Learning Integration with unit step function like u(x - a)

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hello maths experts
is the following true?
http://img9.imageshack.us/img9/4596/int15oe.jpg [Broken]

graphically, this is how i view it
http://img9.imageshack.us/img9/179/int28ut.jpg [Broken]
 
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Answers and Replies

LeonhardEuler
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Yes, that's pretty much correct, but the right hand side is missing a "+C" because it is an indefinite integral.
 
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how about this?

http://img108.imageshack.us/img108/1626/38jk1.jpg [Broken]
is u(x-a) redundant? can i remove it like this?
 
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LeonhardEuler
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hanhao said:
how about this?

http://img108.imageshack.us/img108/1626/38jk1.jpg [Broken]
is u(x-a) redundant? can i remove it like this?
No, the integral is constant for x<a. The u(x-a) keeps the part of the integral that is dependant on x zero for x<a, so it is just the constant of integration before that. The integral should be u(x-a)[F(x)-F(a)]+C.
 
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i want to understand this graphically
how would the graph of u(x-a)[F(x)-F(a)] look like compared to [F(x)-F(a)] ??
am i correct to say that my bottom graph is [F(x)-F(a)] ??
 
LeonhardEuler
Gold Member
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I am assuming you mean the bottom graph in this image, so tell me if I am wrong:
http://img9.imageshack.us/img9/179/int28ut.jpg [Broken]
This is not the graph of [F(x)-F(a)]. The function itself is f(x)u(x-a). The area represents the integral of this, which is u(x)[F(b)-F(a)], where b is the upper limit.

[F(x)-F(a)] represents an antiderivative of f(x) without the step function. Suppose b and c are both less than a. Obviously the integral,I, of f(x)u(x-a) from b to c is zero, but look what happens when you plug this in to the function you proposed:
I=[F(c)-F(a)]-[F(b)-F(a)]=F(c)-F(b)
which is not necessarily zero.
 
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We have limits between infinite to minus infinite how can i compute when i multiply a function with an unit step function. i mean i have an integral the limits of that integral is infinite to minus infinite and inside the integral i have f(t).u(t-a) this. So how can i compute this integral ?
 

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