\( \int_{1}^{3}\displaystyle\frac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}\,dx \) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
By the property \( \int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx \), with a + b = 4, x can be replaced by 4 − x:
\( \displaystyle I=\int_{1}^{3}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}\,dx \)
\( \displaystyle =\int_{1}^{3}\frac{\sqrt{4-x}}{\sqrt{4-x}+\sqrt{x}}\,dx \)
Adding the two forms of I:
\( \displaystyle 2I=\int_{1}^{3}\frac{\sqrt{x}+\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}}\,dx \)
\( \displaystyle =\int_{1}^{3}1\,dx \)
\( \displaystyle =2 \)
\( \displaystyle I=1 \)
Shortcut: For integrals of the form \( \int_{a}^{b}\frac{f(x)}{f(x)+f(a+b-x)}\,dx \), the value is always \( \frac{b-a}{2} \); here \( \frac{3-1}{2}=1 \).