CUET UG 2025 — Mathematics Calculus

The shaded area $A$A is defined by the integral of the function $y = \sqrt{x}$y=x​ from $x = 1$x=1 to $x = 4$x=4.

$$A = \int_{1}^{4} \sqrt{x} \, dx$$A=∫14​x​dx

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The integral of $x^{1/2}$x1/2 is calculated using the power rule:

$$\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$$∫x1/2dx=3/2x3/2​=32​x3/2

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Evaluating the expression at the upper and lower limits:

$$A = [\frac{2}{3}x^{3/2}]_{1}^{4}$$A=[32​x3/2]14​

$$A = \frac{2}{3}(4)^{3/2} - \frac{2}{3}(1)^{3/2}$$A=32​(4)3/2−32​(1)3/2

$$A = \frac{2}{3}(8) - \frac{2}{3}(1)$$A=32​(8)−32​(1)

$$A = \frac{16}{3} - \frac{2}{3}$$A=316​−32​

$$A = \frac{14}{3}$$A=314​

The area is $\frac{14}{3}$314​ square units.

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Note: Yes, the question said "nearest integral value", the answer should have been 6 square units. However, it is NTA's world and we're just living in it.

Here's proof (please don't take it out on us, we're as frustrated as you are):