- #1
EastWindBreaks
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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]
EastWindBreaks said: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]
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 said: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 said: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.
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')PeroK said: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 said: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.
oh I see! thank you! but why up to x instead of b/2?FactChecker said: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)
I got this:PeroK said: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 said:oh I see! thank you! but why up to x instead of b/2?
EastWindBreaks said: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...
wow, thank you, I forgot b/2 is a definite interval on the graph.PeroK said: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.
Integration is a mathematical process that involves finding the area under a curve. It is used to solve problems related to rates of change, such as finding displacement, velocity, or acceleration.
Integration is important because it allows us to solve real-world problems that involve rates of change. It is used in various fields such as physics, engineering, economics, and more.
Integration and differentiation are inverse operations of each other. While integration is used to find the area under a curve, differentiation is used to find the rate of change of a function at a specific point.
There are several methods of integration, including the substitution method, integration by parts, trigonometric substitution, and partial fractions. Each method is used to solve different types of integrals.
The best way to improve your understanding of integration is to practice solving different types of integrals. You can also seek help from a math tutor or use online resources such as videos, tutorials, and practice problems.