Question

Do integration


quality answer to the question ☑☑​

Answer 1

Answer

y = | Log x |  

    Log x is negative for  x = 1/e to 1.     Log 1 is 0.

     Log x is positive for x from 1 to e.

• Integration  of | Log x | from 1/e to e.

• This is same as the area under the graph. see diagram for clarity.

$\int \limits_{1/e}^{e} {| Log \ x |} \, dx\\\\ = \int \limits_{1/e}^{1} {- Log \ x } \, dx + \int \limits_{1}^{e} {Log \ x } \, dx\\\\=- [ x \ Log \ x-x]_{1/e}^{1} + [x \ Log \ x - x ]_{1}^{e}\\\\=1-\frac{1}{e}-\frac{1}{e}+e-e-0+1\\\\=2\frac{(e-1)}{e}$

Answer is   2 (e - 1) / e

Answer 2

Answer:

$Answer</p><p></p><p>y = | Log x |  </p><p></p><p>    Log x is negative for  x = 1/e to 1.     Log 1 is 0.</p><p></p><p>     Log x is positive for x from 1 to e.</p><p></p><p>Integration  of | Log x | from 1/e to e.</p><p></p><p>This is same as the area under the graph. see diagram for clarity.</p><p></p><p>\begin{gathered}\int \limits_{1/e}^{e} {| Log \ x |} \, dx\\\\ = \int \limits_{1/e}^{1} {- Log \ x } \, dx + \int \limits_{1}^{e} {Log \ x } \, dx\\\\=- [ x \ Log \ x-x]_{1/e}^{1} + [x \ Log \ x - x ]_{1}^{e}\\\\=1-\frac{1}{e}-\frac{1}{e}+e-e-0+1\\\\=2\frac{(e-1)}{e}\end{gathered}1/e∫e∣Log x∣dx=1/e∫1−Log xdx+1∫eLog xdx=−[x Log x−x]1/e1+[x Log x−x]1e=1−e1−e1+e−e−0+1=2e(e−1)</p><p></p><p>Answer is   2 (e - 1) / e</p><p></p><p>$

Answer

y = | Log x |

Log x is negative for x = 1/e to 1. Log 1 is 0.

Log x is positive for x from 1 to e.

Integration of | Log x | from 1/e to e.

This is same as the area under the graph. see diagram for clarity.

\begin{gathered}\int \limits_{1/e}^{e} {| Log \ x |} \, dx\\\\ = \int \limits_{1/e}^{1} {- Log \ x } \, dx + \int \limits_{1}^{e} {Log \ x } \, dx\\\\=- [ x \ Log \ x-x]_{1/e}^{1} + [x \ Log \ x - x ]_{1}^{e}\\\\=1-\frac{1}{e}-\frac{1}{e}+e-e-0+1\\\\=2\frac{(e-1)}{e}\end{gathered}

1/e

∫

e

∣Log x∣dx

=

1/e

∫

1

−Log xdx+

1

∫

e

Log xdx

=−[x Log x−x]

1/e

1

+[x Log x−x]

1

e

=1−

e

1

−

e

1

+e−e−0+1

=2

e

(e−1)

Answer is 2 (e - 1) / e