`int_(a)^(b) ( sqrt(x))/( sqrt(x) + sqrt(a+b-x))dx` is equal to

Options

A. `(pi)/( 2)`
B. `pi`
C. `(1)/(2) ( b-a)`
D. `b-a`

Answer

To solve the integral \( I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a+b-x}} \, dx \), we will use a property of definite integrals. Here are the steps:

### Step 1: Define the Integral
Let
\[ I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a+b-x}} \, dx \]

### Step 2: Use the Property of Definite Integrals
We can use the property of definite integrals which states that:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]
In our case, we will substitute \( x \) with \( a+b-x \):
\[ I = \int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x} + \sqrt{x}} \, dx \]

### Step 3: Simplify the New Integral
Now, let's denote this new integral as \( I' \):
\[ I' = \int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x} + \sqrt{x}} \, dx \]
Notice that:
\[ I' = \int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x} + \sqrt{x}} \, dx \]

### Step 4: Add the Two Integrals
Now, we will add \( I \) and \( I' \):
\[ I + I' = \int_{a}^{b} \left( \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a+b-x}} + \frac{\sqrt{a+b-x}}{\sqrt{a+b-x} + \sqrt{x}} \right) \, dx \]
The denominators are the same, so we can combine the numerators:
\[ I + I' = \int_{a}^{b} \frac{\sqrt{x} + \sqrt{a+b-x}}{\sqrt{x} + \sqrt{a+b-x}} \, dx = \int_{a}^{b} 1 \, dx \]

### Step 5: Evaluate the Integral
The integral simplifies to:
\[ I + I' = \int_{a}^{b} 1 \, dx = b - a \]

### Step 6: Relate \( I \) and \( I' \)
Since \( I' \) is equal to \( I \), we can write:
\[ 2I = b - a \]
Thus,
\[ I = \frac{b - a}{2} \]

### Final Answer
Therefore, the value of the integral is:
\[ I = \frac{1}{2}(b - a) \]

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