So many PDF's of old books are unreadable because of this shit...
 It's All Fucked Shirt $22.14 |
 |
It's All Fucked Shirt $22.14 |
FRICK THIS THING
So many PDF's of old books are unreadable because of this shit...
It's All Fucked Shirt $22.14 |
|
It's All Fucked Shirt $22.14 |
It's juſt a ſpace ſaving eſſe
you should be able to solve this
Not OP but I've never been in high school due to family situations (I'm 27) and I've never learned what integrals, calculus are and what they do. Are there any books for uneducated ignorants who have no idea what it is?
just youtube it, it's not that hard and STEM books don't help you learn they only supplement it, talking teaches you
Math textbooks? What kind of moronic question is that
Lmfao, there's no need to learn this shit. You're not going to be some brainiac genius. Stick to regular maths.
Functional Analysis by Rudin
I suffered through engineering school. These shits are useless in reality. They have never been useful in my career.
They are not useless.
t. aerospace R&D chad
Also, that's not an integral sign, it's an old-school "s".
>it's an old-school "s".
lol, stick to stem.
Lol, you are uneducated and don't even know what the sign is or how it came to be. You've only read things with modern typesetting.
It is a long S, not an S. It could also be an esh depending on what OP is reading but most likely a long S.
What engineering lmao?
They are definetly not useless
engineers legitimately calculate integrals with the trapezoid rule, they’re all brainlet bugs
I don't, frick you
math is fake. anything other than basic algebra has no purpose. it's just invented for incels to have something to wank that isn't their left hand. the same types of people will collect postage stamps or memorize baseball statistics just to have something to fill the void.
lmao normies get filtered
https://www.gutenberg.org/files/33283/33283-pdf.pdf
>Tan
>You should be able to
Bollocks. This is a hard integral. Why do so many people spend so much effort trying to make other people feel bad?
I had a think and I can (maybe) solve this with a sort of cheap kludge that will make proper pure mathematicians gasp in horror.
Here goes.
Think about 1 / 1 + u ^ 2019, just for positive u.
This is going to be pretty much a square wave.
If u<1, then raising it to that huge power is going to make it more-or-less vanish. So we have 1/1 which is 1.
If u=1, then u ^ 2019 = 1, so the expression is 1/2.
If u>1, then u ^ 2019 is gigantic, so the answer is going to be more-or-less zero. (It's 1/<gigantic number>.)
Now think about our integral. We're integrating a function that's basically 1 all the way from tan x = 0 up to tan x = 1, and then flips to basically zero for tan x > 1.
Now tan x starts at 0 when x = 0, and gets bigger as x gets bigger, and reaches 1 when x = pi/4.
So the integral is basically a rectangle, of height 1, and width pi/4.
So the answer is pi/4.
The trouble is I've sort of kludged things, but MAYBE they all cancel, and the answer is exactly this. But even if it isn't exactly this, it's good enough for all practical purposes. (Pure mathematicians hate statements like that. Sorry, pure mathematicians.)
OK, I got interested and had a look at what a real mathematician would do.
So it turns out there's a cute trick you can use. Basically you take advantage of two things:
— tan (pi/2 - x) = 1 / tan (x)
— the integral "happens" to be from 0 to pi/2.
So you can re-write the integral in two different ways, using that identity, and just using the substitution y = a-x. Then you add the two things together, and the result is 1, because the fractions cancel. So twice the integral is 1, over those limits. So the integral itself is half of that, which is pi/4.
So I was completely correct with my kludge, although I might have just got lucky. But of course the number 2019 was just picked to be the date, so you know it can't really matter. So you "know", in an exam context, that there is going to be some clever trick like this.
But when you're solving real problems, you can't rely on there being a cute trick like this, so you just do the damn thing numerically, the way I did. (Sorry Alicia Western.)
I wonder if there's some middle course? Something that uses a bit more clever maths than I did, but doesn't use the above trick (which only works for this exact thing)? Any mathematicians here could probably tell us.
Mate i vividly remember this sorta thing from the last year of highschool. And it's not a cute trick, that's how they teach you to solve it. It's called King's Rule if i remember correctly.
Whilst working with definite integrals, there are like 7 or 8 other 'tricks' as you say you can utilize to solve it easily.
I remember all this coz i had the exact same question in my final highschool exam, except the exponential was 2011 or 2012
I have a B.S. in Mathematics.
What I would do is first convert tan to sin/cos
>int 0 to pi/2 : 1 / (1 + sin^2019 x /cos^2019 x)
Then I would get rid of the awkward fraction
>int 0 to pi/2 : cos^2019 x / (cos^2019 x + sin^2019 x)
Then you recognize that on the interval (0, pi/2), sin x and cos x are symmetric. This implies that the above is equal to
>int 0 to pi/2 : sin^2019/(sin^2019+cos^2019)
So adding the two together, you get
>int 0 to pi/2 : 1
Which is pi/2, which is twice the desired integral. So the answer is pi/4.
Right. I guess "tan(pi/2-x) = 1/ tan(x)" is just an obvious consequence of swapping sin and cos & running backwards over the range. So in a way the two methods are quite similar. I remember it can be useful, with definite integrals, to consider summing them "either way" (right to left or left to right).
If you can't easily read other scripts fluently then you're just low IQ.
if you are actually interested, you will become used to it and your brain will gain a new wrinkle
I HATE THOSE ESSES TOO
Fricking hate integrals, luckily I never touched them again after Calc 3
the sign of soros
ENTER: ∮
I'm sorry Anon, but I'm having trouble imagining a scenario where I would be reading multiple PDFs of old calculus books.