Question

The value of the definite integral \( \int_2^7 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9-x}} dx \) is:

• A. \( \frac{7}{2} \)
• B. \( \frac{5}{2} \)
• C. \( 7 \)
• D. \( 2 \)

Hint:

Recognize the special form \( \int_a^b \frac{f(x)}{f(x) + f(a+b-x)} dx \). Integrals of this type evaluate to \( \frac{b-a}{2} \). Here, \( \frac{7-2}{2} = \frac{5}{2} \).

Answer 1

The Correct Option is B

Step 1: Let \( I = \int_2^7 \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9-x}} dx \). Use the property \( \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \). Here, \( a=2, b=7 \), so \( a+b=9 \).
Step 2: Apply the property by replacing \( x \) with \( 9-x \). \[ I = \int_2^7 \frac{\sqrt{9-x}}{\sqrt{9-x} + \sqrt{9-(9-x)}} dx = \int_2^7 \frac{\sqrt{9-x}}{\sqrt{9-x} + \sqrt{x}} dx \] Step 3: Add the original and transformed integrals. \[ 2I = \int_2^7 \frac{\sqrt{x} + \sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} dx = \int_2^7 1 \, dx \] Step 4: Evaluate the simple integral. \[ 2I = [x]_2^7 = 7 - 2 = 5 \implies I = \frac{5}{2} \]