Question

Solve the following : `int_1^3 x^2 log x*dx`

Answer 1

Let I = `int_1^3 x^2 log x*dx`

= `[log x int x^2*dx]_1^3 - int_1^3 [d/dx (log x) int x^2*dx]*dx`

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

= `[9 log 3 - log 1* 1/3] - (1)/(3) int_1^3 x^2*dx`

= `[9 log 3 - 0] - (1)/(3)[x^3/3]_1^3`

= `9 log 3 - (1)/(3)(27/3 - 1/3)`

= `9 log 3 - (1)/(3)(26/3)`

∴ I = `9 log 3 - (26)/(9)`.