Anonymous

# 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)

Relevance

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

• 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