The value of \(\int_{0}^{3}\displaystyle \frac{\sqrt{x}}{\sqrt{x} + \sqrt{3 - x}}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{2}\)
Let
\( \displaystyle I = \int_{0}^{3}\frac{\sqrt{x}}{\sqrt{x} + \sqrt{3 - x}}\,dx \)
By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\), we may replace x by \(3 - x\):
\( \displaystyle I = \int_{0}^{3}\frac{\sqrt{3 - x}}{\sqrt{3 - x} + \sqrt{x}}\,dx \)
Adding the two forms of I, the numerators add up to the denominator:
\( \displaystyle 2I = \int_{0}^{3}\frac{\sqrt{x} + \sqrt{3 - x}}{\sqrt{x} + \sqrt{3 - x}}\,dx \)
\( \displaystyle 2I = \int_{0}^{3} 1\,dx = 3 \)
\( \displaystyle I = \frac{3}{2} \)