NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

The value of the integral $\int_3^6 \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx$ is:

Choose the correct answer:

Explanation

Solution

1. Identify the Property

To solve this integral, we use the property of definite integrals:

\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx

2. Set up the Equations

Let the given integral be $I$ :

I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \dots \text{(Equation 1)}

Applying the property where $a = 3$ and $b = 6$ , we replace $x$ with $(3 + 6 - x)$ , which is $(9 - x)$ :

I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{9-(9-x)} + \sqrt{9-x}} \, dx

I = \int_3^6 \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx \quad \dots \text{(Equation 2)}

3. Add the Equations

Adding Equation 1 and Equation 2:

I + I = \int_3^6 \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx + \int_3^6 \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx

2I = \int_3^6 \frac{\sqrt{x} + \sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx

2I = \int_3^6 1 \, dx

4. Solve for $I$

Now, integrate:

2I = [x]_3^6

2I = 6 - 3

2I = 3

I = \frac{3}{2}

Correct Option:

The correct option is (c) $3/2$ .