Let the function f : [ 0 , 2 ] be defined as f ( x ) = { e min { x 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 0 2 x f ( x ) d x i

Question

Let the function $f : [0, 2] \to ℝ$ be defined as $f (x) =$ ${\begin{matrix} e^{\min {x^{2}, x - [x]}}, & x \in [0, 1) \\ e^{[x - \log_{e} x]}, & x \in [1, 2) \end{matrix}$ where $[t]$ denotes the greatest integer less than or equal to $t$ . Then the value of the integral $\int_{0}^{2} x f (x) d x$ is [2023]

Smart Achievers · Accepted Answer

(2)

$f (x) =$ ${\begin{matrix} e^{\min {x^{2}, x - [x]}}, & x \in [0, 1) \\ e^{[x - \log_{e} x]}, & x \in [1, 2) \end{matrix}$

$For x \in [0, 1);$ $\min {x^{2}, {x}} = x^{2}$

and for $x \in$ [1,2); $[x - \log_{e} x] = 1$

$∴$ $f (x) =$ ${\begin{matrix} e^{x^{2}}, & x \in [0, 1) \\ e, & x \in [1, 2) \end{matrix}$

$Now, \int_{0}^{2} x f (x) = \int_{0}^{1} x e^{x^{2}} d x + \int_{1}^{2} x e d x$

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

Solution

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

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

3 $1 + \frac{3 e}{2}$

4 $2 e - 1$

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Solution

Q. Let the function f:[0,2]→ℝ be defined as f(x)={emin{x2, x-[x]},x∈[0,1)e[x-logex],x∈[1,2) where [t] denotes the greatest integer less than or equal to t. Then the value of the integral ∫02xf(x)dx is [2023]

1 (e-1)(e2+12)

2 2e-12

3 1+3e2

4 2e-1

Ans. (2) f(x)={emin{x2, x-[x]},x∈[0,1)e[x-logex],x∈[1,2) For x∈[0,1); min{x2, {x}}=x2 and for x∈[1,2); [x-logex]=1 ∴ f(x)={ex2,x∈[0,1)e,x∈[1,2) Now, ∫02xf(x)=∫01xex2 dx+∫12xedx =12(e-1)+12(4-1)e=2e-12

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

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

3 $1 + \frac{3 e}{2}$

4 $2 e - 1$