Functions $f(x)=\begin{cases}[x]\{x\}&x\neq 0\\0&x=0\end{cases}$ and $g(x)=\begin{cases}\{x\}[x]&x\neq 0\\0&x=0\end{cases}$. Which statements are true?
A) $\displaystyle\int_0^{n+1}f(x)\,dx - \int_0^n f(x)\,dx = \frac{n-1}{\ln n}$ (... some form)
B) $\displaystyle\int_0^{n+1}g(x)\,dx - \int_0^n g(x)\,dx = \frac{1}{n+1}$ (... some form)
C) $\displaystyle\int_0^t f(x)\,dx = \int_0^t g(x)\,dx$ has at least one solution in $t\in\left(0,\frac{\pi}{2}\right)$
D) $e < \displaystyle\lim_{n\to\infty}\int_0^n g(x)\,dx < e^2$
Step-by-Step Solution
Key Concept: On $[n,n+1)$: $f(x)=[x]\{x\}=n(x-n)$, $g(x)=\{x\}[x]=(x-n)^n$... wait: $g(x)=\{x\}^{[x]}=(x-n)^n$. $\int_n^{n+1}g(x)dx=\int_0^1 t^n dt=1/(n+1)$. B ✓. A involves $f$: $\int_n^{n+1}f(x)dx=n\int_0^1 t\,dt=n/2$... not matching A's form unless different.
A ✓, B ✓. Answer: A, B.
Correct Answer: AB