Question relating to manipulating definite integrals?

Question

If the integral of f(x) between 1 and 5 equals 5 what is the integral between -1 and -5 of f(-x)?

Why is the answer =5? If the integral of f(x) between 1 and 5 equals 5, then the integral of f(-x) between -1 and -5 can be calculated by settin u=-x, and calculating the integral of f(u) between 1 and 5 which should also equal to 5. Anyone see an error in above logic?

If the integral of f(x) between 1 and 5 equals 5, calculate the integral of f(-x) between -1 and -5. I thought answer is also 5 because if you define u = -x the above becomes the integral of f(u) between u=1 and u=5 which should also be equal to 5. Anybody see what's wrong with the logic?

Answer 1

We must cast belief (+5) --- but see P.S. below --- and voting (-5) aside. Calculus has little to do with either religion or democracy, thank Newton (or Leibnitz, if you prefer).

You have to keep a straight head when doing things like this. Let's hope I DO. I'll go through each step in order to clearly PROVE the result(s):

Write the original integral and its value as int (1 to 5) f(x)dx = 5,
and use "--->" as "becomes."

I'm only going to rewrite what's on the LHS (left-hand side).

Change x to -z, that is:

x ---> - z.

Then f(x) ---> f(-z), dx --->d(-z), in other words - dz, and the two limits become z = -1 and -5 respectively. (Keeping this ordering straight is essential at this stage.)

Then int (1 to 5) f(x)dx ---> int (-1 to -5) f(-z) (-dz),

which equals - int (-1 to -5) f(-z)dz; but z, like x, is a "dummy variable" under the integral sign (it doesn't matter what you call it) , so this is in fact:

- int (-1 to -5) f(-x)dx; the value must still be 5, as we've only rewritten the form of the integral itself.

So int (-1 to -5) f(-x)dx = -5, the negative of what you started with.

The LHS immediately above looks like what you asked, IF by integrating it you mean "please put dx after it, then integrate." Only you can know whether that's what you were thinking --- please let us all know.

Note however that integration is normally performed from a SMALLER to a LARGER limit. So, SWAPPING THE LIMITS back to that convention, we get:

int (-5 to -1) f(-x) dx = +5, the original value.

I hope this helps!

Live long and prosper.

P.S. The first responder, sdiver248... has now changed the sign of his/her answer at least two times in the past half-hour, which suggests a rather fluid belief system!

I'm also not clear on to whom your own "additional details" are addressed; perhaps to sdiver248... following the latter's initial response of 5 (before changing it to "-5" and then back again)?

In any case, I'm not sure from your "additional details," exactly what you mean --- that's the problem with language about matters like this. It's possible, in saying "integrate," but not "with respect to what," that there's a negative sign ambiguity. (Will you be integrating with respect to x or to u?) Going through ALL the implications of the change in variable, as I tried to do above, hopefully clears up the possible confusion here.

Thanks for promoting careful thought!

Answer 2

i will use Int to point integration now Int[ dx / squarert( a^2 - x^2 ) = sin^-a million( x/a) is a wide-spread quintessential for a > x right here we've Int[ dx /squarert ( 9 - x^2 ) ] = Int[ dx /squarert ( 3^2 - x^2) ] = sin^-a million( x/3) comparing between 0 and a million supplies [ sin^-a million( a million/3) ] - [ sin^-a million( 0) ] = 19.40 seven - 0 ie 19.40 seven degrees ( 2 dp)

Answer 3

I believe the answer should be 5

Answer 4

i vote for -5