The value of \( \int_{0}^{3}\displaystyle\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{3}{2} \)
By the property \( \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx \) with a = 3, x can be replaced by \( 3-x \):
\( \displaystyle I=\int_{0}^{3}\frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\,dx \)
\( \displaystyle I=\int_{0}^{3}\frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}}\,dx \)
Adding the two forms of I, the numerators add up to the common denominator:
\( \displaystyle 2I=\int_{0}^{3}\frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}}\,dx \)
\( \displaystyle =\int_{0}^{3}1\,dx \)
\( \displaystyle =3 \)
\( \displaystyle I=\frac{3}{2} \)
Tip: For \( \int_{0}^{a}\frac{f(x)}{f(x)+f(a-x)}\,dx \), the answer is always \( \frac{a}{2} \), half the length of the interval.