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kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits2b(1+t)dt\] How does my prof get b(1+t)^2 + C but I got 2bt +bt^2 +C

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.3\[\int\limits2b(1+t)dt \] \[\int\limits2b(1+t)dt \] \[2b\int\limits1dt +2b\int\limits td t\] bt^2+2bt+c

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0It seems like my prof might have done lik a usubstituion on (1+t) but how come it doesn't give the same answer?

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.3\[2b*t+2b*\frac{t^{2}}{2} +c\] \[2bt+bt^2+c\] it is the same..

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits2b(1+t)dt= (1+t)^2 + k\] Leave the 2b alone and put it back later since it's a constant. \[=\frac{1}{2}(1+t)^2 +k\] Put back the 2b. \[=\frac{2b}{2}(1+t)^2 +k\] \[b(1+t)^2 + k\]

sami21
 one year ago
Best ResponseYou've already chosen the best response.1your professor is right . because \[\Large \int\limits_{}^{} (ax+b)=\frac{(ax+b)^2}{a}+C\]

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0k is a constant. Just replace that k with a big C.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2 but this is not the same as 2bt+bt^2 o_O

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0am I missing something?

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0@kirbykirby You meant to differentiate it.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0You expand then differentiate.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Wait sorry you differentiate by not expanding.

Mimi_x3
 one year ago
Best ResponseYou've already chosen the best response.3differentiate?? this is integration!

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0No, he wants to get the answer back to the original expression.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0When you integrate it, you can differentiate it back to the expression that you originally integrated.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0@kirbykirby You differentiate that expanding expression you had, to get back your original expression. Integration is the reverse of differentiation.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0Ok. I'm trying to understand though how the 2 methods though give a different answer though...

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0differentiate b+2bt+bt^2 with respect to t and you get: 2b+2bt=2b(1+t)

sami21
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large \int\limits_{}^{}2b(1+t)dt=2b \frac{(1+t)^2}{2}+C=b(1+t)^2+C\] differentiate \[\Large \frac{d}{dt}(b(1+t)^2)+C=2b(1+t)\] you get original back . means it is correct .

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Which is the original expression you had in the question...

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0oh is it because the b + C gets absorbed into another constant "C" using the other method o_O?

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0isn't it a constant here with respect to t?

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0If it was respect to b, then it's a different story.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2 but this is not the same as 2bt+bt^2 So is this difference because of the b being absorbed??

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0If you want to know why there's a C at the end everytime you integrate, it's because there are many curves that can still produce the same derivative. For example. The derivative of this curve y=x^2+1 is the same as y=x^2

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0That's for you to understand why there's a constant attached everytime you itnegrate. If you don't attach a C or k whatever, you get it wrong.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0No I am not asking why there is a C constant... I am trying to understand why: Integrating it gives either "2bt+bt^2 + C" or "b(t+1)^2 + C"... which gives "b+2bt+bt^2+C" which is not the same as above

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0\[2bt+bt^2 \neq=2b(t+1)\]

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0You're meant to differentiate to get the original result. The word differentiate is the key word here.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Once you integrate, you must differentiate to get back the original, otherwise, it's going to be certain that it's not the same as the original.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0Because One method is using \[2b( \int\limits(1)dt+\int\limits(t)dt)\] = 2bt+bt^2 +C OR \[2b \int\limits(1+t)dt=2b \frac{(1+t)^2}{2}=2b(1+t)^2+C\]

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Integration is just getting the original equation of a curve or function. When you're integrating something, that "something" is the derivative.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0oops forget the extra "2" in the last expression

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0To be frank with you kirby, you shouldn't be using formulas when you're beginning to learn how to integrate.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0I am not learning how to integrate :( I know how to integrate

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0I just don't see how these 2 methods are giving different answers

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Still, these easy indefinite integrals require you to just think about differentiating, but working backwards.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0If you were to differentiate 3x, I would probably first but an indice of 2 beside x giving me x^2 and then when you integrate s^2, you get 2x. but the original equation just has an x without the two. SO then you put a half in front of the x^2. That gives you (1/2)x^2. and then once you put back the 3, you get (3/2)x^2.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0That's how you should do it. If you have that mindset, nothing can stop you from getting good marks in integration.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0I just want to be clear... I know how to differentiate... Factoring 2b and integrating "1" and "t" individually gives "2bt+bt^2+C" But if you factor 2b and integrate (1+t) using a substitution, you get "integral 2bu" = 2bu^2/2 +C= bu^2+C => b(1+t)^2+C

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Why would you do it that way. You rather keep integration as simple as possible. Don't go into that mucky stuff. You will end up confusing yourself in the end.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0I know it is stupid to do that, but shouldn't the answers AGREE? I mean what is the point if the two methods don't agree???

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Then use one method. Simple as that.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0But why does one not give the same answer as the other :( THIS is what I wanna understand

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0No point in pondering on two methods. I'm not using any of the two but integrating through differentiation.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0If you really want to know, then go to your professor and ask him why.

Azteck
 one year ago
Best ResponseYou've already chosen the best response.0Your professor's there to help you understand, not bore you to death.

kirbykirby
 one year ago
Best ResponseYou've already chosen the best response.0is it because the extra "b" in the second method gets absorbed into the constant C?? (which is why they give the same derivative)... but I never saw an answer like "2x+2+C" being absorbed as 2x+C
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