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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
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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