Question

Evaluate:

`int_1^3 log x dx`

Answer 1

Let I = `int_1^3 log x dx`

= `int_1^3 log x * 1 dx`

= `[log x int 1 * dx]_1^3 - int_1^3[d/(dx) (log x) int 1 * dx] dx`

= `[(log x) * x]_1^3 - int_1^3 1/x * x  dx`

= `[x log x]_1^3 - int_1^3 1* dx`

= `(3 log 3 - 1 log 1) - [x]_1^3`

= (3 log 3 – 0) – (3 – 1)

= 3 log 3 – 2

= log 3³ – 2

∴ I = log 27 – 2