Why is integral from 0 to 1 of e^(1-t) dt equal to integral from 1 to 2 of e^(t-1) dt?

Substituting u = 1-t in the first integral and u = t-1 in the second gives integral from 0 to 1 of e^u du in both cases.

JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Minimizing an integral with an absolute-value kernel

Q: Why is the minimum guaranteed to lie in [0, 2]?

Because outside [0, 2], the integrand e^(|x-t|) grows strictly as x moves away from the interval [0, 2]. Moving x closer to the interval always decreases the integrand pointwise.

The minimum value of the function f(x) = integral from 0 to 2 of e^(|x-t|) dt is:

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Ae(e-1)B2(e-1)C2D2e-1

Step-by-step solutionView

Correct answer

The minimum value of $f (x) = \int_{0}^{2} e^{∣ x - t ∣} d t$ is $2 (e - 1)$ , achieved at $x = 1$ .

Option analysis

Why each option works or fails

A · e(e-1)

Evaluating the function at the boundaries $x = 0$ or $x = 2$ where $f (0) = f (2) = e^{2} - 1 = (e - 1) (e + 1)$ , or mistakenly factoring $e^{2} - e$ as the minimum instead of locating the interior stationary point. Differentiate $f (x)$ or inspect symmetry to see that the global minimum lies at the midpoint $x = 1$ , not at the interval boundaries.

B · 2(e-1)

By symmetry or differentiation, the critical point is $x = 1$ , giving $f (1) = \int_{0}^{1} e^{1 - t} d t + \int_{1}^{2} e^{t - 1} d t = (e - 1) + (e - 1) = 2 (e - 1)$ . Correctly identified $x = 1$ as the minimizing point and integrated properly across the split domain.

C · 2

Integrating $e^{∣ x - t ∣}$ as if the exponential and absolute value cancel out to a constant, or setting $e^{0} = 1$ and integrating $1$ over $[0, 2]$ . Keep the exponential term: $\int_{0}^{1} e^{1 - t} d t = e - 1$ , which does not simplify to $1$ .

D · 2e-1

Evaluating only one half of the symmetric integral at $x = 1$ , calculating $\int_{0}^{1} e^{1 - t} d t = e - 1$ and adding $e$ instead of $e - 1$ . Ensure both sub-integrals $\int_{0}^{1} e^{1 - t} d t$ and $\int_{1}^{2} e^{t - 1} d t$ are computed and summed: $(e - 1) + (e - 1) = 2 (e - 1)$ .

Reviewed route

Solution

StepWorking

01Given

$f (x) = \int_{0}^{2} e^{∣ x - t ∣} d t$ , where $t \in [0, 2]$ and $x \in R$ .

02Goal

Find the global minimum value of $f (x)$ over $x \in R$ .

03Approach

Partition the real line into $x \leq 0$ , $0 < x < 2$ , and $x \geq 2$ to split the absolute value $∣ x - t ∣$ . In the middle interval, find the critical point using calculus or AM-GM inequality, and verify that the endpoints match the outer monotonic pieces.

04Execute

For $x \in [0, 2]$ , split at $t = x$ : $f (x) = \int_{0}^{x} e^{x - t} d t + \int_{x}^{2} e^{t - x} d t = e^{x} {[- e^{- t}]}_{0}^{x} + e^{- x} {[e^{t}]}_{x}^{2} = (e^{x} - 1) + (e^{2 - x} - 1) = e^{x} + e^{2 - x} - 2$

05Execute

For $0 \leq x \leq 2$ , apply AM-GM to $e^{x}$ and $e^{2 - x}$ : $e^{x} + e^{2 - x} \geq 2 \sqrt{e^{x} \cdot e^{2 - x}} = 2 \sqrt{e^{2}} = 2 e$ Equality holds when $e^{x} = e^{2 - x} ⟹ x = 1$ . Thus, the minimum value on $[0, 2]$ is $2 e - 2 = 2 (e - 1)$ .

06Execute

For $x < 0$ , $∣ x - t ∣ = t - x$ , so $f (x) = e^{- x} (e^{2} - 1)$ , which is strictly decreasing with infimum at $x \to 0^{-}$ equal to $e^{2} - 1$ . For $x > 2$ , $∣ x - t ∣ = x - t$ , so $f (x) = e^{x} (1 - e^{- 2})$ , which is strictly increasing with infimum at $x \to 2^{+}$ equal to $e^{2} - 1$ . Since $e^{2} - 1 = (e - 1) (e + 1) > 2 (e - 1)$ (as $e + 1 > 2$ ), the global minimum is indeed at $x = 1$ .

✓Verify

Differentiating using Leibniz rule: $f^{'} (x) = \int_{0}^{x} e^{x - t} d t - \int_{x}^{2} e^{t - x} d t = (e^{x} - 1) - (e^{2 - x} - 1) = e^{x} - e^{2 - x}$ . Setting $f^{'} (x) = 0$ gives $x = 1$ , with $f^{''} (1) = e + e > 0$ , confirming a unique global minimum.

✓ 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: 8 September 2026

Quick checks

Students also ask

Why is the minimum guaranteed to lie in

[0, 2]

Because outside $[0, 2]$ , the integrand $e^{∣ x - t ∣}$ grows strictly as $x$ moves away from the interval $[0, 2]$ . Moving $x$ closer to the interval always decreases the integrand pointwise.

Why is

\int_{0}^{1} e^{1 - t} d t

equal to

\int_{1}^{2} e^{t - 1} d t

Substituting $u = 1 - t$ in the first integral and $u = t - 1$ in the second gives $\int_{0}^{1} e^{u} d u$ in both cases.

Answer

The minimum value of $f (x) = \int_{0}^{2} e^{∣ x - t ∣} d t$ is $2 (e - 1)$ , achieved at $x = 1$ .

Why each option works or fails

A: e(e-1) - Evaluating the function at the boundaries $x = 0$ or $x = 2$ where $f (0) = f (2) = e^{2} - 1 = (e - 1) (e + 1)$ , or mistakenly factoring $x = 1$ as the minimum instead of locating the interior stationary point. Differentiate $f (x)$ or inspect symmetry to see that the global minimum lies at the midpoint $x = 1$ , not at the interval boundaries.
B · correct: 2(e-1) - By symmetry or differentiation, the critical point is $x = 1$ , giving $f (1) = \int_{0}^{1} e^{1 - t} d t + \int_{1}^{2} e^{t - 1} d t = (e - 1) + (e - 1) = 2 (e - 1)$ . Correctly identified $x = 1$ as the minimizing point and integrated properly across the split domain.
C: 2 - Integrating $e^{∣ x - t ∣}$ as if the exponential and absolute value cancel out to a constant, or setting $e^{0} = 1$ and integrating $1$ over $[0, 2]$ . Keep the exponential term: $\int_{0}^{1} e^{1 - t} d t = e - 1$ , which does not simplify to $1$ .
D: 2e-1 - Evaluating only one half of the symmetric integral at $x = 1$ , calculating $\int_{0}^{1} e^{1 - t} d t = e - 1$ and adding $e$ instead of $e - 1$ . Ensure both sub-integrals $\int_{0}^{1} e^{1 - t} d t$ and $\int_{1}^{2} e^{t - 1} d t$ are computed and summed: $(e - 1) + (e - 1) = 2 (e - 1)$ .

Step-by-step solution

Given: $f (x) = \int_{0}^{2} e^{∣ x - t ∣} d t$ , where $t \in [0, 2]$ and $x \in R$ .
Goal: Find the global minimum value of $f (x)$ over $x \in R$ .
Approach: Partition the real line into $x \leq 0$ , $0 < x < 2$ , and $x \geq 2$ to split the absolute value $∣ x - t ∣$ . In the middle interval, find the critical point using calculus or AM-GM inequality, and verify that the endpoints match the outer monotonic pieces.
Execute: For $x \in [0, 2]$ , split at $t = x$ : $f (x) = \int_{0}^{x} e^{x - t} d t + \int_{x}^{2} e^{t - x} d t = e^{x} {[- e^{- t}]}_{0}^{x} + e^{- x} {[e^{t}]}_{x}^{2} = (e^{x} - 1) + (e^{2 - x} - 1) = e^{x} + e^{2 - x} - 2$
Execute: For $0 \leq x \leq 2$ , apply AM-GM to $e^{x}$ and $e^{2 - x}$ : $e^{x} + e^{2 - x} \geq 2 \sqrt{e^{x} \cdot e^{2 - x}} = 2 \sqrt{e^{2}} = 2 e$ Equality holds when $e^{x} = e^{2 - x} ⟹ x = 1$ . Thus, the minimum value on $[0, 2]$ is $2 e - 2 = 2 (e - 1)$ .
Execute: For $x < 0$ , $∣ x - t ∣ = t - x$ , so $f (x) = e^{- x} (e^{2} - 1)$ , which is strictly decreasing with infimum at $x \to 0^{-}$ equal to $e^{2} - 1$ . For $x > 2$ , $∣ x - t ∣ = x - t$ , so $f (x) = e^{x} (1 - e^{- 2})$ , which is strictly increasing with infimum at $x \to 2^{+}$ equal to $e^{2} - 1$ . Since $e^{2} - 1 = (e - 1) (e + 1) > 2 (e - 1)$ (as $e + 1 > 2$ ), the global minimum is indeed at $x = 1$ .
Verify: Differentiating using Leibniz rule: $f^{'} (x) = \int_{0}^{x} e^{x - t} d t - \int_{x}^{2} e^{t - x} d t = (e^{x} - 1) - (e^{2 - x} - 1) = e^{x} - e^{2 - x}$ . Setting $f^{'} (x) = 0$ gives $x = 1$ , with $f^{''} (1) = e + e > 0$ , confirming a unique global minimum.

Shortcut:

When to use it: Fastest route: differentiate $f (x)$ directly to find the critical point without computing the full integral beforehand.

Given: $f (x) = \int_{0}^{2} e^{∣ x - t ∣} d t$ . The integrand is convex in $x$ , so $f (x)$ is convex.

Goal: Find the stationary point $x^{*}$ where $f^{'} (x^{*}) = 0$ , then evaluate $f (x^{*})$ .

Approach: Assume the minimum occurs at $x \in (0, 2)$ . Split the integral: $f (x) = \int_{0}^{x} e^{x - t} d t + \int_{x}^{2} e^{t - x} d t$ . Differentiate with respect to $x$ via Leibniz rule.

Execute: $f^{'} (x) = e^{x - x} (1) + \int_{0}^{x} e^{x - t} d t - e^{x - x} (1) - \int_{x}^{2} e^{t - x} d t = (e^{x} - 1) - (e^{2 - x} - 1) = e^{x} - e^{2 - x}$ . Setting $f^{'} (x) = 0 ⟹ e^{x} = e^{2 - x} ⟹ x = 1$ .

Execute: Calculate $f (1)$ : $f (1) = \int_{0}^{2} e^{∣ 1 - t ∣} d t = \int_{0}^{1} e^{1 - t} d t + \int_{1}^{2} e^{t - 1} d t = 2 \int_{0}^{1} e^{u} d u = 2 (e - 1)$

Verify: Symmetry check: $∣ 1 - t ∣$ on $t \in [0, 2]$ is symmetric around $t = 1$ , giving $2 \int_{0}^{1} e^{u} d u = 2 (e - 1)$ .

Minimizing an integral with an absolute-value kernel

What feels right?

The minimum value of $f (x) = \int_{0}^{2} e^{∣ x - t ∣} d t$ is $2 (e - 1)$ , achieved at $x = 1$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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