## kirbykirby 2 years ago Integral 2b(1+t) dt

1. kirbykirby

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

2. Mimi_x3

$\int\limits2b(1+t)dt$ $\int\limits2b(1+t)dt$ $2b\int\limits1dt +2b\int\limits td t$ bt^2+2bt+c

3. kirbykirby

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

4. Mimi_x3

$2b*t+2b*\frac{t^{2}}{2} +c$ $2bt+bt^2+c$ it is the same..

5. Azteck

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

6. sami-21

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

7. sami-21

its 2a !

8. Azteck

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

9. kirbykirby

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

10. Mimi_x3

but im still right!

11. kirbykirby

am I missing something?

12. sami-21

yes you are !

13. Azteck

@kirbykirby You meant to differentiate it.

14. Azteck

You expand then differentiate.

15. Azteck

Wait sorry you differentiate by not expanding.

16. Mimi_x3

differentiate?? this is integration!

17. Azteck

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

18. Azteck

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

19. Azteck

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

20. kirbykirby

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

21. Azteck

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

22. sami-21

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

23. Azteck

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

24. kirbykirby

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

25. kirbykirby

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

26. Azteck

Yes.

27. Azteck

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

28. kirbykirby

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??

29. Azteck

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

30. Azteck

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.

31. Azteck

integrate*

32. kirbykirby

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

33. Azteck

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

34. Azteck

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

35. Azteck

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.

36. kirbykirby

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

37. Azteck

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

38. kirbykirby

oops forget the extra "2" in the last expression

39. Azteck

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

40. kirbykirby

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

41. kirbykirby

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

42. Azteck

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

43. Azteck

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.

44. Azteck

first put*

45. Azteck

x^2*

46. Azteck

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

47. kirbykirby

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

48. Azteck

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.

49. kirbykirby

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???

50. Azteck

Then use one method. Simple as that.

51. kirbykirby

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

52. Azteck

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

53. Azteck

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

54. Azteck

55. Azteck