tired glacier
Just plug in the definition of the Riemann sum for left-rectangular regular partitions (i.e. Riemann not Lebesgue):
Integral over [a,b] f = lim n to inf of sum from i = 1 to n of delta x * f(a+delta x i) where delta x = (b-a)/n.
Clearly, one is correct. So is two, because it is merely a shift. So is 4.
polar cobalt
This is what I got for the right and left riemann summs
tired glacier
I don’t know if they want you to look for definition of Riemann sum or merely what will compute the Riemann sum
The question seems vague to someone who does any form of analysis
tired glacier
Option 1 and 2 r correct. Do u agree?
I will trust you
Yes
Option 1,2 give standard Riemann sum
polar cobalt
Why did u think option 4 is correct?
tired glacier
Why did u think option 4 is correct?
Look at option four. 2^(3(i-1)/n) = 8^((i-1)/n)
Four also is correct
No matter what
If you choose 1,2, you must also choose 4
It’s just a manipulation of the original sum
Replacing 2^(3…) with 8^(…)
tired glacier
Yes, 1,2,4
polar cobalt
Because it's simply the left riemann sum right?
tired glacier
Because it's simply the left riemann sum right?
Yes
tired glacier
Yes
I will explain why these are true or false
Statement one is true.
Nvm it’s wrong
Wrong wrong wrong
See that if we choose g(t) = h(t)*0.99, assuming g(t) and h(t) strictly positive
And f(t) = 0.99g(t)
We have 0.99h(t) + 0.99^2h(t) integral from a to b is less than h(t)
Since they are positive this is clearly a contradiction
For part two, merely consider f(t) = 0.
It is clearly wron
polar cobalt
Yeah cuz 0 < 0 is what we would end up with which is wrong.
tired glacier
For part 3, consider just a function that is increasing but very negative
So say f(x) = -a + b/a*x
Just integrate that
You’ll realize that this integral is negative
But clearly f’(t) = b/a
If we chose b,a > 0, we are good as a counter example
For 4, just apply FTA
Consider F(x)
F(b) - F(a) = -|F(a) - F(b)|
Yes.
4 is true I believe
Only 4
polar cobalt
Hmm ok
For 1.
If we split the integral
Since f(t) < g(t), it follows that ∫[a to b] f(t) dt < ∫[a to b] g(t) dt.
also because
g(t) < h(t)
we have ∫[a to b] g(t) dt < ∫[a to b] h(t) dt.
then if we combine the inequalities
we will end up with ∫[a to b] f(t) dt + ∫[a to b] g(t) dt < ∫[a to b] g(t) dt + ∫[a to b] h(t) dt
which would just become this:
∫[a to b] (f(t) + g(t)) dt < ∫[a to b] h(t) dt
So isn't 1. true or is my logic wrong?
tired glacier
∫[a to b] (f(t) + g(t)) dt < ∫[a to b] h(t) dt
No you deleted g(t)
int f(t) + g(t) < int h(t) + g(t)
polar cobalt
ah
tired glacier
If there was a 2 as a coefficient on the right hand side it would be true
polar cobalt
but there is no way we can get a 2 right?
tired glacier
No, we can’t here, because g < h
Here is a good example to consider.
Consider h(t) = t
g(t) = 0.99h
f(t) = 0.99g
We will have f(t) + g(t) = 0.99^2 h(t) + 0.99 h(t)
Thus, we have now
Int a^b 0.99^2*t + 0.99t < int a^b t
Which is wrong for a,b > 0 and a > b
tired glacier
C is false because we can just choose a negative value to start for f(t)
tired glacier
So 1. would be wrong then
Yes
polar cobalt
C is false because we can just choose a negative value to start for f(t)
but for c f'(t) has to be > 0
tired glacier
Yes
f(a) is negative
But f(t) is increasing
So like -100000000 + x say
This has derivative 1
Everywhere
tired glacier
Yes
tired glacier
f’(t) gives the shape of f(x), not so much it’s location on the y-axis
tired glacier
but how would the integral be negative?
Because the value is negative the entire time
It’s just increasing
Take integral of -a + b/a x from a to b
You’ll find the integral is always negative for positive a,b
tired glacier
Yes
tired glacier
option 4 feels like it's wrong lol
It’s not
Just use FTC
It’s a trick they are pulling
Hmm let me check though
F(a) - F(b) = -|F(b) - F(a)|
F(a) - F(b) = -|-(F(a) - F(b))|
Hmm yeah
It’s wrong
Those silly bastards
@polar cobalt absolute value fucks it up
tired glacier
Let’s see
Choose easier one
Just choose a constant
f(t) = 1
a = 1
b = 2
int a to b of f = 1
polar cobalt
f(t) = 1 is a continuous function right?
tired glacier
-|int b to a of f| = -|1-2| = -|-1| = -1
tired glacier
f(t) = 1 is a continuous function right?
Yes
polar cobalt
yes cuz with ur exampl
the integral from 1to2
is just 1
lol
and 1 does not equal negative 1
which is the result of the right side of the equation
tired glacier
Yes
polar cobalt
wow so answer is none of the above
tired glacier
Yes
polar cobalt
I just realized for 1.
we could have taken
f = 1
g = 3/2
h = 2
right? to use as counter example
tired glacier
Yes
polar cobalt
wow
tired glacier
You see? It only tells you the shape not the location
tired glacier
Yes
polar cobalt
wow
I had one last question
for the first question
how is option 4 the same as option 2?
i'm not seeing the algebra for some reason
tired glacier
Oh
First remove the constant outside the sum
Then convert $8^{\frac{i-1}{n}}$ to $2^{3\frac{i-1}{n}}$.
obtuse loomBOT
Magma <3 (SL_2(R) Group)
tired glacier
You should see the rest
tired glacier
Yes
polar cobalt
and question 2 is none of the above
polar cobalt
You taught me more than my prof ever taught in his life 😂
tired glacier
lol let’s hope
polar cobalt
Thank you so much. I really appreciate it.
polar cobalt
I will submit it now lol
tired glacier
Ok
polar cobalt
I will let u know of the results
tired glacier
Alr
tired glacier
Aight excellent
tired glacier
we went through so many counter examples lol
Yes lol
polar cobalt
Thanks again. Have a great day!