JEE Main 2025 Shift 1: The floor-function definite integral
The question is remembered because the obvious approach — treating the floor function as a constant and integrating — produces a number that is wrong by a small but definite amount, and the smallness is what makes it dangerous.
A full worked solution for the JEE Main 2025 January shift 1 calculus question, where the integrand combined the floor function with fractional-part notation and the only correct approach was to split the range at every integer.
How this actually works
Reading the question
The question gave a definite integral whose integrand was built from the floor function and the fractional-part notation. The first trap is to try to integrate the floor function as if it were a smooth polynomial. It is not. The floor part is a step function, and the only correct move is to split the integral at the jump points, which are the integers inside the integration range.
Splitting the range
Over each open interval between two consecutive integers, the floor value is constant. That turns the integral into a finite sum of rectangles, one per unit-width strip, plus the fractional-part contribution, which is continuous on each strip. The work is then mechanical: find the integers in the range, write one term per strip, and add the continuous remainder on each strip.
∫ab f(x) dx = Σ (one term per integer strip) + (continuous remainder per strip)
Work the bounds in order and keep the last strip separate if the upper limit is not an integer. That last point is where marks went: the answer depends on whether the upper limit itself is included in the final strip, and the JEE expects the strip boundaries to be handled explicitly rather than assumed.
The common error
Students who carried the floor symbol outside the integral treated it as a constant for the whole range and got one number too high or too low. The floor is not constant across a multi-unit range — that is the entire point of the question. Splitting is the exercise; splitting correctly, with the right bounds, is what separates a 4-mark answer from a 0.
The final check
After summing the strips, verify against the symmetry of the original integrand. The fractional-part pieces should telescope or pair off, and the net should be a simple rational number. A messy irrational answer is the signal that a strip was counted twice or dropped — the same signal you would get from an arithmetic slip, but the fix is different, so it is worth checking before moving on.
How to practise this pattern
Floor-function integrals are rare enough that you will meet two or three across a full JEE Main attempt. The efficient preparation is to solve three variants of the same shape, not fifteen different questions: one with integer bounds, one with a fractional upper limit, and one where the integrand has a negative term inside the floor. That covers every bound case the exam uses, and it takes about forty minutes.
JEE Main 2025 Shift 1: The floor-function definite integral: the facts to keep
| Question type | Definite integration, Section A (4 marks) |
|---|---|
| The step function | The floor function jumps at every integer, so the integral is a sum over unit-width strips |
| Where the marks go | Splitting the range correctly (2 marks), then evaluating the sum (2 marks) |
| The classic error | Treating the floor value as constant across a multi-unit range |
| Self-check | The answer should reduce to a clean rational number; a messy irrational usually means a strip was double-counted |
Questions students ask about this
Why can the floor function not be integrated directly?
It is a step function with jump discontinuities at every integer, so it has no elementary antiderivative. The integral has to be rewritten as a sum over the intervals where the floor value is constant.
How many marks did this question carry?
It carried 4 marks in Section A. Recognising the split and setting it up correctly fetches the first marks; the arithmetic of the sum fetches the rest.
What if the integration range crosses zero?
Zero is an integer, so it is a jump point like any other. The floor is negative on the interval below zero and zero at zero itself, which is exactly the case where the "floor is a constant" shortcut fails most visibly.