ml.pow.2026.001UndergraduateReal Analysis

Problem of the Week #1 --- Solution

A Rolle's theorem argument via an antiderivative construction

Published April 13, 2026Issue 1Volume 1Sequence 001

Problem

Let $f : [0, 1] \to \mathbb{R}$ be a continuous function satisfying

\int_0^1 f(x)\,dx = 1.

Solution

Step 1 — Construction of the auxiliary function

Define $g : [0,1] \to \mathbb{R}$ by

g(x) = \int_0^x f(t)\,dt - x^2

Abstract

Every continuous function $f : [0,1] \to \mathbb{R}$ satisfying $\int_0^1 f(x)\,dx = 1$ admits a point with .

Publication Metadata

Public ID: ml.pow.2026.001
Internal ID: ML-POW-2026-001
Series: Problem of the Week Series
Series Code: POW
Volume: 1
Issue: 1
Sequence: 001
Topic: Real Analysis
Difficulty: Undergraduate
Publication Date: April 13, 2026
Accepted: April 13, 2026
Received: April 13, 2026
Week Of: April 7, 2026
Deadline: April 13, 2026
Version: 1
Status: published

Authors

)

d

x

=

1.

Prove that there exists a point $c \in (0, 1)$ such that

f(c) = 2c.

0

x

f

(

t

)

d

t

−

x2

Since $f$ is continuous on $[0,1]$ , the Fundamental Theorem of Calculus ensures that the function

x \mapsto \int_0^x f(t)\,dt

is continuous on $[0,1]$ .

The function $x \mapsto x^2$ is a polynomial, hence continuous on $[0,1]$ and differentiable on $(0,1)$ .

Therefore, $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$ .

Step 2 — Verification of Rolle's theorem hypotheses

We compute the endpoint values:

g(0) = \int_0^0 f(t)\,dt - 0^2 = 0

g(1) = \int_0^1 f(t)\,dt - 1^2

From the given condition:

\int_0^1 f(t)\,dt = 1

we get:

g(1) = 1 - 1 = 0

Hence, $g(0) = g(1) = 0$ .

Rolle's Theorem

Let $h : [a,b] \to \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$ , and suppose $h(a) = h(b)$ .
Then there exists $\xi \in (a,b)$ such that $h'(\xi) = 0$ .

Applying this to $g$ :

$g$ is continuous on $[0,1]$
$g$ is differentiable on $(0,1)$
$g(0) = g(1) = 0$

Therefore, there exists $c \in (0,1)$ such that:

g'(c) = 0

Step 3 — Conclusion via the derivative

Differentiating:

g(x) = \int_0^x f(t)\,dt - x^2

Using the Fundamental Theorem of Calculus:

g'(x) = f(x) - 2x

Evaluating at $c$ :

0 = g'(c) = f(c) - 2c

Thus,

f(c) = 2c

which is the required result. ∎

d

x

=

1

c \in (0,1)

f(c) = 2c

The proof constructs the auxiliary function $g(x) = \int_0^x f(t)\,dt - x^2$ and applies Rolle's theorem after establishing the endpoint condition $g(0) = g(1) = 0$ .

Akhilesh Yadav

Corresponding

editorial@mathlumen.com

MathLumen, New Delhi, India

Keywords

real analysisRolle's theoremfundamental theorem of calculusauxiliary functioncontinuity

Solver Acknowledgements

Solver acknowledgements have not been published for this issue.

Suggested Citation

Akhilesh Yadav. "Problem of the Week #1 --- Solution." Problem of the Week Series, vol. 1, no. 1 (April 13, 2026). https://mathlumen.com/pow/ml.pow.2026.001