Question

The value of $\underset{0}{\overset{4042}{\int }}\frac{\sqrt{x}dx}{\sqrt{x}+\sqrt{4042-x}}$ is equal to

• A. 4042
• B. 2021
• C. 8084
• D. 1010

Answer 1

Answer: (B)

$\underset{a}{\overset{b}{\int }}\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}{2}$