Question

Let \[ f:[0,1]\rightarrow\mathbb R \] be a function defined by \[ f(x)+f(1-x)=1. \] Then \[ \int_0^1 f(x)\,dx= \ ? \]

• A. 0
• B. 1
• C. 2
• D. \(\frac12\)

Hint:

If a function satisfies \[ f(x)+f(a-x)=k, \] then \[ \int_0^a f(x)\,dx=\frac{ka}{2}. \] This is a standard result.

Answer 1

The Correct Option is D

Concept: Whenever a function satisfies a relation involving \[ f(x) \quad\text{and}\quad f(a-x), \] the substitution \[ x=a-t \] inside the integral is usually the most effective method.

Step 1: Let the required integral be \(I\).
Define \[ I = \int_0^1 f(x)\,dx. \]

Step 2: Apply the substitution \(x=1-t\).
Then \[ dx=-dt. \] Hence \[ I = \int_0^1 f(1-t)\,dt. \] Renaming \(t\) as \(x\), \[ I = \int_0^1 f(1-x)\,dx. \]

Step 3: Add the two integrals.
\[ 2I = \int_0^1 \Big[f(x)+f(1-x)\Big] dx. \] Using \[ f(x)+f(1-x)=1, \] we get \[ 2I = \int_0^1 1\,dx. \] \[ 2I=1. \] Therefore \[ I=\frac12. \] Hence \[ \boxed{\frac12}. \]