`int_(2)^(7) sqrt(x)/ (sqrt(x) + sqrt(9-x))dx=`

Options

A. `9/2`
B. `(-9)/2`
C. `5/2`
D. `(-5)/2`

Answer

To solve the integral

\[ I = \int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} \, dx, \]

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

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

then

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

In our case, \( a = 2 \) and \( b = 7 \), so \( a + b = 9 \). We can rewrite the integral as:

\[ I = \int_{2}^{7} f(9 - x) \, dx, \]

where

\[ f(x) = \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}}. \]

Now, we need to find \( f(9 - x) \):

\[ f(9 - x) = \frac{\sqrt{9 - x}}{\sqrt{9 - x} + \sqrt{x}}. \]

Notice that

\[ f(9 - x) = \frac{\sqrt{9 - x}}{\sqrt{9 - x} + \sqrt{x}} = \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}}. \]

Now we can express \( I \) as:

\[ I = \int_{2}^{7} f(9 - x) \, dx = \int_{2}^{7} \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} \, dx. \]

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

1. \( I = \int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} \, dx \)
2. \( I = \int_{2}^{7} \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} \, dx \)

Adding these two equations gives:

\[ 2I = \int_{2}^{7} \left( \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} + \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} \right) \, dx. \]

The expression inside the integral simplifies to:

\[ \frac{\sqrt{x} + \sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} = 1. \]

Thus, we have:

\[ 2I = \int_{2}^{7} 1 \, dx. \]

Calculating the integral:

\[ \int_{2}^{7} 1 \, dx = [x]_{2}^{7} = 7 - 2 = 5. \]

So we have:

\[ 2I = 5 \implies I = \frac{5}{2}. \]

Thus, the final answer is:

\[ \int_{2}^{7} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} \, dx = \frac{5}{2}. \]