find integral x* (3-x)^(1/2)?

Answer 1

So you want to find

Integral (x (3 - x)^(1/2) dx)

One of the ways you can do this is through substitution, but used in a slightly nonconventional method. Normally during U substitution, you immediately solve for du; here, we're going to take a few steps prior to solving for du.

Let u = 3 - x. Then, moving the -x to the left hand side and moving the u to the right hand side,

x = 3 - u

Differentiating, we get

dx = -du

And now we substitute accordingly and get

Integral [(3 - u) [u]^(1/2) (-du) ]

It's always a good idea to pull out constants; in this case, we can pull out the (-1) in front of that du, to obtain,

(-1) * Integral [ (3 - u)u^(1/2) du]

Now, let's distribute that u^(1/2) over the brackets, to obtain

(-1) * Integral [ (3u^(1/2) - u*u^(1/2) du ]

Note that u is the same as u^(1), and u^(1/2) is multiplied to it. That means we're multiplying two things with the same base but different exponent; when we do this, we add the exponents, so
u*u^(1/2) = u^(1 + 1/2) = u^(3/2). Thus, we get

(-1) * Integral [ (3u^(1/2) - u^(3/2) du ]

Now, we integrate normally using the power rule.

(-1) * [ 3 (2/3)u^(3/2) - (2/5) u^(5/2) ] + C

Let's clean this up a bit, to obtain,

(-1) * [ u^(3/2) - (2/5) u^(5/2) ] + C

And note the impact of that (-1) on the outside; that's just going to reverse the terms, while keeping it a difference of terms.

(2/5) u^(5/2) - u^(3/2) + C

Now we replace u = 3 - x.

(2/5) (3 - x)^(5/2) - (3 - x)^(3/2) + C

Answer 2

Let u = x and dv/dx = (3-x)^(1/2)
So du/dx = 1 and v = (-2/3)*(3-x)^(3/2)

Since
Integral of (u)(dv/dx) dx = uv - integral of (v)(du/dx) dx
Integral of x* (3-x)^(1/2)
= (2x/3)*(3-x)^(3/2) - integral of (-2/3)*(3-x)^(3/2) dx + C
= (2x/3)*(3-x)^(3/2) - (-2/5)(-2/3)*(3-x)^(5/2) + C
= (2x/3)*(3-x)^(3/2) - (4/15)*(3-x)^5/2 + C

Answer 3

I'd rather skin my own legs with a potato peeler!

Answer 4

Stop posting homework questions.