Question

Evaluate `int_1^3 x^2*log x  "d"x`

Answer 1

Let I = `int_1^3 x^2*log x  "d"x`

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

= `[log x* x^3/3]_1^3 - int_1^3 1/x*x^3/3  "d"x`

= `[9log3 - log1*1/3] - 1/3 int_1^3 x^2  "d"x`

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

= `9log3 - 1/3(27/3 - 1/3)`

= `9log3 - 1/3(26/3)`

∴ I = `9log 3 - 26/9`