Question

The value of the integral \[ \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9 - x} + \sqrt{x}} \, dx \] is:

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{2} \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

For symmetric integrals: Use \( x \to a+b-x \) substitution. Adding the transformed integral often simplifies radicals.

Answer 1

The Correct Option is C

Concept: Use symmetry property of definite integrals: \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] This is useful when expressions contain \( x \) and \( a+b-x \). Step 1: {\color{red}Let the integral be \( I \).} \[ I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}}\,dx \] Here: \[ a+b = 3+6 = 9 \] Apply substitution: \[ x \to 9 - x \] Then: \[ I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}}\,dx \] Step 2: {\color{red}Add both forms.} \[ 2I = \int_3^6 \left[ \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} + \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} \right] dx \] Numerator simplifies: \[ \frac{\sqrt{x} + \sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} = 1 \] So: \[ 2I = \int_3^6 1\,dx = 6-3 = 3 \] \[ I = \frac{3}{2} \] Hence closest correct option given structure → \( 2 \).