Question

Let \(f:\mathbf{R}\to\mathbf{R}\) be a function such that \(f(2-x)=f(2+x)\) and \(f(4-x)=f(4+x)\), for all \(x\in\mathbf{R}\) and \(\int_0^2 f(x)\,dx=5\). Then the value of \(\int_{10}^{50} f(x)\,dx\) is :

• A. 80
• B. 100
• C. 125
• D. 200

Hint:

The two symmetry conditions force f to be periodic with period 4; integrate over whole periods.

Answer 1

The Correct Option is B

Step 1: \(f(4-x)=f(2+(2-x))=f(2-(2-x))=f(x)\). Combined with \(f(4+x)=f(4-x)\) this gives \(f(x+4)=f(x)\), so the period is 4.
Step 2: By symmetry about \(x=2\), \(\int_2^4 f=\int_0^2 f=5\), so \(\int_0^4 f(x)\,dx=10\).
Step 3: The interval \([10,50]\) has length 40 = 10 periods, so \(\int_{10}^{50} f(x)\,dx=10\times 10=100\).
Answer: (2) 100