`int _(2) ^(7) (sqrtx)/( sqrtx + sqrt ( 9-x)) dx =`

Options

A. `7/2`
B. `5/2`
C. `7`
D. 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 for a function \( f(x) \),

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

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

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

Now, let's denote this new integral as \( J \):

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

Now we can add the two integrals \( I \) and \( J \):

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

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

\[ I + J = \int_{2}^{7} \frac{\sqrt{x} + \sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} \, dx = \int_{2}^{7} 1 \, dx. \]

The integral of 1 from 2 to 7 is simply:

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

Thus, we have:

\[ I + J = 5. \]

Since \( I \) and \( J \) are equal (as shown by the symmetry of the integrals), we can set \( I = J \). Therefore, we have:

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

So the value of the integral is:

\[ \boxed{\frac{5}{2}}. \]