Integration Help

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

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

All help appreciated

It's equivalent to $F(x)-F(a)$ which is the solution to $\displaystyle \int_a^x f(t)dt$

In your screen shot $\dfrac{1}{b-a}$ is a constant.
You integrate it by using the rule $\displaystyle \int t^n dt=\dfrac{t^{n+1}}{n+1}+\mathcal{C}$.
It's a definite integral so you don't have to worry about the the C as it cancels and $n=0$.

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) ?

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?

You're not replacing the 1, you're replacing the t.

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

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).

I don't think you understand the fact I have no clue what integration even is..

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).

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/integration-introduction.html

'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 $\int \frac{1}{b-a} \, \mathrm{d}t = \frac{1}{b-a} \times t$.

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

So $\bigg[\frac{t}{b-a}\bigg]_a^{x} = \frac{x}{a-b} - \frac{a}{a-b}$

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?