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

(C) $2 \mathrm{e}-\frac{1}{2}$

(D) $(\mathrm{e}-1)\left(\mathrm{e}^{2}+\frac{1}{2}\right)$

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Definite Integrals

Let the function

f : [0, 2] \to R

be defined as

f (x) = {e^{m i n {x^{2}, x - [x]}}, e^{[x - l o g_{e} x]}, x \in [0, 1) x \in [1, 2]

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

Answer: (C)

For

x \in [0, 1)

min {x^{2}, {x}} = x^{2}

(since

{x} = x

and

x^{2} < x

in this range), so

f (x) = e^{x^{2}}

.

For

x \in [1, 2]

[x - lo g_{e} x] = 1

(since

x - ln x

ranges from 1 to approximately 1.31), so

f (x) = e

.

Thus:

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

For the first integral, let

u = x^{2}

d u = 2 x d x

\int_{0}^{1} x e^{x^{2}} d x = \frac{1}{2} [e^{x^{2}}]_{0}^{1} = \frac{1}{2} (e - 1)

For the second:

\int_{1}^{2} e x d x = e [\frac{x ^{2}}{2}]_{1}^{2} = e \cdot \frac{3}{2}

Total:

\frac{e - 1}{2} + \frac{3 e}{2} = \frac{4 e - 1}{2} = 2 e - \frac{1}{2}

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option