Effin' integrals.

[eLiu]

I read the hint. It doesn't make much sense if you're not already very fluent in Calc stuff. :|

I'm good at doing the regular types of problems, not these ones that are based more off the language.

Regular calc stuff? LoL so you're nothing more than a human calculator. And a bad one at that, b/c a computer knows how to integrate sqrt(1+x^2).

Instead of "I don't get it," why don't you at least try? Seriously the hint GIVES you the answer. Ok fine, I admit that 1 additional step is required for part 1, but part 2 is free. If the problem didn't have a hint, and someone wrote down the words in the hint, I'd be tempted to give them full credit.

So you read the hint. You don't know how to proceed. We all know you're not hte fastest zergling in the control group, but go ahead and describe exactly what you do understand, and exactly what you don't understand. Post your attempt to apply the hint (or what part(s) of it you understand) to the problem.

edit: credit to Sean "Day[9]" Plott for the zergling humorz.

 

[eLiu]

What do you mean by regular type of problems? My point is, if you can't understand the hint.. its better you not solve the problem.

I have to ask, do you know the concept of additivity of integral?

c'mon now, giving away too much again

OP, this is pretty much the "1 additional step" I referred to.

 

[TecHNooB]

Regular calc stuff? LoL so you're nothing more than a human calculator. And a bad one at that, b/c a computer knows how to integrate sqrt(1+x^2).

Instead of "I don't get it," why don't you at least try? Seriously the hint GIVES you the answer. Ok fine, I admit that 1 additional step is required for part 1, but part 2 is free. If the problem didn't have a hint, and someone wrote down the words in the hint, I'd be tempted to give them full credit.

So you read the hint. You don't know how to proceed. We all know you're not hte fastest zergling in the control group, but go ahead and describe exactly what you do understand, and exactly what you don't understand. Post your attempt to apply the hint (or what part(s) of it you understand) to the problem.

kekekekekekekekekekekekekekekekekee

 

[Mr. Pedantic]

It's funny how you're getting sarcastic and annoyed with all these other posters TridenT, when they are the ones who know the answer and (some) are trying to help you. Humility isn't a bad quality in a student.

I'm good at doing the regular types of problems, not these ones that are based more off the language.

Maths is about thinking. If you suck at thinking, then you suck at maths. And you haven't exactly shown that you're a brilliant logician.

 

[Fayd]

Calc 2 yes?

are you kidding me?

i've seen that in calc 1.

 

[thescreensavers]

Good luck to you, I barley made it out alive on calc, I have calc 2 next semester.

Who did you have for Calc?

 

[TridenT]

It's funny how you're getting sarcastic and annoyed with all these other posters TridenT, when they are the ones who know the answer and (some) are trying to help you. Humility isn't a bad quality in a student.


Maths is about thinking. If you suck at thinking, then you suck at maths. And you haven't exactly shown that you're a brilliant logician.

I don't understand the language because most of my teachers teach math as in how to do the problems, not how to do proofs.

 

[TridenT]

What do you mean by regular type of problems? My point is, if you can't understand the hint.. its better you not solve the problem.

I have to ask, do you know the concept of additivity of integral?

Maybe? I know int(f(x) + g(x))dx is same as int(f(x))dx + int(g(x))dx

 

[Fayd]

if i might take a stab at it.

ignore the fact it's under a square root... for now.

it's a positive constant + a positive variable, between 0 and 1.

on open interval between 0 and 1, x is always > than x^2.

so on that entire interval, the real value of 1+x^2 will always be less than 1+x. (and therefore the square root of that will always be greater..)

so... since the real value of sqrt(1+x^2) is always less than sqrt(1+x), the definite integral between 0 and 1 will always have the first one being smaller than the second.

boom. i solved the question, i didn't evaluate the integral. this is basic arithmetic shit. i can't believe you're in calc2 and can't solve this.

 

[eLiu]

Maybe? I know int(f(x) + g(x))dx is same as int(f(x))dx + int(g(x))dx

That is true with indefinite integrals (can you prove it?). What about definite integrals?

int(f(x), x=a..b) (interpreted as the integral of f(x) from a to b)

int(f(x),x=a..b) + int(g(x),x=a..b) = ?
int(f(x),x=a..b) + int(g(x),x=c..d) = ?

Why? Prove it.

 

[TridenT]

That is true with indefinite integrals (can you prove it?). What about definite integrals?

int(f(x), x=a..b) (interpreted as the integral of f(x) from a to b)

int(f(x),x=a..b) + int(g(x),x=a..b) = int(f(x) + g(x), x=a..b)
int(f(x),x=a..b) + int(g(x),x=c..d) = They're different equations with different coordinates. If the equations were same and coordinates were within reach of each-other then that's something I guess.

Why? Prove it.

I don't know if I can prove it because I don't know what that entails.

 

[Modular]

I don't know if I can prove it because I don't know what that entails.

Keep at it bud! You can do this.

 

[eLiu]

I don't know if I can prove it because I don't know what that entails.

Yeah that's what I figured.

Ok, I tell you that:
int(f(x),x=a..b) + int(g(x),x=c..d) = int(f(x)+g(x),x=a..b) + int(g(x),x=?) + int(g(x),x=??)

How about now? What ranges go in "?" and "??"

edit: as for the 'proving' part, go back and look in your book. You must've learned the definition of integrals. Or a certain theorem of calculus is that is so critical/important that its name tells you that everything stems from it. Christ, now I'm being as generous as busydude D: If you didn't learn these things in class, crack open that book/internet/whatever and start reading.

 

[TridenT]

Yeah that's what I figured.

Ok, I tell you that:
int(f(x),x=a..b) + int(g(x),x=c..d) = int(f(x)+g(x),x=a..b) + int(g(x),x=?) + int(g(x),x=??)

How about now? What ranges go in "?" and "??"

edit: as for the 'proving' part, go back and look in your book. You must've learned the definition of integrals. Or a certain theorem of calculus is that is so critical/important that its name tells you that everything stems from it. Christ, now I'm being as generous as busydude D: If you didn't learn these things in class, crack open that book/internet/whatever and start reading.

Yeah.. I never learned that two totally unrelated equations with totally unrelated intervals have any relation to each other whatsoever...

Like saying these two should pan out as your equation is:

 

[eLiu]

Yeah.. I never learned that two totally unrelated equations with totally unrelated intervals have any relation to each other whatsoever...

Like saying these two should pan out as your equation is:

I doubt you would have learned that in class. They can't possibly spoonfeed you every derived result of calculus.

I'm not asking you to recite shit from the book. I assume that part 1 of that problem involved shit you could just recite, we wouldn't be here.

If and when you figure out how to prove that int(f(x),x=a..b) + int(g(x),x=a..b) = int(f(x) + g(x), x=a..b), maybe you'll have a chance of seeing what's going on. Don't hold your breath.

I've given you this form:
int(f(x),x=a..b) + int(g(x),x=c..d) = int(f(x)+g(x),x=a..b) + int(g(x),x=?) + int(g(x),x=??)

It is possible to get f & g under the same integral sign with the same bounds, but then you get 2 extra integrals involving only g. I'm not saying this is simpler or more useful than having them separate from each other. But it is possible... and useful for your homework.

I mean at this point you could practically just try a bunch of shit & guess what ? and ?? are, then work backwards. This is dumb, but it could work. ? is a1..b1. ?? is a2..b2. a1,b1,a2,b2 are probably at worst simple relations of a,b,c,d.

edit: once again, start w/the proof of the simpler statement. Preferably using that important theorem.

 

[Fenixgoon]

They ain't so bad normally, but give me a problem like this:

"Use the properties of integrals to verify the inequality without evaluating the integrals."

I'm not too aware of the properties of integrals when it relates to shit being under a squareroot that isn't able to be evaluated.

i haven't done integrals in a while, but the first looks like a prime candidate for a trig substitution.

the second, i don't see why you couldn't use a "u" substitution.

it's been a while since i've used calc

 

[TridenT]

i haven't done integrals in a while, but the first looks like a prime candidate for a trig substitution.

the second, i don't see why you couldn't use a "u" substitution.

it's been a while since i've used calc

Not supposed to evaluate the integrals.

It's been solved in the thread already.

f(x)<=g(x), a<=x<=b if int(f(x))dx<=int(g(x))dx, a<=x<=b

Take the f(x) and g(x)... show how f(x) is <= than the g(x) on [0,1] and blah.

 

[DrPizza]

NOOOOOOOOoOOOoooooOO you gave it away

edit: though mad propz to anyone who can work out hte integral on the left by hand. I guessed (=bullshitted) the solution... it's really similar to integrating ln(x), something I assumed after integrating by parts. In this case, it helps to know that derivative of asinh(x) = 1/sqrt(1+x^2). It comes out to 1/2(x*sqrt(1+x^2) - asinh(x)). Not sure how that would've helped the OP even if he could derive it, lol.

The integral on the left is very similar to an arc length problem (of a parabola) that I assign to my calc I students every year. It's one of my favorite problems because it starts with knowing the formula for finding the arclength, then they need to make a "tricky" trig substitution (replace x with tan(theta), dx with sec²(theta)). Then, to integrate sec^3, they have to do an integration by parts. During the integration by parts, they have to integrate secant, which involves picking "multiply by sec(theta)+tan(theta) over sec(theta)+tan(theta)" out of thin air. (Plus there are some coefficients/constants in there, since it's not simply the length of y=.5x^2 from 0 to 1.)

When I teach them how to integrate sec(x), they always ask "how did you figure out to multiply by that??"

Back when we used chalkboards, it was the best problem in the world to do before an Algebra class - I'd make the solution wrap around the room & would leave it up. "What's all that stuff?" "Oh, that's the answer to ONE of their homework problems last night. Now, you were whining again about how much work the quadratic formula is??"

 

[MovingTarget]

I don't understand the language because most of my teachers teach math as in how to do the problems, not how to do proofs.

Your high school Geometry teacher is going to come back to haunt you for saying that. I certainly would do that to my students if they ever uttered such blasphemy.

<-- Currently teaches high school Geometry

 

[TridenT]

Your high school Geometry teacher is going to come back to haunt you for saying that. I certainly would do that to my students if they ever uttered such blasphemy.

<-- Currently teaches high school Geometry

Proofs like those was around 9th grade. It's been around 6 years since I did any of that... We never even used them after Geometry from what I can recall.

I've taken Calc I and Calc II before. It was most definitely not this difficult/confusing in high school. The Calc I teacher I had recently was a lot like the high school teacher I had in that the work revolved around the typical problems and not the proofs.

 

[eLiu]

The integral on the left is very similar to an arc length problem (of a parabola) that I assign to my calc I students every year. It's one of my favorite problems because it starts with knowing the formula for finding the arclength, then they need to make a "tricky" trig substitution (replace x with tan(theta), dx with sec&#178;(theta)). Then, to integrate sec^3, they have to do an integration by parts. During the integration by parts, they have to integrate secant, which involves picking "multiply by sec(theta)+tan(theta) over sec(theta)+tan(theta)" out of thin air. (Plus there are some coefficients/constants in there, since it's not simply the length of y=.5x^2 from 0 to 1.)

When I teach them how to integrate sec(x), they always ask "how did you figure out to multiply by that??"

Back when we used chalkboards, it was the best problem in the world to do before an Algebra class - I'd make the solution wrap around the room & would leave it up. "What's all that stuff?" "Oh, that's the answer to ONE of their homework problems last night. Now, you were whining again about how much work the quadratic formula is??"

Hmmm, cool. That's where I was hung up (see a few posts down; another poster suggested something that amounts to integrating -csc^3, I gave a suggestion that results in sec^3, but I had no idea after that either way). Integrating trig stuff is usually a trick I leave to Maple. Though I guess googling could've shown me how to integrate secant easily enough. I'm guessing that to end up with the asinh(x) term, after you do the trig magic, you'll have to gather terms to find the definition of asinh() using natural log?

Mmmm guessing it was still the slickest option Integrating by parts immediately gives you x*sqrt(1+x^2); I figured then that I needed something whose derivative is 1/sqrt(1+x^2); guessing out some constant factors, and tada. Just like the integral of lnx is x*lnx - ?. ? Must have derivative = 1.


edit: OP, nothing you've posted here has actually been hard or even "not easy." You haven't learned calculus before. You just learned some mechanical steps without any real understanding.

 

[elwood]

Just use a calculator. Damn y'all is stupid!

 

[MovingTarget]

Proofs like those was around 9th grade. It's been around 6 years since I did any of that... We never even used them after Geometry from what I can recall.

I've taken Calc I and Calc II before. It was most definitely not this difficult/confusing in high school. The Calc I teacher I had recently was a lot like the high school teacher I had in that the work revolved around the typical problems and not the proofs.

The specific "stuff" in Geometry that you learned isn't as important as the process of dealing with that information, i.e. proof. Logical reasoning skills are what you are supposed to take away from Geometry. Will you ever use the Hinge theorem again? Maybe, maybe not. But you had better be able to reason through a problem step by step.

It is sounding more and more like this professor is trying to do you a favor by not relying so much on specific examples. Learn the process, but don't memorize it. This is why the Calc you took before in high school did not stick.

 

[MovingTarget]

Just use a calculator. Damn y'all are stupid!

Fixed.

Calculators are dumb this I know,
Linear Algebra told me so!

 

[Gibson486]

and you want to be a CS major? Dam son, suck it up.