\(\int_{0}^{a}\displaystyle\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{a}{2}\)
Let I be the integral. By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\) (King’s property), replace x by a − x:
\( \displaystyle I = \int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx \)
Adding the two forms of I, the numerators add up to the common denominator:
\( \displaystyle 2I = \int_{0}^{a}\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx = \int_{0}^{a}1\,dx = a \)
\( \displaystyle I = \frac{a}{2} \)
Tip: A result worth remembering: \(\int_{0}^{a}\frac{f(x)}{f(x)+f(a-x)}\,dx = \frac{a}{2}\) for any such f, so these questions can be answered at sight.