0}^{3} = \sqrt{30} \cdot (3 - 0) = 3\sqrt{30} \]

### Step 4: Combine the results
From the inequalities, we have:
\[ 3\sqrt{3} \leq I \leq 3\sqrt{30} \]

### Conclusion
Thus, the value of the integral \( I = \int_{0}^{3} \sqrt{3 + x^3} \, dx \) lies in the interval:
\[ [3\sqrt{3}, 3\sqrt{30}] \]