Question

\( \int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4033 - x}} \, dx \) is equal to:

• A. \( \frac{1}{4} \)
• B. \( \frac{3}{2} \)
• C. \( \frac{2017}{2} \)
• D. \( \frac{1}{2} \)
• E. \( 508 \)

Hint:

For any integral of the form \( \int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx \), the result is always \( \frac{b-a}{2} \). Here, \( a=2016, b=2017 \), so result \( = (2017-2016)/2 = 1/2 \).

Answer 1

The Correct Option is D

Concept: This problem uses the specific property of definite integrals: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \). When the integrand is of the form \( \frac{g(x)}{g(x) + g(a+b-x)} \), applying this property creates a system that simplifies to half the length of the interval.

Step 1: Apply the integral property.
Let \( I = \int_{2016}^{2017} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4033 - x}} \, dx \quad \cdots (1) \) Using the property \( x \to (2016 + 2017) - x = 4033 - x \): \[ I = \int_{2016}^{2017} \frac{\sqrt{4033 - x}}{\sqrt{4033 - x} + \sqrt{4033 - (4033 - x)}} \, dx \] \[ I = \int_{2016}^{2017} \frac{\sqrt{4033 - x}}{\sqrt{4033 - x} + \sqrt{x}} \, dx \quad \cdots (2) \]

Step 2: Combine the two integrals.
Adding (1) and (2): \[ 2I = \int_{2016}^{2017} \left( \frac{\sqrt{x}}{\sqrt{x} + \sqrt{4033 - x}} + \frac{\sqrt{4033 - x}}{\sqrt{x} + \sqrt{4033 - x}} \right) \, dx \] \[ 2I = \int_{2016}^{2017} \frac{\sqrt{x} + \sqrt{4033 - x}}{\sqrt{x} + \sqrt{4033 - x}} \, dx \] \[ 2I = \int_{2016}^{2017} 1 \, dx \]

Step 3: Solve for \( I \).
\[ 2I = [x]_{2016}^{2017} = 2017 - 2016 = 1 \] \[ I = \frac{1}{2} \]