does integral calculus make sense?

Question

(a)i ran int[x/(4-x^2)^2]dx from
x=0 to x=3 through a computer
and it came up with infinity-

however, i did a bit of juggling
and by putting u=4-x^2 i came
up with -9/40 for x between 0
and 3-

which version(if either) do you
believe to be correct and why?

(b)what is a singularity in
mathematics?
does it exist?

if you want to be brave,be
careful!

juggle it a little
integral becomes
-1/2int[1/(4-x^2)^2(-2x)dx
(put u=4-x^2) and change
the x limits to u limits
it works!

Answer 1

At x = 2 your integrand becomes infinite. That is probably why your computer is flagging it. I'm not certain if your substitution steps are correct.

A singularity is a point where some mathematical expression is not defined (as int he point x =0 for the expression 1/x), or a point where a function is no longer "well behaved" in the context of being finite, or differentiable. These singularities do exist.

In mathematics in particular, it is very true that "if you want to be brave, be careful!".

Answer 2

Despite the above posts, the answer is indeed -9/40. The reason the computer had difficulty was because there is an isolated singularity at x = 2. This will be the reason for the computer having difficulty integrating.

A singularity is a point at which the function is undefined (in this case at x = 2 the function equals 2/0 which is of course undefined). Look at a standard textbook on complex analysis, and you will find a more technical definition/explanation.

Answer 3

When x=2, you have a denominator of 0. Therefore the area under the curve between 0 and 3 is infinity. You must have done your substitution wrong.

Or you can work it through:

x(4- x^2 )^-2= x(16 - 8x^2 + x^4)^-1
=(16/x - 8x + x^3)^-1
=x/16 - 1/8(x^-1) + x^-3

integrate...

x^2/32 - ln(x)/8 - (x^-2)/2

evaluate at 0 and 3...
ln 0= infinity.

Answer 4

The integral between x=0 and x=3 for this equation is 3/25.

The fact that the denominator is 0 at x=2, making the answer infinity will be why your computer doesn't like that.

Answer 5

x(x^2-a million)^5 dx you should enable u = (x^2-a million) because in case you enable u = x, you're merely replacing each and every of the x's contained in the question right into a u. enable u = x^2 - a million du/dx = 2x --> multiply dx yo both area du = 2xdx --> divide both area by technique of 2x dx = du/2x Now sub in u = (x^2-a million) and dx = du/2x into the question: ?x(x^2-a million)^5 dx = ?x(u)^5 du/2x --> cope with du/2x as a volume, so that you may cancel out the x's = ?(u)^5 du/2 --> convey a million/2 onto the exterior of the fundamental signal = a million/2 ?(u)^5 du --> combine as widely used = a million/2 ((u^6)/6) + c --> multiply ((u^6)/6) by technique of one million/2 = (u^6)/12 + c--> sub decrease back (x^2-a million) somewhat than u = ((x^2-a million)^6)/12 +c <-- your answer

Answer 6

Integral calculus did make sense when i did it at uni, but now its all just evil

Answer 7

Errr...

Answer 8

*pulling out hair*

UGHHHH

Answer 9

no