Value of \(\int_{2}^{3}\displaystyle\frac{\sqrt{x}}{\sqrt{x}+\sqrt{5-x}}\,dx\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
Let I be the integral. By the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx\), with \(a + b = 5\), x is replaced by \(5 - x\):
\(\displaystyle I = \int_{2}^{3} \frac{\sqrt{5 - x}}{\sqrt{5 - x} + \sqrt{x}}\,dx\)
Adding the two forms of I:
\(\displaystyle 2I = \int_{2}^{3} \frac{\sqrt{x} + \sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}}\,dx\)
\(\displaystyle = \int_{2}^{3} 1\,dx = 1\)
\(\displaystyle I = \frac{1}{2}\)
Tip: A result worth remembering: \(\int_{a}^{b} \frac{f(x)}{f(x) + f(a + b - x)}\,dx = \frac{b - a}{2}\). Here it gives \(\frac{3 - 2}{2} = \frac{1}{2}\) at once.