`int_(0)^(1)sqrt(x+1)` `dx =`

Options

A. `(2)/(3)(sqrt(8)+1)`
B. `(3)/(2)(sqrt(8)-1)`
C. `(2)/(3)(sqrt(8)-1)`
D. `(3)/(2)(sqrt(8)+1)`

Answer

To solve the definite integral \( \int_{0}^{1} \sqrt{x+1} \, dx \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{1} \sqrt{x+1} \, dx \]

### Step 2: Substitution
We will use the substitution \( t = x + 1 \). Then, we differentiate both sides:
\[ dt = dx \]
This means that \( dx = dt \).

### Step 3: Change the limits of integration
Now, we need to change the limits of integration according to our substitution:
- When \( x = 0 \), \( t = 0 + 1 = 1 \)
- When \( x = 1 \), \( t = 1 + 1 = 2 \)

Thus, the new limits of integration are from \( t = 1 \) to \( t = 2 \).

### Step 4: Rewrite the integral in terms of \( t \)
Now we can rewrite the integral in terms of \( t \):
\[ I = \int_{1}^{2} \sqrt{t} \, dt \]

### Step 5: Integrate
We know that \( \sqrt{t} = t^{1/2} \). The integral of \( t^{n} \) is given by:
\[ \int t^{n} \, dt = \frac{t^{n+1}}{n+1} + C \]
For our case, \( n = \frac{1}{2} \):
\[ \int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \]

### Step 6: Evaluate the integral with limits
Now we evaluate the integral from \( 1 \) to \( 2 \):
\[ I = \left[ \frac{2}{3} t^{3/2} \right]_{1}^{2} \]
Calculating this gives:
\[ I = \frac{2}{3} \left( 2^{3/2} - 1^{3/2} \right) \]

### Step 7: Simplify
Calculating \( 2^{3/2} \):
\[ 2^{3/2} = \sqrt{8} = 2\sqrt{2} \]
Thus, we have:
\[ I = \frac{2}{3} \left( 2\sqrt{2} - 1 \right) \]

### Final Answer
The final answer for the integral \( \int_{0}^{1} \sqrt{x+1} \, dx \) is:
\[ I = \frac{2}{3} (2\sqrt{2} - 1) \]

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