Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Calculus: U-substitution?

Please solve this problem letting u=√(4-x)

∫x√(4-x) dx

thanks

Update:

please try and explain the concept of changing dx to du

Update 2:

u must equal sqrt(4-x)

3 Answers

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  • 1 decade ago
    Favorite Answer

    u=√(4-x)

    u² = 4-x

    x = 4-u²

    dx = -2u du

    So subbing in the various components of your original integral, you now have:

    ∫(4-u²)(u)(-2u du)

    = -2 ∫4u² - u^4) du

    = -2 (4u³/3 - u^5/5) + C

    = -8u³/3 + 2u^5/5 + C

    = -8(4-x)^(3/2) / 3 + 2(4-x)^(5/2)/5 + C

  • Jared
    Lv 7
    1 decade ago

    I'm just going to trudge through this.

    u = √(4 - x) --> u² = 4 - x --> x = 4 - u²

    dx = (dx/du) * du = -2u * du

    substitute for √(4 - x), the x, and the dx:

    ∫(4 - u²) * u * (-2u * du) --> -2∫(4u² - u⁴)du

    this integral is pretty simple:

    -2 * ((4 / 3) * u³ - (1 / 5)u⁵)

    plugging back in with x:

    ∫(x√(4 - x))dx = (-8/3) * (4 - x)^(3/2) + (2/5) * (4 - x)^(5/2) + C

    This is a BAD example of u-substitution, this is the GENERAL way to do u-substitution:

    1. Create substitution: u = f(x) --> find inverse x = g(u)

    2. Find dx --> dx = g'(u)du, sub. into equation:

    However, most u-substitutions are motivated by the chain rule, i.e., something that looks liked this:

    ∫g'(x)f(g(x))dx

    Because then, u-substitution is:

    u = g(x)

    du = g'(x)dx <-- but this appears in system, so substitute g'(x)dx, in the original, with JUST du:

    ∫du * f(u) --> ∫f(u)du

    When dealing with an integral, especially, variables are "dummy variable" (meaning they can be anything):

    ∫f(u)du = ∫f(x)dx = ∫f(y)dy = ∫f(ξ)dξ = ∫f(⌘)d⌘

    (it doesn't matter WHAT symbol we use, they are ALL equivalent)

    Usually f(x) is easy to integrate (thus the need for u-substitution):

    Here's a common example:

    ∫(2x√(1 + x²))dx --> u = 1 + x², du = 2xdx

    becomes:

    ∫√(u)du = (2/3) * u^(3/2) + C = (2/3) * (1 + x²)^(3/2) + C

    Here's something to think about. What if the 2, in 2x, hadn't been there? Would this still have worked?

  • Anonymous
    1 decade ago

    u = 4-x

    du = -1 dx

    so, ∫ (4-u)√(u) -du

    = -∫ (4-u)√(u) du

    If this is a definite integral, make sure to change the parameters to be compliant with u instead of x.

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