Let the function f :[0,2] R be defined as f(x)= cc e^ [x^2, x-[x]… - Vidyadip

MCQ

Let the function $f :[0,2] \rightarrow R$ be defined as

$f(x)=\left\{\begin{array}{cc}e^{\min \left[x^2, x-[x]\right\}}, & x \in[0,1) \\e^{\left[x-\log _e x\right]}, & x \in[1,2]\end{array}\right.$

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

A
$2 e -1$
B
$1+\frac{3 e }{2}$
✓
$2 e -\frac{1}{2}$
D
$(e-1)\left(e^2+\frac{1}{2}\right)$

✓

Answer

Correct option: C.

$2 e -\frac{1}{2}$

c
Minimum

$m \left\{ x ^2,\{ x \}\right\}= x ^2 ; x \in[0,1)$

${\left[ x -\log _{ e } x \right]=1 ; x \in[1,2)}$

$\therefore f ( x )=\left\{\begin{array}{l} e ^{ x ^2} ; x \in[0,1) \\ e ; x \in[1,2)\end{array}\right.$

$\int \limits_0^2 x f(x) d x=\int \limits_0^1 x e^{x^2} d x+\int \limits_1^2 e x d x$

$=\frac{1}{2}( e -1)+\frac{1}{2}(4-1) e$

$=2 e -\frac{1}{2}$

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