JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 13)
Let the function $$f:[0,2] \rightarrow \mathbb{R}$$ be defined as
$$f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}$$
where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_\limits{0}^{2} x f(x) d x$$ is :
$$2 e-1$$
$$2 e-\frac{1}{2}$$
$$1+\frac{3 e}{2}$$
$$(e-1)\left(e^{2}+\frac{1}{2}\right)$$
Explanation
$$
\begin{aligned}
\operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\
{\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2)
\end{aligned}
$$
$$ \therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\ \mathrm{e} ; \mathrm{x} \in[1,2) \end{array}\right. $$
$$ \begin{aligned} & \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\ & x^2=t \Rightarrow 2 x d x=d t \end{aligned} $$
$$ =\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2 $$
$$ \begin{aligned} & =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\ & =2 \mathrm{e}-\frac{1}{2} \end{aligned} $$
$$ \therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l} \mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\ \mathrm{e} ; \mathrm{x} \in[1,2) \end{array}\right. $$
$$ \begin{aligned} & \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\ & x^2=t \Rightarrow 2 x d x=d t \end{aligned} $$
$$ =\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2 $$
$$ \begin{aligned} & =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\ & =2 \mathrm{e}-\frac{1}{2} \end{aligned} $$
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