Integral Question

[AgaBoogaBoo]

We have to evaluate the integral of the function (2t-1)^2 dt from 0 to 1. When I'm trying to solve this, how will it differ compared to finding it for the function (2t-1)dt?

I'm basically asking how we deal with it being squared compared to when it isn't if that makes sense.

 

[UncleWai]

foil it out and int it.

 

[AgaBoogaBoo]

Originally posted by: UncleWai
foil it out and int it.

wow, why didn't I think of that? :Q

 

[Random Variable]

or make a substitution

 

[chuckywang]

Let y=2t-1 and note dy = 2dt. Also don't forget to change your limits of integration.

 

[Syringer]

Both of you guys are winners.

 

[UncleWai]

Originally posted by: Random Variable
or make a substitution

I thought about that, but you can't right with integrals?
2t-1 will become t^2-t. There's no U sub in here.

 

[Syringer]

Originally posted by: UncleWai

Originally posted by: Random Variable
or make a substitution

I thought about that, but you can't right with integrals?
2t-1 will become t^2-t. There's no U sub in here.

u = 2t -1
du = 2

Though I just tried both methods and got different answers 😕

 

[Syringer]

(2t-1)^2 = 4t^2 - 4t + 1

Integral of that is (4/3)t^3 - 2t^2 + t

And if u = 2t - 1, du = 2

Integral = u^3/3, multiplied by 1/2

= (2t-1)^3/6 = (4/3)t^3 - 2t^2 + t - 1/6

Zah?

I know you're supposed to add a constant to the end of each term, but why does the -1/6 appear?

 

[Random Variable]

Originally posted by: Syringer

Originally posted by: UncleWai

Originally posted by: Random Variable
or make a substitution

I thought about that, but you can't right with integrals?
2t-1 will become t^2-t. There's no U sub in here.

u = 2t -1
du = 2dt

Though I just tried both methods and got different answers 😕

 

[Random Variable]

Originally posted by: Syringer
(2t-1)^2 = 4t^2 - 4t + 1

Integral of that is (4/3)t^3 - 2t^2 + t

And if u = 2t - 1, du = 2

Integral = u^3/3, multiplied by 1/2

= (2t-1)^3/6 = (4/3)t^3 - 2t^2 + t - 1/6

Zah?

I know you're supposed to add a constant to the end of each term, but why does the -1/6 appear?

It's just part of the constant. When you evaluate the definite integral, the answer is the same in both cases.