Question

What is the value of \[\int_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{{{\log }_e}x}}{x}} \right|} dx\]?
A. \[\dfrac{3}{2}\]
B. \[\dfrac{5}{2}\]
C. 3
D. 5

Answer 1

Hint: First we will break the integration interval by positive and negative values of \[\dfrac{{{{\log }_e}x}}{x}\]. Then we will use the substitution method. Integrate it to find the value of the integration.


Formula Used:Definite integration:
\[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \], where \[b < c < a\]
Integration formula:
\[\int {xdx} = \dfrac{{{x^2}}}{2} + c\]



Complete step by step solution:Given definite integral is \[\int_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{{{\log }_e}x}}{x}} \right|} dx\].
Assume that, \[I = \int_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{{{\log }_e}x}}{x}} \right|} dx\].
We know that, when \[0 < x < 1\], \[{\log _e}x < 0\]
When \[x > 1\], \[{\log _e}x > 0\]
So that \[{e^{ - 1}} < x < 1\] implies \[{\log _e}x < 0\].
When \[1 < x < {e^2}\], \[{\log _e}x > 0\]
Applying the property \[\int_b^a {f\left( x \right)dx} = \int_b^c {f\left( x \right)dx} + \int_c^a {f\left( x \right)dx} \] in the integration:
Here \[a = {e^2}\], \[b = {e^{ - 1}}\], and c = 1
\[I = \int_{{e^{ - 1}}}^{{e^2}} {\left| {\dfrac{{{{\log }_e}x}}{x}} \right|} dx\]
\[ \Rightarrow I = \int_{{e^{ - 1}}}^1 { - \dfrac{{{{\log }_e}x}}{x}} dx + \int_1^{{e^2}} {\dfrac{{{{\log }_e}x}}{x}} dx\] ……(i)
Assume that, \[{\log _e}x = z\]
Differentiate with respect to x:
\[\dfrac{1}{x}dx = dz\]
When \[x = {e^{ - 1}}\], \[z = {\log _e}{e^{ - 1}} = - 1\]
When \[x = 1\], \[z = {\log _e}1 = 0\]
When \[x = {e^2}\], \[z = {\log _e}{e^2} = 2\]
Substitute \[\dfrac{1}{x}dx = dz\] and \[{\log _e}x = z\] and also change the limit in (i)
\[ \Rightarrow I = \int_{ - 1}^0 { - zdz} + \int_0^2 {zdz} \]
\[ \Rightarrow I = - \left[ {\dfrac{{{z^2}}}{2}} \right]_{ - 1}^0 + \left[ {\dfrac{{{z^2}}}{2}} \right]_0^2\]
Now putting the limits:
\[ \Rightarrow I = - \left[ {\dfrac{{{0^2}}}{2} - \dfrac{{{{\left( { - 1} \right)}^2}}}{2}} \right] + \left[ {\dfrac{{{2^2}}}{2} - \dfrac{{{0^2}}}{2}} \right]\]
\[ \Rightarrow I = \dfrac{1}{2} + 2\]
\[ \Rightarrow I = \dfrac{5}{2}\]



Option ‘B’ is correct



Note: Students often mistake when they solve a definite integral using the substitution method. When they solve the integration \[I = \int_{{e^{ - 1}}}^1 { - \dfrac{{{{\log }_e}x}}{x}} dx + \int_1^{{e^2}} {\dfrac{{{{\log }_e}x}}{x}} dx\] by substitution method, they forgot to change the limits and solve \[I = \int_{{e^{ - 1}}}^1 { - zdz} + \int_1^{{e^2}} {zdz} \]. For this reason, they got incorrect answers. The correct integration is \[I = \int_{ - 1}^0 { - zdz} + \int_0^2 {zdz} \].