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Find the value of the integral $\int\limits_1^4 {\mathop {\log }\nolimits_e } [x]dx$
  from the options given below:
A. $\mathop {\log }\nolimits_e 2$
B. $\mathop {\log }\nolimits_e 3$
C. $\mathop {\log }\nolimits_e 6$
D. None of the above

Last updated date: 13th Jul 2024
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Hint-We will make use of the formula of integration by parts and solve it.
We have the integral I= $\int\limits_1^4 {\mathop {\log }\nolimits_e } [x]dx$
Since we have the limit of integral from 1 to 4,we will split the limits and write the integral
So, we get
I=\[\] \[\;\;\int\limits_1^2 {\mathop {\log }\nolimits_e } [x]dx\] +\[\;\;\int\limits_2^3 {\mathop {\log }\nolimits_e } [x]dx\] +\[\;\;\int\limits_3^4 {\mathop {\log }\nolimits_e } [x]dx\]
Greatest integer function is discontinuous at all integers. So we should write the definition of
 the function in each of the smaller limits.So,we can write the integral I as
I=\[\;\;\int\limits_1^2 {\mathop {\log }\nolimits_e } 1dx + \;\;\int\limits_2^3 {\mathop {\log }\nolimits_e } 2dx + \;\;\int\limits_3^4 {\mathop {\log }\nolimits_e } 3dx\]
Let us solve this integral by putting the value of ${\log _e}1 = 0$ ,since \[{\log _e}2\] and \[{\log _e}3\] are constants take it out of the integral and solve
I=$\int\limits_1^2 0 dx$ +${\log _e}2\int\limits_2^3 {dx} $ +${\log _e}3\int\limits_3^4 {dx} $

We know that integral 1 dx is equal to x
So, on solving the integral further ,we get
I= 0+${\log _e}2[x]_1^2 + {\log _e}3[x]_3^4$
On applying limits, we get
I=${\log _e}2$$[2 - 1]$ +${\log _e}3[4 - 3]$
So , we get I= ${\log _e}2 + {\log _e}3$
But we know the formula which says
${\log _e}a + {\log _e}b = {\log _e}ab$
Therefore, we can write
${\log _e}2 + {\log _e}3$=${\log _e}6$
So, therefore the value of the integral I=${\log _e}6$=ln6
So, option C is the correct answer

Note: It is possible to integrate to the greatest integer functions only when the limits are
 given, if the limits are not given, we cannot solve the problem.