I dont understand this integration

[EastWindBreaks]

Homework Statement

in the textbook, it says:

how does the integral of y'' gives you y' like that?

Homework Equations

given C_a is a dimensionless peak acceleration factor.

The Attempt at a Solution

[/B]

 

[PeroK]

Homework Statement

in the textbook, it says:View attachment 215838
how does the integral of y'' gives you y' like that?

Homework Equations

View attachment 215839
given C_a is a dimensionless peak acceleration factor.

The Attempt at a Solution

View attachment 215840[/B]

You seem to be confusing definite and indefinite integrals. The indefinite integral is an antiderivative of the function, but a definite integral is a number.

 

[EastWindBreaks]

You seem to be confusing definite and indefinite integrals. The indefinite integral is an antiderivative of the function, but a definite integral is a number.

but if I do it as indefinite integral, the result would be more confusing to me: y'= -C*(b/π)*cos(πx/b), which is totally different from what the book has.

 

[PeroK]

but if I do it as indefinite integral, the result would be more confusing to me: y'= -C*(b/π)*cos(πx/b), which is totally different from what the book has.

I'm not sure what the problem is. If you differentiate $y'$ you get $y'' = C\sin(\pi x/b)$.

And if you take the indefinite integral of that you get $y'$, up to an the arbitrary constant of integration.

 

[EastWindBreaks]

I'm not sure what the problem is. If you differentiate $y'$ you get $y'' = C\sin(\pi x/b)$.

And if you take the indefinite integral of that you get $y'$, up to an the arbitrary constant of integration.

the question is to integrate the given y'' (the acceleration function) to get the velocity function y'. (assuming that's how the book get y')
its not an actual problem or anything, I am just reading the textbook and I don't understand how they got the velocity function.

 

[FactChecker]

Your integral is the definite integral up to x=b/2 and it looks like their integral is the integral up to a variable x (0≤x≤b/2)

 

[PeroK]

the question is to integrate the given y'' (the acceleration function) to get the velocity function y'. (assuming that's how the book get y')
its not an actual problem or anything, I am just reading the textbook and I don't understand how they got the velocity function.

So, what do you get if you integrate $y''$? Remember you need a boundary condition of some sort to quantify the constant of integration.

 

[EastWindBreaks]

Your integral is the definite integral up to x=b/2 and it looks like their integral is the integral up to a variable x (0≤x≤b/2)

oh I see! thank you! but why up to x instead of b/2?

 

[EastWindBreaks]

So, what do you get if you integrate $y''$? Remember you need a boundary condition of some sort to quantify the constant of integration.

I got this:

but it seems like I should be integrating up to x instead of b/2, I don't know why though, I must have forgot something important in integrals...

 

[PeroK]

oh I see! thank you! but why up to x instead of b/2?

If you integrate a function on a definite interval, you simply get a number. Take a simple example like:

$\int_0^{\pi/2} \cos(x) dx = 1$

But:

$\int_0^{x} \cos(u) du = \sin(x)$

Note the need for a new dummy variable for the integrand.

Altertnatively:

$\int \cos(x) dx = \sin(x) + C$

Where $C$ is specified by some boundary condition.

 

[PeroK]

I got this: View attachment 215841
but it seems like I should be integrating up to x instead of b/2, I don't know why though, I must have forgot something important in integrals...

Yeah, you forgot that a definite integral is a number.

 

[EastWindBreaks]

If you integrate a function on a definite interval, you simply get a number. Take a simple example like:

$y''(x) = \cos(x)$

$\int_0^{\pi/2} \cos(x) dx = 1$

But:

$\int_0^{x} \cos(u) du = \sin(x)$

Note the need for a new dummy variable for the integrand.

Altertnatively:

$\int \cos(x) dx = \sin(x) + C$

Where $C$ is specified by some boundary condition.

wow, thank you, I forgot b/2 is a definite interval on the graph.

 

[FactChecker]

Right. And if they will do more integrals of this result, they need the formula for the general integral, not just the value of the definite integral. So this is not a definite integral.