The value of the integral I= displaystyle ∫ 12t[ t ] + t(1 + ln t) dt is equal to ( [.] and . denotes the greatest integer and fractional part function respectively)

Question

The value of the integral I= displaystyle ∫ 12t[ t ] + t(1 + ln t) dt is equal to ( [.] and . denotes the greatest integer and fractional part function respectively) (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

As [ t ]=0 , we get, I= displaystyle ∫ 12tt(1 + ln t)dt = displaystyle ∫ 12 d (tt) =[tt]12 =22-11=3

Q. The value of the integral $I = \int_{1}^{2} t^{[{t}] + t} (1 + l n t) d t$ is equal to

( $[.]$ and ${.}$ denotes the greatest integer and fractional part function respectively)

$0$

$1$

$2$

$3$

As $[{t}] = 0$ , we get,
$I = \int_{1}^{2} t^{t} (1 + l n t) d t$
$= \int_{1}^{2} d (t^{t})$
$= [t^{t}]_{1}^{2}$
$= 2^{2} - 1^{1} = 3$

Q. The value of the integral I=∫12​t[{t}]+t(1+lnt)dt is equal to ( [.] and {.} denotes the greatest integer and fractional part function respectively)

0

1

2

3

As [{t}]=0 , we get, I=∫12​tt(1+lnt)dt =∫12​d(tt) =[tt]12​ =22−11=3

Q. The value of the integral $I = \int_{1}^{2} t^{[{t}] + t} (1 + l n t) d t$ is equal to

( $[.]$ and ${.}$ denotes the greatest integer and fractional part function respectively)

$0$

$1$

$2$

$3$

As $[{t}] = 0$ , we get,
$I = \int_{1}^{2} t^{t} (1 + l n t) d t$
$= \int_{1}^{2} d (t^{t})$
$= [t^{t}]_{1}^{2}$
$= 2^{2} - 1^{1} = 3$