∫_{{2}}²f(x)dx is equal to ?

Correct Answer is : 6−3−226-{3}-2{2}6−3−22

Correct Answer is : 6−3−226-{3}-2{2}6−3−22

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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 99-100

Consider the following for the two (02) items that follow:

Let

f(x) = [x^{2}]

where

[.]

is the greatest integer function.

Question : 100 of 120

Marks: +1, -0

∫_{\sqrt{2}}^{2}f(x)dx

is equal to ?

Solution:

Concept:

The integral of a floor function is found by splitting the interval where the floor value is constant.

Explanation:

For

x

from

\sqrt{2}

2

x^2

lies between

2

and

4

When

\sqrt{2} \le x < \sqrt{3}

x^2

is in

[2,3)

, so

\lfloor x^2 \rfloor = 2

When

\sqrt{3} \le x \le 2

x^2

is in

[3,4]

, so

\lfloor x^2 \rfloor = 3

Thus, the integral splits:

\int_{\sqrt{2}}^{2} f(x)\,dx = \int_{\sqrt{2}}^{\sqrt{3}} 2\,dx + \int_{\sqrt{3}}^{2} 3\,dx

Compute each part:

2 \times (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}

3 \times (2 - \sqrt{3}) = 6 - 3\sqrt{3}

Add them:

(2\sqrt{3} - 2\sqrt{2}) + (6 - 3\sqrt{3}) = 6 - \sqrt{3} - 2\sqrt{2}

Answer:

Option A:

6-\sqrt{3}-2\sqrt{2}

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