#integral
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Not like that, no.
What you can do is u substitution.
No try to imagine it like a function , you have a function and have area between two limits diveded by 2 but you half both the limits . Will that be true ?
Like I said, u substitution.
i tried it as you can see from the first pic . do you mean substitute a’s with u too. if so i don’t know how to.
wait
ok
a is actually x
Okay, you know how when you do a u substitution, you re-substitute x back in at the end?
No.
i dont know how to do a u substitution here
Just, look. You know that part, right?
what part
The part I just explained.
The part you just replied to.
But didn't actually respond to.
you mean like in the end you have a function like for example f (u) du and since everything is in u form you write it like f(x) dx?
...no.
I mean.
You have a function of x.
You do a u substitution.
You integrate.
And then you reverse your substitution.
Because your initial integrand was a function of x.
Therefore your final answer must be a function of x.
Do you know how to do u substitution at all?
yes
Okay. Integrate (2x + 3)^2 using u substitution.
Why is your result a function of u? You integrated a function of x.
A function of x.
I mistyped.
okay i can write it like 1/2.x3/3 +c
...okay, so differentiate that.
x2/2
Which is not (2x + 3)^2.
Yes.
where did i do it wrong
You can't just replace u with the letter x.
You have to replace u with the equivalent value in terms of x.
so 2x+3
Yes.
yes yes i get it
So here's the thing.
When you're doing a definite integral, you could do that.
Or, equivalently, you could recalculate the bounds of integration to be in terms of u instead of x.
yes i know that actually
...given that you didn't even know to plug back in the value of u after u substitution, I doubt it.
But let's look at our first integral here. u = x + 1. So then our bounds of integration in terms of u are...?
i actually knew it really… but used to always do it in definite integral and since there is only u , i converted it to x without thinking about it
...if you knew it, I don't know how you're having trouble on this problem, because it literally is just recalculating the bounds of integration in terms of u.
i dont know because the bounds does not even have x it has a
...so then you don't know how to recalculate the bounds of integration.
ok i dont know z i actually know but i guess you wont tell me how to do it unless i say “i dont know”
This is not something that anyone who actually knows what I'm talking about would ever say. This is not a confusion you would have if you knew how to do this.
okay
Look.
We're integrating from x = a - 1 to x = a + 3. If u = x + 1, then when x = a - 1, u equals what?
a
Right. And when x = a + 3, u equals what?
a+4
Right.
It's okay to not know things. We're here to teach you. You just need to admit to what you don't know so that we can teach you it.
yes of course, but i just felt so humiliated because of the brain in your profile pic and the assertiveness in the way you write
but its okay thanks
thank you so much i solved it and would never forget how to substitute thanks to you :)))