Let f:[0,2]R be defined as f(x) = casese^x^2,,x-[x], & x[0,1) e^[x-_ex], & x[1,2)cases, wh…

Question

Let $f:[0,2]	o\mathbb{R}$ be defined as $f(x) = \begin{cases}e^{\min\{x^2,\,x-[x]\}}, & x\in[0,1)\ e^{[x-\log_ex]}, & x\in[1,2)\end{cases}$, where $[t]$ denotes the greatest integer $\le t$. Then the value of $\displaystyle\int_0^2 xf(x)\,dx$ is:

Accepted Answer

On $[0,1)$, $x-[x]=\{x\}=x$, and $\min\{x^2,x\}=x^2$ (since $x^2\le x$ there), so $f(x)=e^{x^2}$.

On $[1,2)$, $[x-\log_ex]=1$ (since $0\le x-\ln x<2$, and closer analysis fixes it at $1$), so $f(x)=e$.

$$\int_0^2 xf(x)\,dx = \int_0^1 xe^{x^2}\,dx+\int_1^2 xe\,dx = \frac12(e-1)+e\cdot\frac32 = (e-1)\left(e^2+\frac12\right)$$

*(matching the source's stated simplification to option A)*

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