Integral Calculus JEE 2026: Floor-Function Integral

What is the answer to this integral calculus JEE 2026 floor-function question?

Option B, 36, is correct for this integral calculus JEE 2026 question, but the supplied working contains two incorrect intermediate values and an unjustified final adjustment. The solution below keeps its substitution, interval-splitting and symmetry method.

The supplied question-bank record identifies this as JEE Main, January 2026, slot 1, tagged hard. The record gives 180 seconds as the expected solve time, not measured student performance.

The floor of a number is the largest integer not exceeding it. Subtracting the floor gives its fractional part. First calculate:

α = \int_{0}^{64} (x^{1 / 3} - ⌊ x^{1 / 3} ⌋) d x .

Then evaluate:

I = \frac{1}{π} \int_{0}^{α π} \frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ .

The supplied options are:

How do you substitute and split the floor-function integral?

Use the cube root as the new variable, then split at integers. This turns the floor-function jumps into unit-interval boundaries. Keep the derivative factor: the fractional part rises linearly on each interval, but the transformed integration weight is not constant.

t = x^{1 / 3}, x = t^{3}, d x = 3 t^{2} d t .

The new limits are:

x = 0 \Rightarrow t = 0, x = 64 \Rightarrow t = 4 .

On each half-open unit interval, the floor is fixed:

⌊ t ⌋ = n for t \in [n, n + 1) .

Changing an integrand at finitely many endpoints does not change its integral. We can therefore use closed limits for each piece:

α = \sum_{n = 0}^{3} \int_{n}^{n + 1} 3 (t - n) t^{2} d t .

Keep the supplied solution’s antiderivative and evaluate both limits:

\int 3 (t - n) t^{2} d t = \frac{3 t^{4}}{4} - n t^{3} .

The four contributions are:

n = 0 : {[\frac{3 t^{4}}{4}]}_{0}^{1} = \frac{3}{4} - 0 = \frac{3}{4} .

n = 1 : {[\frac{3 t^{4}}{4} - t^{3}]}_{1}^{2} = (12 - 8) - (\frac{3}{4} - 1) = \frac{17}{4} .

n = 2 : {[\frac{3 t^{4}}{4} - 2 t^{3}]}_{2}^{3} = (\frac{243}{4} - 54) - (12 - 16) = \frac{43}{4} .

n = 3 : {[\frac{3 t^{4}}{4} - 3 t^{3}]}_{3}^{4} = (192 - 192) - (\frac{243}{4} - 81) = \frac{81}{4} .

Hence:

α = \frac{3 + 17 + 43 + 81}{4} = \frac{144}{4} = 36 .

Accuracy note: the supplied working prints the following incorrect last contribution and total:

\frac{93}{4}, α = 39 .

The displayed antiderivative instead gives:

\frac{81}{4}, α = 36 .

This is an arithmetic correction, not a replacement method. Keep the interval contributions separate until this checkpoint is verified.

How do periodicity and reflection reduce the trigonometric integral?

Count 36 full periods, then use reflection to halve the integral over one period. These are separate operations. Keeping them separate prevents a factor-of-two error: equal integrals over two quarter-cycles do not establish a quarter-cycle period.

Define:

f (θ) = \frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ}, J = \int_{0}^{π / 2} f (θ) d θ .

Periodicity gives:

f (θ + π) = f (θ), \int_{0}^{36 π} f (θ) d θ = 36 \int_{0}^{π} f (θ) d θ .

Reflection about the midpoint of one period gives:

f (π - θ) = f (θ), I = \frac{36}{π} (2 J) = \frac{72}{π} J .

Now apply the complementary-angle substitution within the quarter-cycle integral:

θ ⟼ \frac{π}{2} - θ, J = \int_{0}^{π / 2} \frac{\cos^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ .

Adding the two expressions gives:

2 J = \int_{0}^{π / 2} \frac{1}{\sin^{6} θ + \cos^{6} θ} d θ .

The original integrand does not have a quarter-cycle period:

f (0) = 0, f (π / 2) = 1 .

Complementary-angle reflection exchanges the numerator rather than leaving it unchanged. It gives equal definite integrals after substitution, not equal function values at every point.

How does tangent substitution give option B exactly?

Tangent substitution followed by a second substitution gives the quarter-cycle integral exactly. Factor the denominator first, then match the remaining numerator to a derivative. This retains every factor needed to obtain option B, 36.

Start with:

u = \tan θ, d θ = \frac{d u}{1 + u^{2}}, 0 \leq u < \infty .

Since:

\sin^{6} θ + \cos^{6} θ = \frac{u^{6} + 1}{(1 + u^{2})^{3}},

we obtain:

2 J = \int_{0}^{\infty} \frac{(1 + u^{2})^{2}}{1 + u^{6}} d u .

Factor and cancel:

1 + u^{6} = (1 + u^{2}) (u^{4} - u^{2} + 1),

2 J = \int_{0}^{\infty} \frac{1 + u^{2}}{u^{4} - u^{2} + 1} d u .

For positive values of the substitution variable, divide numerator and denominator by its square:

\frac{1 + u^{2}}{u^{4} - u^{2} + 1} = \frac{1 + 1 / u^{2}}{(u - 1 / u)^{2} + 1} .

The zero endpoint is handled as an improper limit. Now set:

v = u - \frac{1}{u}, d v = (1 + \frac{1}{u^{2}}) d u .

This substitution is strictly increasing, with limits:

u \to 0^{+} \Rightarrow v \to - \infty, u \to \infty \Rightarrow v \to \infty .

Therefore:

2 J = \int_{- \infty}^{\infty} \frac{d v}{1 + v^{2}} = {[\arctan v]}_{- \infty}^{\infty} = π, J = \frac{π}{2} .

Finally:

I = \frac{72}{π} \cdot \frac{π}{2} = 36 (option B) .

The supplied claim below is incorrect: $J = \frac{π}{4} .$

Its subsequent jump from 19.5 to 36 is not a valid simplification. The corrected calculations recover the stated answer without an answer-key adjustment.

α = 36, J = \frac{π}{2}, I = 36 .

How can treating a bound as an equality produce option D?

0 \leq x^{1 / 3} - ⌊ x^{1 / 3} ⌋ < 1 .

Replacing the fractional part by 1 gives an upper-bound estimate, not its exact integral. This is one possible reasoning error that leads to option D. It is not evidence about student responses or how the options were designed.

The false equality would be:

α \overset{wrong}{=} \int_{0}^{64} 1 d x = 64 .

Carrying that incorrect integer through the correctly evaluated trigonometric stage gives:

I \overset{wrong}{=} \frac{64}{π} \int_{0}^{π} f (θ) d θ = \frac{64}{π} (2 J) = 64 .

That produces option D, but the actual conclusion from the bound is:

α < 64, not α = 64 .

Interval-by-interval integration is necessary here to obtain the value. The bound only limits how large it can be.

Which three practice questions reinforce these techniques?

Use these questions to practise cubic substitution, retaining the Jacobian under square-root substitution, and period counting. They are original related practice questions, not additional verified PYQs. Attempt each before reading its short solution.

Question 1: Evaluate the cube-root fractional-part integral.

\int_{0}^{8} (x^{1 / 3} - ⌊ x^{1 / 3} ⌋) d x .

The transferred skill is cubic substitution. Only the first two unit intervals are needed:

t = x^{1 / 3}, d x = 3 t^{2} d t,

\int_{0}^{8} (x^{1 / 3} - ⌊ x^{1 / 3} ⌋) d x = \frac{3}{4} + \frac{17}{4} = 5 .

Question 2: Evaluate the square-root fractional-part integral.

\int_{0}^{9} (\sqrt{x} - ⌊ \sqrt{x} ⌋) d x .

The transferred skill is retaining the Jacobian, the derivative factor in the differential. Each unit interval has a different weight:

t = \sqrt{x}, d x = 2 t d t,

\int_{n}^{n + 1} 2 t (t - n) d t = n + \frac{2}{3} .

\sum_{n = 0}^{2} (n + \frac{2}{3}) = \frac{2}{3} + \frac{5}{3} + \frac{8}{3} = 5 .

Question 3: Evaluate the five-period trigonometric integral.

\frac{1}{π} \int_{0}^{5 π} \frac{\sin^{2} θ}{\sin^{6} θ + \cos^{6} θ} d θ .

The transferred skill is period counting. Reuse the established quarter-cycle result rather than integrating again:

\frac{5}{π} (2 J) = \frac{5}{π} (2 \cdot \frac{π}{2}) = 5 .

Redo any question you missed. Write the transformed limits and differential before taking the next step.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the answer to the integral calculus JEE 2026 floor-function question?

The correct answer is option B, 36. The cube-root fractional-part integral gives α = 36, and the subsequent normalised trigonometric integral also evaluates to 36.

How do I integrate a cube-root fractional part from 0 to 64?

Set t = x^(1/3), so dx = 3t² dt and the limits become 0 and 4. Split the integral at t = 1, 2 and 3, using floor(t) = n on each interval from n to n + 1. The four contributions are 3/4, 17/4, 43/4 and 81/4, which sum to 36.

Is the period of sin²θ/(sin⁶θ + cos⁶θ) equal to π/2?

No: the integrand has period π, not π/2, since its values at 0 and π/2 are 0 and 1. Complementary-angle substitution exchanges the sine-squared numerator with cosine-squared and gives equal definite integrals over 0 to π/2. This equality of integrals does not establish a π/2 period.

What errors are corrected in the supplied solution?

The last floor-function interval contributes 81/4, not 93/4, so α is 36 rather than 39. The quarter-cycle trigonometric integral J is π/2, not π/4. These corrections give I = 36 directly, without the supplied working's unjustified jump from 19.5 to 36.