Calculus

Please verify my answer.

Find the average value of the function f(x) = x√(1+2x) on the interval [0,4]
Favg = 1/(b-a) ∫ from a to b of f(x) dx

= 1/(4-0) ∫ x√(1+2x) dx
= 1/4 ∫ x(1+2x)^(1/2) dx
= 1/4 [ (x^(2))/2 (2/3) (1+2x)^(3/2)]
= 1/4 [ ∫b -∫a]
= 1/4 [ 144]
= 36 (The average value)

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  3. 👁 186
  1. Not quite. For
    ∫ x√(1+2x) dx
    Let u=1+2x, so du = 2dx and x = (u-1)/2 -- then you have
    ∫ (u-1)/2 * √u * 1/2 du = 1/4 ∫(u^(3/2) - u^(1/2)) du
    See where that takes you

    Recall that ∫f(x)*g(x) ∫f(x) * ∫g(x)
    also you forgot to work with the (1+2x) properly

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  2. Alright so,

    = 1/(4-0) ∫ x√(1+2x) dx

    u = 1+2x
    du = 2 dx
    1/2 du = x

    = 1/4 [ (u-1)/2 (u)^(1/2) (2)] du
    = 1/4 [ u^(3/2) - u^(1/2)] du
    = 1/4 [ (2/5)(1+2x)^(5/2) - (2/3)(1+2x)^(3/2)]
    = (1/10)(1+2x)^(5/2) - (1/6)(1+2x)^(3/2) + C

    Then, [ ∫b -∫a]
    198/10 + 2/30 = 298/15

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  3. not quite
    du = 2 dx, so dx = 1/2 du
    you're not watching things.

    1/4 ∫[0,4] x√(1+2x) dx = 1/4 (∫[1,9] (u-1)/2 * u^(1/2) * 1/2 du)
    = 1/16 ∫[1,9] u^(3/2) - u^(1/2) du

    are you familiar with changing the limits of integration? It saves having to convert everything back to x's before evaluating.

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