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# Integration Help

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1. We have been asked to find the CDF of the continuos uniform distribution.. I have no idea how integration works..

like that does t|xa mean? [t]xa in screenshot

All help appreciated
2. (Original post by JKITFC)
We have been asked to find the CDF of the continuos uniform distribution.. I have no idea how integration works..

like that does t|xa mean? [t]xa in screenshot

All help appreciated
It's equivalent to which is the solution to

In your screen shot is a constant.
You integrate it by using the rule .
It's a definite integral so you don't have to worry about the the C as it cancels and .
3. (Original post by JKITFC)
We have been asked to find the CDF of the continuos uniform distribution.. I have no idea how integration works..

like that does t|xa mean? [t]xa in screenshot

All help appreciated
It means (plug the number at the top of the square bracket into the function in the square bracket) - (plug the number at the bottom of the square bracket into the function in the square bracket). In this case, the function is just t.

So it's just (x) - (a).
4. (Original post by Kvothe the arcane)
It's equivalent to which is the solution to

In your screen shot is a constant.
You integrate it by using the rule .
It's a definite integral so you don't have to worry about the the C as it cancels and .
That makes no sense.
5. (Original post by Zacken)
It means (plug the number at the top of the square bracket into the function in the square bracket) - (plug the number at the bottom of the square bracket into the function in the square bracket). In this case, the function is just t.

So it's just (x) - (a).
Why does it replace the 1? To 'plug it into' the function - why does it replace the numerator? Surely to replace the function is to replace the whole 1/(b-a) ?
6. (Original post by JKITFC)
Why does it replace the 1? To 'plug it into' the function - why does it replace the numerator? Surely to replace the function is to replace the whole 1/(b-a) ?
No, 1/(b-a) is a constant.

It's like saying "plug x=2 into the number 5" - doesn't make sense.
7. (Original post by Zacken)
No, 1/(b-a) is a constant.

It's like saying "plug x=2 into the number 5" - doesn't make sense.
Why does being a constant mean only the 1 is replaced with the x-a - why is the denominator left alone?
8. It helps if you think of integration like area under curve. f(x) - f(a) is the area between a and x on the curve
9. (Original post by JKITFC)
Why does being a constant mean only the 1 is replaced with the x-a - why is the denominator left alone?
You're not replacing the 1, you're replacing the t.
10. (Original post by Zacken)
You're not replacing the 1, you're replacing the t.
Right, finally getting somewhere.

Why is the b-a left alone/remain - as there is only [t]xa with no b there?
11. (Original post by rajanp1)
It helps if you think of integration like area under curve. f(x) - f(a) is the area between a and x on the curve
Thanks for the input, doesn't help me though i'm afraid. What does a curve have to do with replacing t with x-a/b-a?
12. (Original post by JKITFC)
Right, finally getting somewhere.

Why is the b-a left alone/remain - as there is only [t]xa with no b there?
You could but the b-a in there if you want, it'll make no difference.

[t/(b-a)]_a^x is still x/(b-a) - a/(b-a) = (x-a)/(b-a).
13. (Original post by Zacken)
You could but the b-a in there if you want, it'll make no difference.

[t/(b-a)]_a^x is still x/(b-a) - a/(b-a) = (x-a)/(b-a).
Think I've got it..

There is no t there? It is a one?
I need some more logic explained.
To me it still just looks like the 1 is replaced with x-a..
Someone else described it as.. when _ is x what is.. I don't think you understand the fact I have no clue what integration even is..
14. (Original post by JKITFC)
I don't think you understand the fact I have no clue what integration even is..
Perhaps you should then look it up and understand what it is, I'm afraid that I can't explain how to do this when you don't know the very basics.

Here's a good article on it: https://www.mathsisfun.com/calculus/...roduction.html
15. (Original post by Zacken)
You could but the b-a in there if you want, it'll make no difference.

[t/(b-a)]_a^x is still x/(b-a) - a/(b-a) = (x-a)/(b-a).
'Plug it into the function' is familiar..
But why is there a minus, why if the 1/(b-a) is a constat.. is only the top line replaced?
Why does (b-a) remain?

How about I ask it this way.. What is the generic thinking that goes into it.. like.. Times the numerator by x then minus the bottom part of the symbol outside of the square brackets?
16. (Original post by Zacken)
Perhaps you should then look it up and understand what it is, I'm afraid that I can't explain how to do this when you don't know the very basics.

Here's a good article on it: https://www.mathsisfun.com/calculus/...roduction.html
It is useless.
Looked through all of these - and not once is there a | for example.. the [t]xa in the screenshot
17. (Original post by JKITFC)
'Plug it into the function' is familiar..
But why is there a minus, why if the 1/(b-a) is a constat.. is only the top line replaced?
Why does (b-a) remain?

How about I ask it this way.. What is the generic thinking that goes into it.. like.. Times the numerator by x then minus the bottom part of the symbol outside of the square brackets?
Generic thinking:

Integral of a constant with respect to t is (t times that constant)

So .

When you put limits on, you do the integral evaluated at the top number minus integral evaluated at the bottom number.

So
18. (Original post by Zacken)
Generic thinking:

Integral of a constant with respect to t is (t times that constant)

So .

When you put limits on, you do the integral evaluated at the top number minus integral evaluated at the bottom number.

So
Thanks, Looks good - I'm nearly there, I understand the minus and I understand "you do the integral evaluated at the top number minus integral evaluated at the bottom number."

The thing I still don't get.. why does nothing happen to the b-a? Why is the b-a not effected by the x or a limits?
19. It's because you're integrating 1/b-a with respect to t. So you'll increase the power of t by one. in this case, you'll make t^0 into t^1. The square brackets mean you do f(top value) - f(bottom value). when you've plugged the values in you'll get x/b-a - a/b-a which is x-a/b-a
20. (Original post by JKITFC)
Thanks, Looks good - I'm nearly there, I understand the minus and I understand "you do the integral evaluated at the top number minus integral evaluated at the bottom number."

The thing I still don't get.. why does nothing happen to the b-a? Why is the b-a not effected by the x or a limits?
Because it's just a constant.

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