If I = ∫_0^4(√x)/(√x+√(4-x)) dx , then 8I is:

The correct answer is Option (3) → 16

Let

    \( \displaystyle I = \int_{0}^{4}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}\,dx \qquad (1) \)

By the property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\), replace x by 4 − x:

    \( \displaystyle I = \int_{0}^{4}\frac{\sqrt{4-x}}{\sqrt{4-x}+\sqrt{x}}\,dx \qquad (2) \)

Adding (1) and (2):

    \( \displaystyle 2I = \int_{0}^{4} 1\,dx = 4 \)

    \(I = 2\)

So

    \(8I = 16\)