\(\int_{1}^{2}\displaystyle\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)
Let
\( \displaystyle I = \int_{1}^{2}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx \)
By the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx\), with \(a + b = 3\), replace x by \(3 - x\):
\( \displaystyle I = \int_{1}^{2}\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}\,dx \)
Adding the two forms of I, the numerators add up to the common denominator:
\( \displaystyle 2I = \int_{1}^{2} 1\,dx = 1 \)
\( \displaystyle I = \frac{1}{2} \)
Tip: A result worth remembering: \(\int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)}\,dx = \frac{b-a}{2}\), half the length of the interval.