UPSC NDA I 2025 Mathematics Paper

Given,
The function is f(x)=|x−3|, and we need to find the area bounded by the curve and the line y=3.
To find the points of intersection, we set the function equal to 3 :
|x−3|=3
Solving for x :
- ForX ≥3,x−3=3, which gives x=6.
- For (x<3),3−x=3, which gives x=0.
Therefore, the points of intersection are x=0 and x=6.
The area can be calculated by integrating the difference between the curve and the line from x=0 to x=6. The integral is split into two parts due to the absolute value function:
A=

3(3−x)

∫

0

dx+

6(x−3)

∫

3

dx
For x in [0,3],(f(x)=3−x), and for (x in [3,6]),(f(x)=x−3).
Compute both integrals:
- For x in [0,3] :

3(3−x)

∫

0

dx=[3x−‌

x2

2

]03=9−4.5=4.5
- For x in [3,6] :

6(x−3)

∫

3

dx=[‌

x2

2

−3x]36=4.5
Step 4: The total area is the sum of the two areas:
A=4.5+4.5=9 square units
∴ The total area bounded by the curve and the line is 9 square units.