Question

Integral 2b(1+t) dt

Answer 1

\[\int\limits2b(1+t)dt\]
How does my prof get b(1+t)^2 + C but I got 2bt +bt^2 +C

Answer 2

\[\int\limits2b(1+t)dt \]
\[\int\limits2b(1+t)dt \]

\[2b\int\limits1dt +2b\int\limits td t\]
bt^2+2bt+c

Answer 3

It seems like my prof might have done lik a usubstituion on (1+t) but how come it doesn't give the same answer?

Answer 4

\[2b*t+2b*\frac{t^{2}}{2} +c\]

\[2bt+bt^2+c\]
it is the same..

Answer 5

\[\int\limits2b(1+t)dt= (1+t)^2 + k\]
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.
\[=\frac{2b}{2}(1+t)^2 +k\]
\[b(1+t)^2 + k\]

Answer 6

your professor is right .
because
\[\Large \int\limits_{}^{} (ax+b)=\frac{(ax+b)^2}{a}+C\]

Answer 7

its 2a !

Answer 8

k is a constant. Just replace that k with a big C.

Answer 9

b(1+t)^2 = b(1+2t+t^2)=b+2bt+bt^2
but this is not the same as 2bt+bt^2 o_O

Answer 10

but im still right!

Answer 11

am I missing something?

Answer 12

yes you are !

Answer 13

@kirbykirby You meant to differentiate it.

Answer 14

You expand then differentiate.

Answer 15

Wait sorry you differentiate by not expanding.

Answer 16

differentiate?? this is integration!

Answer 17

No, he wants to get the answer back to the original expression.

Answer 18

When you integrate it, you can differentiate it back to the expression that you originally integrated.

Answer 19

@kirbykirby You differentiate that expanding expression you had, to get back your original expression. Integration is the reverse of differentiation.

Answer 20

Ok. I'm trying to understand though how the 2 methods though give a different answer though...

Answer 21

differentiate b+2bt+bt^2 with respect to t and you get:
2b+2bt=2b(1+t)

Answer 22

\[\Large \int\limits_{}^{}2b(1+t)dt=2b \frac{(1+t)^2}{2}+C=b(1+t)^2+C\]
differentiate
\[\Large \frac{d}{dt}(b(1+t)^2)+C=2b(1+t)\] you get original back . means it is correct .

Answer 23

Which is the original expression you had in the question...

Answer 24

oh is it because the b + C gets absorbed into another constant "C" using the other method o_O?

Answer 25

isn't it a constant here with respect to t?

Answer 26

Yes.

Answer 27

If it was respect to b, then it's a different story.

Answer 28

b(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??

Answer 29

If 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

Answer 30

That'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.

Answer 31

integrate*

Answer 32

No 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

Answer 33

\[2bt+bt^2 \neq=2b(t+1)\]

Answer 34

You're meant to differentiate to get the original result. The word differentiate is the key word here.

Answer 35

Once 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.

Answer 36

Because One method is using \[2b( \int\limits(1)dt+\int\limits(t)dt)\] = 2bt+bt^2 +C
OR
\[2b \int\limits(1+t)dt=2b \frac{(1+t)^2}{2}=2b(1+t)^2+C\]

Answer 37

Integration is just getting the original equation of a curve or function. When you're integrating something, that "something" is the derivative.

Answer 38

oops forget the extra "2" in the last expression

Answer 39

To be frank with you kirby, you shouldn't be using formulas when you're beginning to learn how to integrate.

Answer 40

I am not learning how to integrate :( I know how to integrate

Answer 41

I just don't see how these 2 methods are giving different answers

Answer 42

Still, these easy indefinite integrals require you to just think about differentiating, but working backwards.

Answer 43

If 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.

Answer 44

first put*

Answer 45

x^2*

Answer 46

That's how you should do it. If you have that mindset, nothing can stop you from getting good marks in integration.

Answer 47

I 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

Answer 48

Why 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.

Answer 49

I 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???

Answer 50

Then use one method. Simple as that.

Answer 51

But why does one not give the same answer as the other :( THIS is what I wanna understand

Answer 52

No point in pondering on two methods. I'm not using any of the two but integrating through differentiation.

Answer 53

If you really want to know, then go to your professor and ask him why.

Answer 54

Your professor's there to help you understand, not bore you to death.

Answer 55

*ask him/her

Answer 56

is 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