CUET UG 2025 — Mathematics Calculus

Let $\displaystyle I = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx$I=∫0a​x​+a−x​x​​dx ... (1)

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Using the property $\displaystyle \int_0^a f(x)dx = \int_0^a f(a-x)dx$∫0a​f(x)dx=∫0a​f(a−x)dx, replace $x$x with $(a-x)$(a−x):

$$I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{a-(a-x)}} dx$$I=∫0a​a−x​+a−(a−x)​a−x​​dx

Since $a-(a-x) = x$a−(a−x)=x:

$\displaystyle I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} dx$I=∫0a​a−x​+x​a−x​​dx ... (2)

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Adding equations (1) and (2):

$$I + I = \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx + \int_0^a \frac{\sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}} dx$$I+I=∫0a​x​+a−x​x​​dx+∫0a​x​+a−x​a−x​​dx

$$2I = \int_0^a \left[\frac{\sqrt{x} + \sqrt{a-x}}{\sqrt{x} + \sqrt{a-x}}\right] dx$$2I=∫0a​[x​+a−x​x​+a−x​​]dx

$$2I = \int_0^a 1 \, dx$$2I=∫0a​1dx

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$$2I = [x]_0^a$$2I=[x]0a​

$$2I = a - 0$$2I=a−0

$$2I = a$$2I=a

$$I = \dfrac{a}{2}$$I=2a​

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Therefore, $\displaystyle \int_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} dx = \dfrac{a}{2}$∫0a​x​+a−x​x​​dx=2a​