Integration

• Audi

  2 Aug 09, 20:00

  Differentiate x^2 ln x with respect to x

  Hence, evaluate x ln x between range 3 and 2, giving your answer correct to 2 decimal places

  I do know how to differentiate but because the integration term is similar to the term for differentiation, I don't see how I can make use of my answer to get the integration part. Unless they tell me to integrate the answer i got after diffentiating.

  Many thanks

• Forbiddensinner

  2 Aug 09, 20:20

  Originally posted by Audi:

  Differentiate x^2 ln x with respect to x

  Hence, evaluate x ln x between range 3 and 2, giving your answer correct to 2 decimal places

  I do know how to differentiate but because the integration term is similar to the term for differentiation, I don't see how I can make use of my answer to get the integration part. Unless they tell me to integrate the answer i got after diffentiating.

  Many thanks

  y = x^2.Inx

  dy/dx = x + 2x.Inx

  x.Inx = x.Inx + (1/2)x - (1/2)x

          = 1/2 [ 2x.Inx + x - x ]

          = 1/2 [ x + 2x.Inx ] - (1/2)x

  Integrate this to get [(1/2)(x^2)(Inx) - (1/4)x^2 + c]

  You should be able to continue from here.

   

• wee_ws

  3 Aug 09, 09:00

  Hi,

    We know that d/dx [ x^2 ln x ] = x + 2x ln x.

    Taking the definite integral of both sides with respect to x, we obtain:

    [ x^2 ln x ] over 2 and 3 = [x^2 / 2] over 2 and 3

                                               + 2 (integral of x ln x) over 2 and 3.

    Learn, from this typical question, that there is a close relationship between differentiation and integration.

    Thanks!

  Cheers,
  Wen Shih