The value of the integral, $\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} dx$ is

JEE Main · 2006 · 30 Apr 2006 · Offline · Definite Integrals

The value of the integral,

\int_{3}^{6} \frac{x}{9 - x + x} d x

Answer: (B)

Call the integral

I = \int_{3}^{6} \frac{x}{9 - x + x} d x

. Replacing

x

3 + 6 - x = 9 - x

, which leaves the limits unchanged, turns the integrand into

\frac{9 - x}{x + 9 - x}

, so

I = \int_{3}^{6} \frac{9 - x}{x + 9 - x} d x

as well. Adding the two forms, the numerators combine over the common denominator to give

1

, so

2 I = \int_{3}^{6} 1 d x = 3

. Hence

I = \frac{3}{2}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option

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