Question

Let \(a\) be a positive real number such that
\(\displaystyle\int_0^a e^{x-[x]}\,dx = 10e-9\)
where \([x]\) is the greatest integer less than or equal to \(x\). Then \(a\) is equal to:

• A. \(10-\log_e(1+e)\)
• B. \(10+\log_e 2\)
• C. \(10+\log_e 3\)
• D. \(10+\log_e(1+e)\)

Hint:

The integrand \(e^{x-[x]}=e^{\{x\}}\) is periodic with period 1, so each unit interval contributes \(e-1\).

Answer 1

The Correct Option is B

Let \(a=n+f\) with \(n=[a]\) a non-negative integer and \(0\le f\lt 1\).
Because \(e^{x-[x]}\) has period 1, \(\int_0^n e^{x-[x]}dx = n\int_0^1 e^x dx = n(e-1)\).
On the last partial interval \(\int_n^{n+f} e^{x-n}dx = e^f-1\).
So \(n(e-1)+e^f-1 = 10e-9 = 10(e-1)+1\).
Try \(n=10\): \(e^f-1=1\), so \(e^f=2\) and \(f=\log_e 2\lt 1\), which is valid.
Try \(n=9\): \(e^f-1 = 10e-9-9e+9 = e+1\), so \(e^f=e+2\gt e\) and \(f\gt 1\), which is impossible. Smaller \(n\) fails similarly.
Hence \(a=10+\log_e 2\).
Answer: (2) \(10+\log_e 2\)