Evaluating Definite Integrals with Absolute Values

The problem asks us to evaluate the definite integral \(\mathop \smallint \nolimits_{{{\rm{e}}^{ - 1}}}^{{{\rm{e}}^2}} \left| {\frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right|{\rm{dx}}\).

The integrand is \(\left| {\frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right|\). To handle the absolute value, we need to determine the sign of the expression inside it, which is \(\frac{{\ln {\rm{x}}}}{{\rm{x}}}\), over the interval of integration \([{\rm{e}}^{ - 1}, {\rm{e}}^2]\).

For \({\rm{x}} \in [{\rm{e}}^{ - 1}, {\rm{e}}^2]\), \({\rm{x}}\) is always positive. Therefore, the sign of \(\frac{{\ln {\rm{x}}}}{{\rm{x}}}\) is determined by the sign of \(\ln {\rm{x}}\).

• \(\ln {\rm{x}} \gt 0\) when \({\rm{x}} \gt {\rm{e}}^0 = 1\).
• \(\ln {\rm{x}} \lt 0\) when \({\rm{x}} \lt {\rm{e}}^0 = 1\).
• \(\ln {\rm{x}} = 0\) when \({\rm{x}} = 1\).

The interval of integration is \([{\rm{e}}^{ - 1}, {\rm{e}}^2]\). We note that \({\rm{e}}^{ - 1} \approx 0.368\), \(1\), and \({\rm{e}}^2 \approx 7.389\). The point \({\rm{x}}=1\) is within the interval \([{\rm{e}}^{ - 1}, {\rm{e}}^2]\) and is where \(\ln {\rm{x}}\) changes sign.

Thus, we must split the integral into two parts based on where \(\ln {\rm{x}}\) is negative and where it is positive:

• For \({\rm{x}} \in [{\rm{e}}^{ - 1}, 1]\), \(\ln {\rm{x}} \le 0\), so \(\left| {\frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right| = - \frac{{\ln {\rm{x}}}}{{\rm{x}}}\).
• For \({\rm{x}} \in [1, {\rm{e}}^2]\), \(\ln {\rm{x}} \ge 0\), so \(\left| {\frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right| = \frac{{\ln {\rm{x}}}}{{\rm{x}}}\).

The integral becomes:

\(\mathop \smallint \nolimits_{{{\rm{e}}^{ - 1}}}^{{{\rm{e}}^2}} \left| {\frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right|{\rm{dx}} = \mathop \smallint \nolimits_{{{\rm{e}}^{ - 1}}}^1 \left( { - \frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right){\rm{dx}} + \mathop \smallint \nolimits_1^{{{\rm{e}}^2}} \frac{{\ln {\rm{x}}}}{{\rm{x}}}{\rm{dx}}\)

Now, let's evaluate the indefinite integral \(\smallint \frac{{\ln {\rm{x}}}}{{\rm{x}}}{\rm{dx}}\). We can use a substitution method.

Let \({\rm{u}} = \ln {\rm{x}}\). Then, the differential \({\rm{du}} = \frac{1}{{\rm{x}}}{\rm{dx}}\).

The integral becomes \(\smallint {\rm{u}} \, {\rm{du}}\), which integrates to \(\frac{{{\rm{u}}^2}}{2} + {\rm{C}}\).

Substituting back \({\rm{u}} = \ln {\rm{x}}\), the indefinite integral is \(\frac{{{{(\ln {\rm{x}})}^2}}}{2} + {\rm{C}}\).

Now, we evaluate the definite integrals:

First Integral: \( - \mathop \smallint \nolimits_{{{\rm{e}}^{ - 1}}}^1 \frac{{\ln {\rm{x}}}}{{\rm{x}}}{\rm{dx}}\)

\( - \left[ {\frac{{{{(\ln {\rm{x}})}^2}}}{2}} \right]_{{{\rm{e}}^{ - 1}}}^1 = - \left( {\frac{{{{(\ln 1)}^2}}}{2} - \frac{{{{(\ln {{\rm{e}}^{ - 1}})}^2}}}{2}} \right)\)

We know that \(\ln 1 = 0\) and \(\ln {\rm{e}}^{ - 1} = -1\).

\( = - \left( {\frac{{0^2}}{2} - \frac{{{{( - 1)}^2}}}{2}} \right) = - \left( {0 - \frac{1}{2}} \right) = \frac{1}{2}\)

Second Integral: \( \mathop \smallint \nolimits_1^{{{\rm{e}}^2}} \frac{{\ln {\rm{x}}}}{{\rm{x}}}{\rm{dx}}\)

\( \left[ {\frac{{{{(\ln {\rm{x}})}^2}}}{2}} \right]_1^{{{\rm{e}}^2}} = \left( {\frac{{{{(\ln {{\rm{e}}^2})}^2}}}{2} - \frac{{{{(\ln 1)}^2}}}{2}} \right)\)

We know that \(\ln {\rm{e}}^2 = 2\) and \(\ln 1 = 0\).

\( = \left( {\frac{{{2^2}}}{2} - \frac{{0^2}}{2}} \right) = \left( {\frac{4}{2} - 0} \right) = 2\)

Finally, we sum the results of the two integrals:

\(\mathop \smallint \nolimits_{{{\rm{e}}^{ - 1}}}^{{{\rm{e}}^2}} \left| {\frac{{\ln {\rm{x}}}}{{\rm{x}}}} \right|{\rm{dx}} = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}\)

The value of the definite integral is \(\frac{5}{2}\).

Adding the results from the intervals gives the total value:

\(\frac{1}{2} + 2 = \frac{5}{2}\).

Revision Table: Definite Integral Evaluation

Additional Information: Integrating \(\frac{{\ln {\rm{x}}}}{{\rm{x}}}\)

The integral of \(\frac{{\ln {\rm{x}}}}{{\rm{x}}}\) is a common type that is solved using substitution. The key observation is that \(\frac{1}{{\rm{x}}}\) is the derivative of \(\ln {\rm{x}}\). So, if we let \({\rm{u}} = \ln {\rm{x}}\), then \({\rm{du}} = \frac{1}{{\rm{x}}}{\rm{dx}}\).

The integral transforms from \(\smallint \ln {\rm{x}} \cdot \frac{1}{{\rm{x}}}{\rm{dx}}\) to \(\smallint {\rm{u}} \, {\rm{du}}\). This is a basic power rule integral:

\(\smallint {\rm{u}} \, {\rm{du}} = \frac{{{\rm{u}}^{1+1}}}{{1+1}} + {\rm{C}} = \frac{{{\rm{u}}^2}}{2} + {\rm{C}}\)

Substituting back \({\rm{u}} = \ln {\rm{x}}\) gives the result \(\frac{{{{(\ln {\rm{x}})}^2}}}{2} + {\rm{C}}\).

When evaluating definite integrals using substitution, you can either:

• Find the indefinite integral and then evaluate it at the original limits, or
• Change the limits of integration according to the substitution variable and evaluate the new integral directly.

In this solution, we used the first method (evaluating the indefinite integral at original limits). For the integral \(\mathop \smallint_{{{\rm{e}}^{ - 1}}}^1 \frac{{\ln {\rm{x}}}}{{\rm{x}}}{\rm{dx}}\) with \({\rm{u}}=\ln {\rm{x}}\), the limits change from \({\rm{x}} = {\rm{e}}^{ - 1}\) to \({\rm{u}} = \ln({\rm{e}}^{ - 1}) = -1\) and from \({\rm{x}} = 1\) to \({\rm{u}} = \ln(1) = 0\). The integral becomes \(\mathop \smallint \nolimits_{ - 1}^0 {\rm{u}} \, {\rm{du}}\). For the integral \(\mathop \smallint_1^{{{\rm{e}}^2}} \frac{{\ln {\rm{x}}}}{{\rm{x}}}{\rm{dx}}\), the limits change from \({\rm{x}} = 1\) to \({\rm{u}} = \ln(1) = 0\) and from \({\rm{x}} = {\rm{e}}^2\) to \({\rm{u}} = \ln({\rm{e}}^2) = 2\). The integral becomes \(\mathop \smallint \nolimits_0^2 {\rm{u}} \, {\rm{du}}\). So the original integral is \( - \mathop \smallint \nolimits_{ - 1}^0 {\rm{u}} \, {\rm{du}} + \mathop \smallint \nolimits_0^2 {\rm{u}} \, {\rm{du}}\).

• \( - \left[ \frac{{\rm{u}}^2}{2} \right]_{-1}^0 = - \left( \frac{0^2}{2} - \frac{(-1)^2}{2} \right) = - \left( 0 - \frac{1}{2} \right) = \frac{1}{2}\).
• \( \left[ \frac{{\rm{u}}^2}{2} \right]_0^2 = \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = \left( \frac{4}{2} - 0 \right) = 2\).

Summing these gives \(\frac{1}{2} + 2 = \frac{5}{2}\), confirming the result.