Question

Evaluate the following definite integral:

`int_1^3 logx.dx`

Answer 1

Let I = `int_1^3 logx.dx`

= `int_1^3 logx.1.dx`

= `|logx int1.dx|_1^3 - int_1^3 [d/dx (logx) int1.dx].dx`

= `|logx.x int_1^3 - int_1^3 1/x. x. dx|`

= `|logx. x int_1^3 - int_1^3 1.dx|`

=`[x.log x]_1^3 - [x]_1^3`

= (3 . log 3 − 1 . log 1) − (3 − 1)

= (3 log 3 – 0) – 2

= 3 log 3 – 2

= log 3³ – 2

∴ I = log 27 –  2