CUET UG 2025 — Mathematics Calculus

The integral to evaluate is:

$$\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx$$∫12​3−x​+x​x​​dx

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For definite integrals, the following property holds:

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$∫ab​f(x)dx=∫ab​f(a+b−x)dx

Here $a = 1$a=1 and $b = 2$b=2, so $a + b = 3$a+b=3.

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Let $\displaystyle I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx$I=∫12​3−x​+x​x​​dx

Replacing $x$x with $3-x$3−x:

$$I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{3 - (3-x)} + \sqrt{3-x}} dx$$I=∫12​3−(3−x)​+3−x​3−x​​dx

$$I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} dx$$I=∫12​x​+3−x​3−x​​dx

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Adding both expressions for $I$I:

$$I + I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3 - x} + \sqrt{x}} dx + \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{3-x} + \sqrt{x}} dx$$I+I=∫12​3−x​+x​x​​dx+∫12​3−x​+x​3−x​​dx

$$2I = \int_{1}^{2} \frac{\sqrt{x} + \sqrt{3-x}}{\sqrt{3 - x} + \sqrt{x}} dx$$2I=∫12​3−x​+x​x​+3−x​​dx

$$2I = \int_{1}^{2} 1 \, dx$$2I=∫12​1dx

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$$2I = [x]_{1}^{2}$$2I=[x]12​

$$2I = 2 - 1$$2I=2−1

$$2I = 1$$2I=1

$$I = \frac{1}{2}$$I=21​

Therefore, the integral equals $\frac{1}{2}$21​.