Let the function f:[0,2] → R be defined as f(x)= e^(min ², x-[x]) x [0,1) e^[x- log e x], x [1,2] where [t] denotes the greatest integer less than or equal to t. Then the value of the integral ∫₀² x f(x) d x is:

Integral Calculus: JEE Main 2023 Mathematics question

JEE Main 2023 · 11 Apr, Evening shift · Q3463

JEE Main 202311 Apr, Evening shiftMathematicsMediumIntegral Calculus369× since 2002

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 :

JEE Main 2023 Integral Calculus Question with Solution (11 Apr Evening shift, Q3463) | JEE Arc