Integration (reduction) (1 Viewer)

synthesisFR · Dec 20, 2022

Can someone help out with part d pls?
I have the solutions but idk what is happening and I want to know the thought process for solving it.

cossine · Dec 21, 2022

synthesisFR said:
Can someone help out with part d pls?
I have the solutions but idk what is happening and I want to know the thought process for solving it.

Provide the working out and the part you are stuck on.

synthesisFR · Dec 21, 2022

Its not very clear so i would prefer if anyone could do the question itself

synthesisFR · Dec 22, 2022

Average Boreduser · Dec 22, 2022

@Drongoski ?

5uckerberg · Dec 22, 2022

Q6a)
$\int{ln\left(1+x\right)}\;dx=x\;ln(1+x)-\int{\frac{x}{1+x}}\;dx\;\left(integration\;by\;parts\right)$
$\int{\frac{x}{1+x}}\;dx=\int{1-\frac{1}{1+x}}\;dx\;\left(simplifying\;the\;fraction\right)$
$\int{ln\left(1+x\right)}\;dx=x\;ln(1+x)-\left(\int{1-\frac{1}{1+x}}\;dx\right)$
$\int{ln\left(1+x\right)}\;dx=x\;ln(1+x)-\left(x-ln\left(1+x\right)\right)$
$\int{ln\left(1+x\right)}\;dx=x\;ln(1+x)-x+ln\left(1+x\right)$
$\int{ln\left(1+x\right)}\;dx=\left(1+x\right)\;ln(1+x)-x+C$
b)
$I_{n}=\int_{0}^{1}x^{n}ln\left(1+x\right)\;dx$
using part a
$I_{n}=\int_{0}^{1}x^{n}ln\left(1+x\right)\;dx=[x^{n}\left(\left(1+x\right)\;ln(1+x)-x\right)]_{0}^{1}-\left(\int_{0}^{1}nx^{n-1}\left(1+x\right)ln\left(1+x\right)\;dx-\int_{0}^{1}nx^{n}dx\right)$
$[x^{n}\left(\left(1+x\right)\;ln(1+x)-x\right)]_{0}^{1}-\left(\int_{0}^{1}nx^{n-1}\left(1+x\right)ln\left(1+x\right)\;dx-\int_{0}^{1}nx^{n}dx\right)=2ln2-1-n\int_{0}^{1}x^{n-1}ln\left(1+x\right)\;dx-n\int_{0}^{1}x^{n}ln\left(1+x\right)dx+n\int_{0}^{1}x^{n}dx$
$I_{n}=2ln2-1-n\int_{0}^{1}x^{n-1}ln\left(1+x\right)\;dx-n\int_{0}^{1}x^{n}ln\left(1+x\right)dx+n\int_{0}^{1}x^{n}dx=2ln2-1-nI_{n-1}-nI_{n}+\frac{n}{n+1}=2ln2-1-nl_{n-1}-nl_{n}+1-\frac{1}{1+n}$
$I_{n}=2ln2-nI_{n-1}-nI_{n}-\frac{1}{1+n}$
$\left(n+1\right)I_{n}=2ln2-nI_{n-1}-\frac{1}{1+n},\;n\in\;\mathbb{N}$

5uckerberg · Dec 22, 2022

c)
This is the important part students, please watch carefully.
We start with $I_{0}=\int_{0}^{1}ln\left(1+x\right)dx=[\left(1+x\right)ln\left(1+x\right)-x]_{0}^{1}=2ln2-1$
Now, onto the build-up
$2I_{1}=\frac{1}{1}-\frac{1}{2}\;\;\;\;\;2I_{1}=2ln2-\frac{1}{2}-I_{0}=2ln2-\frac{1}{2}-\left(2ln2-1\right)=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)\;\;\;\;\;3I_{2}=2ln2-\frac{1}{3}-2I_{1}=2ln2-\frac{1}{3}-\left(\frac{1}{1}-\frac{1}{2}\right)=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\;\;\;\;\;2I_{1}=2ln2-\frac{1}{4}-3I_{2}=2ln2-\frac{1}{4}-\left(2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)\right)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .
To sum it up.
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .

d)
For reference

$I_{0}=2ln2-\left(\frac{1}{1}\right)$
$2I_{1}=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .

Now students, sit back and watch the pattern unfold.
Continuing with the pattern
$I_{0}=2ln2-\left(\frac{1}{1}\right)$
$2I_{1}=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$
$5I_{4}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)$
$6I_{5}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}$
$7I_{6}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}\right)$
$8I_{7}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}$

Verify that in your working and then solve the problem. What do you notice?

synthesisFR · Dec 22, 2022

5uckerberg said:
c)
This is the important part students, please watch carefully.
We start with $I_{0}=\int_{0}^{1}ln\left(1+x\right)dx=[\left(1+x\right)ln\left(1+x\right)-x]_{0}^{1}=2ln2-1$
Now, onto the build-up
$2I_{1}=\frac{1}{1}-\frac{1}{2}\;\;\;\;\;2I_{1}=2ln2-\frac{1}{2}-I_{0}=2ln2-\frac{1}{2}-\left(2ln2-1\right)=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)\;\;\;\;\;3I_{2}=2ln2-\frac{1}{3}-2I_{1}=2ln2-\frac{1}{3}-\left(\frac{1}{1}-\frac{1}{2}\right)=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\;\;\;\;\;2I_{1}=2ln2-\frac{1}{4}-3I_{2}=2ln2-\frac{1}{4}-\left(2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)\right)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .
To sum it up.
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .

d)
For reference

$I_{0}=2ln2-\left(\frac{1}{1}\right)$
$2I_{1}=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .

Now students, sit back and watch the pattern unfold.
Continuing with the pattern
$I_{0}=2ln2-\left(\frac{1}{1}\right)$
$2I_{1}=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$
$5I_{4}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)$
$6I_{5}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}$
$7I_{6}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}\right)$
$8I_{7}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}$

Verify that in your working and then solve the problem. What do you notice?

wait how do u write equations here

5uckerberg · Dec 22, 2022

Press the f(x) button for LaTeX and then something_{...} for superscript, \left( for left bracket, \right) for right bracket like this \left(something\right). \frac{...}{...} for fraction, there might be others but you have to search up a lot of the LaTeX features when you first started. Once you start using it then communicating mathematics will be very interesting. The first step is to learn how to do so.

synthesisFR · Dec 22, 2022

5uckerberg said:
Verify that in your working and then solve the problem. What do you notice?

For this part, do I just state the pattern and make and extend the pattern upto the "nth term"?

Also in the solutions they did it differently and it was weird..
Like they took the formula from part b)
then they did this: lemme just write it out:
not sure where the multiplication by (-1)^n-1 or adding the equations came from
(the since odd, since even arrows are from the series for the ln2 above)

5uckerberg · Dec 22, 2022

synthesisFR said:
For this part, do I just state the pattern and make and extend the pattern upto the "nth term"?

Cuz in the solutions they did it differently and it was weird..
Like they took the formula from part b)
then they did this: lemme just write it out:
not sure where the multiplication by (-1)^n-1 or adding the equations came from

"For this part, do I just state the pattern and make and extend the pattern upto the "nth term"?"
I mean you can do that but why waste time on that pattern when you can clearly see what is happening.

synthesisFR · Dec 22, 2022

Gotcha,
Your method is way easier than however the solutions did it!

Thanks for the help!!!

synthesisFR · Jan 15, 2023

5uckerberg said:
c)
This is the important part students, please watch carefully.
We start with $I_{0}=\int_{0}^{1}ln\left(1+x\right)dx=[\left(1+x\right)ln\left(1+x\right)-x]_{0}^{1}=2ln2-1$
Now, onto the build-up
$2I_{1}=\frac{1}{1}-\frac{1}{2}\;\;\;\;\;2I_{1}=2ln2-\frac{1}{2}-I_{0}=2ln2-\frac{1}{2}-\left(2ln2-1\right)=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)\;\;\;\;\;3I_{2}=2ln2-\frac{1}{3}-2I_{1}=2ln2-\frac{1}{3}-\left(\frac{1}{1}-\frac{1}{2}\right)=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\;\;\;\;\;2I_{1}=2ln2-\frac{1}{4}-3I_{2}=2ln2-\frac{1}{4}-\left(2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)\right)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .
To sum it up.
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .

d)
For reference

$I_{0}=2ln2-\left(\frac{1}{1}\right)$
$2I_{1}=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$ .

Now students, sit back and watch the pattern unfold.
Continuing with the pattern
$I_{0}=2ln2-\left(\frac{1}{1}\right)$
$2I_{1}=\frac{1}{1}-\frac{1}{2}$
$3I_{2}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}\right)$
$4I_{3}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$
$5I_{4}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)$
$6I_{5}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}$
$7I_{6}=2ln2-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}\right)$
$8I_{7}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}$

Verify that in your working and then solve the problem. What do you notice?

Yo so I was told that u can't just sub in the terms to get it but u have to actually go prove it full on...

Average Boreduser · Jan 15, 2023

Bro remembered the whole ass question since dec20

synthesisFR · Jan 15, 2023

Pethmin said:
Bro remembered the whole ass question since dec20

I can't move on

synthesisFR · Sep 12, 2023

synthesisFR said:
Can someone help out with part d pls?
I have the solutions but idk what is happening and I want to know the thought process for solving it.

its been 9 months and no one has answer this question
random couples have gotten married, and are probably giving birth rn and no one had the time to help me out on this one @carrotsss

carrotsss · Sep 12, 2023

synthesisFR said:
its been 9 months and no one has answer this question
random couples have gotten married, and are probably giving birth rn and no one had the time to help me out on this one @carrotsss

this was before I learnt integration ok I answer all ur questions nowadays

scaryshark09 · Sep 12, 2023

synthesisFR said:
its been 9 months and no one has answer this question
random couples have gotten married, and are probably giving birth rn and no one had the time to help me out on this one @carrotsss

With great regret, I inform you that part d is really simple.
If you are struggling with it, I suggest that you consider dropping to ext 1 or even adv.

synthesisFR · Sep 12, 2023

scaryshark09 said:
With great regret, I inform you that part d is really simple.
If you are struggling with it, I suggest that you consider dropping to ext 1 or even adv.

is it actually i should probably try the question d now since its been 9 months

scaryshark09 · Sep 12, 2023

synthesisFR said:
is it actually i should probably try the question d now since its been 9 months

lol i have no idea, i havn't tried.
i probs wouldnt be able to do it at first glance

Integration (reduction) (1 Viewer)

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