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Ok, I know the correct answer is 8/3. But I CANNOT figure out how to do it. Can someone please help?!

Just to clarify, the question wants the integral of:
(x) (4-x^2)^2 OR
"x times the square root of 4 minus x squared", where b=2 and a=0.

2007-08-20 19:35:13 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

This problem can be done using the u-substitution method.

Let u = (4 - x^2). Then, du = -2x dx, or (-1/2) du = x dx.

Now, substitute that into your integral to get the integral of:

(-1/2) * (u ^(1/2)) du

When you take the integral, you get

(2/3) * (-1/2) * (u ^ (3/2)).

Now, substitute u = (4 - x^2):

(-1/3) * ((4 - x^2) ^ (3/2))

Suppose we call the above function f(x). Since the limits of integration go from 0 to 2, the value of the integral = f(2) - f(0) =

[(-1/3) * ((4 - 2^2) ^ (3/2))] - [-(1/3) * ((4 - 0^2) ^ (3/2))]
= (1/3) * (4 ^ (3/2))
= (1/3) * (8) = 8/3.

2007-08-20 19:52:20 · answer #1 · answered by Susan B 2 · 0 0

Take the integral from 0 to 2 of

∫{x√(4 - x²)}dx
Let
u = 4 - x²
du = -2x dx
-du/2 = x dx

∫{x√(4 - x²)}dx = -(1/2)∫{√u} du

= -(1/2)(2/3)u^(3/2) = (-1/3)(4 - x²)^(3/2) | [Eval from 0 to 2]

= (-1/3)[0 - 8] = 8/3

2007-08-20 19:46:32 · answer #2 · answered by Northstar 7 · 0 0

Make a substitution,
Let, u = 4 - x^2
du = -2x dx
-du/2 = x dx

-1/2 int [ (u)^1/2 du]
= -1/2 [ 2/3 (u^3/2) ]
= -1/2 [ 2/3 (4 - x^2)^3/2]
= -1/3 [ (4 - x^2)^3/2 ]
substitute the limits, upper limit=2, lower = 0,
= -1/3 [ 0 - (4)^3/2 ]
= -1/3 (-8)
= 8/3

2007-08-20 19:46:02 · answer #3 · answered by Anonymous · 0 0

Let u = 4-x^2. then du = -2x dx.

The integral beomes (-1/2)u^(1/2) du, now you can integrate by power rule to get (-1/2)(2/3)u^(3/2) = (-1/3)u^(3/2), which equals (-1/3)(4-x^2)^(3/2). Evaluating at 2 and 0:

(-1/3)(0-8) = 8/3.

2007-08-20 19:47:17 · answer #4 · answered by Derek C 3 · 0 0

observe how the by-manufactured from x^2 - a million = 2x. consequently, in case you permit the numerator equivalent 2x after which placed a consistent of a million/2 on the front (the two and the a million/2 will cancel if simplified), then you definately might have an fundamental the place the by-manufactured from the denominator is the numerator. fundamental expressions of this variety evaluate to be log_e |denominator|. consequently, for this fundamental expression, the respond may well be (a million/2) log_e |x^2 -a million| and you purely sub interior the limitations. wish this permits!

2016-12-12 08:18:52 · answer #5 · answered by ? 4 · 0 0

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