JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectSingle correctPrevious-year question

Definite integral involving greatest integer function with substitution

Q: What about the upper limit x = 2, where [x] = 2 and [x^3] = 8?

A definite integral is independent of the value of the integrand at individual isolated points (a set of measure zero). So we can take [x] = 1 over the entire interval [1, 2) without affecting the integral.

If [t] denotes the greatest integer ≤ t, then the value of (3(e-1))/e integral from 1 to 2 of x^2 e^([x]+[x^3]) dx is:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Ae^8 - 1Be^7 - 1Ce^8 - eDe^9 - e

Step-by-step solutionView

Correct answer

Using the substitution $u = x^{3}$ , the integral simplifies to a telescoping geometric series of constant-interval integrals that evaluates to $e^{8} - e$ .

Option analysis

Why each option works or fails

A · e^8 - 1

The student forgets that the first term in the geometric sum starts at $e^{2}$ instead of $e$ , subtracting 1 instead of $e$ after multiplying out. Check the lower bound of integration carefully: $u$ runs from 1 to 8, so the lowest exponent of $e$ is $[x] + [u] = 1 + 1 = 2$ , which gives a factor of $e^{2}$ in the first sub-interval.

B · e^7 - 1

The student miscounts the upper bound of $u = x^{3}$ as 7 instead of $2^{3} = 8$ , leading to an exponent of 7. Calculate the upper limit by evaluating $u = x^{3}$ at $x = 2$ , which yields $2^{3} = 8$ , not 7.

C · e^8 - e

This is the correct option. Evaluating $[x] = 1$ on $(1, 2)$ and substituting $u = x^{3}$ transforms the integral into $\frac{e}{3} \sum_{k = 1}^{7} e^{k}$ , which equals $\frac{e^{2} (e^{7} - 1)}{3 (e - 1)}$ ; multiplying by $\frac{3 (e - 1)}{e}$ produces $e (e^{7} - 1) = e^{8} - e$ .

D · e^9 - e

The student includes an extra term at the upper limit (summing up to $k = 8$ instead of $k = 7$ ), forgetting that $[u] = 8$ only holds at the single point $u = 8$ . Since $u < 8$ almost everywhere on the open interval $(1, 8)$ , the integer values for $[u]$ are $1, 2, \dots, 7$ , so the summation terminates at $k = 7$ .

Reviewed route

Solution

StepWorking

01Given

The expression to evaluate is $E = \frac{3 (e - 1)}{e} I$ , where $I = \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ and $[t]$ denotes the greatest integer function.

02Approach

On the interval of integration $(1, 2)$ , $x \in (1, 2)$ so $[x] = 1$ almost everywhere. We factor out $e^{[x]} = e^{1} = e$ , then make the substitution $t = x^{3}$ so that $d t = 3 x^{2} d x$ . The integral then splits into unit intervals where $[t]$ is constant, forming a geometric progression.

03Execute

Since $x \in [1, 2)$ , $[x] = 1$ (the single endpoint $x = 2$ does not change the integral value). Hence, $I = \int_{1}^{2} x^{2} e^{1 + [x^{3}]} d x = e \int_{1}^{2} x^{2} e^{[x^{3}]} d x$

04Execute

Substitute $t = x^{3}$ , giving $d t = 3 x^{2} d x$ or $x^{2} d x = \frac{d t}{3}$ . When $x = 1$ , $t = 1$ ; when $x = 2$ , $t = 8$ . Thus, $I = \frac{e}{3} \int_{1}^{8} e^{[t]} d t$

05Execute

Split the integral across unit intervals where $[t]$ is constant: $\int_{1}^{8} e^{[t]} d t = \sum_{k = 1}^{7} \int_{k}^{k + 1} e^{k} d t = \sum_{k = 1}^{7} e^{k} (1) = e + e^{2} + e^{3} + e^{4} + e^{5} + e^{6} + e^{7}$ Using the formula for the sum of a geometric series: $\sum_{k = 1}^{7} e^{k} = \frac{e (e^{7} - 1)}{e - 1}$

06Execute

Substitute this back into the original expression: $E = \frac{3 (e - 1)}{e} \times I = \frac{3 (e - 1)}{e} \times [\frac{e}{3} \cdot \frac{e (e^{7} - 1)}{e - 1}] = e (e^{7} - 1) = e^{8} - e$

✓Verify

The prefactor $\frac{3 (e - 1)}{e}$ perfectly cancels $\frac{e}{3 (e - 1)}$ , leaving $e \times (e^{7} - 1) = e^{8} - e$ . Matches option C.

Hints that build this answer step by step

On the interval of integration $(1, 2)$ , what value does $[x]$ take, and what is the best substitution to simplify $x^{2} e^{[x^{3}]} d x$ ?
$[x] = 1$ ; substitute $u = x^{3}$ with $d u = 3 x^{2} d x$
After substituting $u = x^{3}$ and pulling out constants, what does the integral $\int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ become?
$\frac{e}{3} \sum_{k = 1}^{7} \int_{k}^{k + 1} e^{k} d u = \frac{e}{3} \sum_{k = 1}^{7} e^{k}$
What is the value of the full expression $\frac{3 (e - 1)}{e} (\frac{e}{3} \sum_{k = 1}^{7} e^{k})$ ?
$e^{8} - e$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

What about the upper limit

x = 2

, where

[x] = 2

and

[x^{3}] = 8

A definite integral is independent of the value of the integrand at individual isolated points (a set of measure zero). So we can take $[x] = 1$ over the entire interval $[1, 2)$ without affecting the integral.

Answer

Using the substitution $u = x^{3}$ , the integral simplifies to a telescoping geometric series of constant-interval integrals that evaluates to $e^{8} - e$ .

Why each option works or fails

A: e^8 - 1 - The student forgets that the first term in the geometric sum starts at $e^{2}$ instead of $e$ , subtracting 1 instead of $e$ after multiplying out. Check the lower bound of integration carefully: $u$ runs from 1 to 8, so the lowest exponent of $e$ is $[x] + [u] = 1 + 1 = 2$ , which gives a factor of $e^{2}$ in the first sub-interval.
B: e^7 - 1 - The student miscounts the upper bound of $u = x^{3}$ as 7 instead of $2^{3} = 8$ , leading to an exponent of 7. Calculate the upper limit by evaluating $u = x^{3}$ at $x = 2$ , which yields $2^{3} = 8$ , not 7.
C · correct: e^8 - e - This is the correct option. Evaluating $[x] = 1$ on $(1, 2)$ and substituting $u = x^{3}$ transforms the integral into $\frac{e}{3} \sum_{k = 1}^{7} e^{k}$ , which equals $\frac{e^{2} (e^{7} - 1)}{3 (e - 1)}$ ; multiplying by $\frac{3 (e - 1)}{e}$ produces $e (e^{7} - 1) = e^{8} - e$ .
D: e^9 - e - The student includes an extra term at the upper limit (summing up to $k = 8$ instead of $k = 7$ ), forgetting that $[u] = 8$ only holds at the single point $u = 8$ . Since $u < 8$ almost everywhere on the open interval $(1, 8)$ , the integer values for $[u]$ are $1, 2, \dots, 7$ , so the summation terminates at $k = 7$ .

Step-by-step solution

Given: The expression to evaluate is $E = \frac{3 (e - 1)}{e} I$ , where $I = \int_{1}^{2} x^{2} e^{[x] + [x^{3}]} d x$ and $[t]$ denotes the greatest integer function.
Approach: On the interval of integration $(1, 2)$ , $x \in (1, 2)$ so $[x] = 1$ almost everywhere. We factor out $e^{[x]} = e^{1} = e$ , then make the substitution $t = x^{3}$ so that $d t = 3 x^{2} d x$ . The integral then splits into unit intervals where $[t]$ is constant, forming a geometric progression.
Execute: Since $x \in [1, 2)$ , $[x] = 1$ (the single endpoint $x = 2$ does not change the integral value). Hence, $I = \int_{1}^{2} x^{2} e^{1 + [x^{3}]} d x = e \int_{1}^{2} x^{2} e^{[x^{3}]} d x$
Execute: Substitute $t = x^{3}$ , giving $d t = 3 x^{2} d x$ or $x^{2} d x = \frac{d t}{3}$ . When $x = 1$ , $t = 1$ ; when $x = 2$ , $t = 8$ . Thus, $I = \frac{e}{3} \int_{1}^{8} e^{[t]} d t$
Execute: Split the integral across unit intervals where $[t]$ is constant: $\int_{1}^{8} e^{[t]} d t = \sum_{k = 1}^{7} \int_{k}^{k + 1} e^{k} d t = \sum_{k = 1}^{7} e^{k} (1) = e + e^{2} + e^{3} + e^{4} + e^{5} + e^{6} + e^{7}$ Using the formula for the sum of a geometric series: $\sum_{k = 1}^{7} e^{k} = \frac{e (e^{7} - 1)}{e - 1}$
Execute: Substitute this back into the original expression: $E = \frac{3 (e - 1)}{e} \times I = \frac{3 (e - 1)}{e} \times [\frac{e}{3} \cdot \frac{e (e^{7} - 1)}{e - 1}] = e (e^{7} - 1) = e^{8} - e$
Verify: The prefactor $\frac{3 (e - 1)}{e}$ perfectly cancels $\frac{e}{3 (e - 1)}$ , leaving $e \times (e^{7} - 1) = e^{8} - e$ . Matches option C.

Definite integral involving greatest integer function with substitution

What feels right?

Using the substitution $u = x^{3}$ , the integral simplifies to a telescoping geometric series of constant-interval integrals that evaluates to $e^{8} - e$ .

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

05Execute

06Execute

✓Verify

We think you should solve this next.

The same chapter, from another paper — Integral Calculus, 2023

The value of the integral…

The same trap, in a different chapter — Binomial Theorem, 2023

Students also ask

Definite integral involving greatest integer function with substitution

Using the substitution u=x3u=x3, the integral simplifies to a telescoping geometric series of constant-interval integrals that evaluates to e8−ee8−e.

01Given

02Approach

03Execute

04Execute

05Execute

06Execute

✓Verify

The same chapter, from another paper — Integral Calculus, 2023

The value of the integral…

The same trap, in a different chapter — Binomial Theorem, 2023

Students also ask

Using the substitution $u = x^{3}$ , the integral simplifies to a telescoping geometric series of constant-interval integrals that evaluates to $e^{8} - e$ .