`int_(1)^(2) sqrt(x)/ (sqrt(3-x) + sqrt(x))dx=`

Options

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

Answer

To solve the integral

\[ I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} \, dx, \]

we can use a property of definite integrals. The property states that if

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx, \]

then we can replace \(x\) with \(3 - x\) in our integral.

### Step 1: Apply the property

Let’s calculate \(I\) using the property:

\[ I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{3-(3-x)} + \sqrt{3-x}} \, dx = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} \, dx. \]

### Step 2: Combine the two integrals

Now we have two expressions for \(I\):

1. \(I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} \, dx\)
2. \(I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} \, dx\)

Adding these two integrals gives:

\[ 2I = \int_{1}^{2} \left( \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} + \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} \right) \, dx. \]

### Step 3: Simplify the expression

Notice that the denominators are the same, so we can combine the fractions:

\[ 2I = \int_{1}^{2} \frac{\sqrt{x} + \sqrt{3-x}}{\sqrt{3-x} + \sqrt{x}} \, dx = \int_{1}^{2} 1 \, dx. \]

### Step 4: Evaluate the integral

Now, we can evaluate the integral:

\[ \int_{1}^{2} 1 \, dx = [x]_{1}^{2} = 2 - 1 = 1. \]

### Step 5: Solve for \(I\)

Now we have:

\[ 2I = 1 \implies I = \frac{1}{2}. \]

### Final Answer

Thus, the value of the integral is

\[ \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} \, dx = \frac{1}{2}. \]

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