Question

\[ \int_3^5 \frac{\sqrt{x} \mathrm{~d} x}{\sqrt{8-x}+\sqrt{x}}= \]

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Use the property that replaces x by (3 + 5 - x), then add the two forms of the integral.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For \(\displaystyle\int_a^b f(x)dx\), we may replace \(x\) by \(a + b - x\). Here \(a + b = 8\), which is exactly the number under the root.

Step 2: Apply the property
\(I = \displaystyle\int_3^5\dfrac{\sqrt{x}}{\sqrt{8-x} + \sqrt{x}}dx = \int_3^5\dfrac{\sqrt{8-x}}{\sqrt{x} + \sqrt{8-x}}dx\).

Step 3: Add the two forms
\(2I = \displaystyle\int_3^5 1\,dx = 2\).

Step 4: Result
Hence \(I = 1\).

Final Answer:
The integral equals 1. \[ \boxed{\text{(B) }1} \]