Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

evaluate the indefinite integral of x^2sqt(6-x^3)dx?

7 Answers

Relevance
  • Anonymous
    1 decade ago
    Best Answer

    ∫ x² √(6 - x^3) dx

    let u = 6 - x^3, then du = -3x²dx <==> x²dx = (-1/3)du

    substitute

    ∫√u * (-1/3)du

    (-1/3) ∫u^(1/2) du

    (-1/3) (2/3) u^(3/2) + C

    (-2/9) (6 - x^3)^(3/2) + C <== answer

  • 1 decade ago

    let u=6-x^3 then you get u'=-3x^2

    you already have x^2 in the equation so you just need neg. 3, so multiply neg.3 in the front of the integral sign and multiply the whole thing by -1/3 so that they make one (so they cancel out) then you have integral of u^1/2= 2/3u^(3/2)=2/3(6-x^3)^(3/2) and that's the answer.

  • J D
    Lv 5
    1 decade ago

    ∫x^2sqt(6-x^3)dx

    let u = 6-x^3; du = -3x^2 dx

    -1/3 ∫sqt(u) du

    = -1/3 (2/3)u^(3/2) + C

    = -2/9(6-x^3)^(3/2) + C

  • 1 decade ago

    This is my stab at it:

    (-2/9)(6-x^3)^(3/2)+c

  • How do you think about the answers? You can sign in to vote the answer.
  • Anonymous
    1 decade ago

    Try

    -(2/3) (6 - x^3)^(3/2)

    Source(s): Longtime college math teacher
  • 1 decade ago

    (-2/9)[(6-x^3)]^(3/2)

Still have questions? Get your answers by asking now.