JEE MainMathematics

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Integral Calculus: JEE Main Mathematics Question with Solution

Q: What about the upper limit , where and ?

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

[t]

denotes the greatest integer

\leq t

, then the value of

\frac{3(e-1)}{e} \int_1^2 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.

e^8 - 1

e^7 - 1

e^8 - e

e^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 $I = \int_1^2 x^2 e^{[x]+[x^3]} \, dx$ 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 $dt = 3x^2 \, dx$ . 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, $I = \int_1^2 x^2 e^{1 + [x^3]} \, dx = e \int_1^2 x^2 e^{[x^3]} \, dx$

04execute

Substitute $t = x^3$ , giving $dt = 3x^2 \, dx$ or $x^2 \, dx = \frac{dt}{3}$ . When $x = 1$ , $t = 1$ ; when $x = 2$ , $t = 8$ . Thus, $I = \frac{e}{3} \int_1^8 e^{[t]} \, dt$

05execute

Split the integral across unit intervals where $[t]$ is constant: $\int_1^8 e^{[t]} \, dt = \sum_{k=1}^7 \int_k^{k+1} e^k \, dt = \sum_{k=1}^7 e^k (1) = e + e^2 + e^3 + e^4 + e^5 + e^6 + e^7$ 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: $E = \frac{3(e-1)}{e} \times I = \frac{3(e-1)}{e} \times \left[ \frac{e}{3} \cdot \frac{e(e^7 - 1)}{e - 1} \right] = e(e^7 - 1) = e^8 - e$

✓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 (2).

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 $x^2 e^{[x^3]} dx$ ?
$[x] = 1$ ; substitute $u = x^3$ with $du = 3x^2 dx$
After substituting $u = x^3$ and pulling out constants, what does the integral $\int_1^2 x^2 e^{[x]+[x^3]} dx$ become?
$\frac{e}{3} \sum_{k=1}^7 \int_k^{k+1} e^k du = \frac{e}{3} \sum_{k=1}^7 e^k$
What is the value of the full expression $\frac{3(e-1)}{e} \left( \frac{e}{3} \sum_{k=1}^7 e^k \right)$ ?
$e^8 - e$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $e$ , 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 $I = \int_1^2 x^2 e^{[x]+[x^3]} \, dx$ 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 $dt = 3x^2 \, dx$ . 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, $I = \int_1^2 x^2 e^{1 + [x^3]} \, dx = e \int_1^2 x^2 e^{[x^3]} \, dx$
execute: Substitute $t = x^3$ , giving $dt = 3x^2 \, dx$ or $x^2 \, dx = \frac{dt}{3}$ . When $x = 1$ , $t = 1$ ; when $x = 2$ , $t = 8$ . Thus, $I = \frac{e}{3} \int_1^8 e^{[t]} \, dt$
execute: Split the integral across unit intervals where $[t]$ is constant: $\int_1^8 e^{[t]} \, dt = \sum_{k=1}^7 \int_k^{k+1} e^k \, dt = \sum_{k=1}^7 e^k (1) = e + e^2 + e^3 + e^4 + e^5 + e^6 + e^7$ 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: $E = \frac{3(e-1)}{e} \times I = \frac{3(e-1)}{e} \times \left[ \frac{e}{3} \cdot \frac{e(e^7 - 1)}{e - 1} \right] = e(e^7 - 1) = e^8 - e$
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 (2).

Integral Calculus: JEE Main Mathematics Question with Solution

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

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Let $g: R \rightarrow R$ be a non constant twice differentiable such that…

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