# 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: 24th Mar 2023

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Answer

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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.

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.

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