Integral of 1/x
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- Опубликовано: 19 дек 2020
- A quick afternoon integral, ep4.
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Хобби
I feel like the significance of the natural log and e isn't all that clear to a lot of people.
The derivative of e^x is itself
Tbh really wasn't until this video
Whatever base you choose for an exponential function, its derivative will always contain natural log, so it really is "natural"
Why is that?
Its a derivation. Derivative of a^x is ln(a) a^x. You can find it by using algebra which is probably not easy but whatever. If you are asking it in a deep meaning like why pi is irrational Im sorry I dont have an answer, neither i am not sure there is a sensible answer.
"That's it!"
Dont care about the math...just look at his skill to write with two marker in one hand!!!
I noticed that ,and yes I do write with two pen in hand 🥲(blank/blue) because over again you don't need to change your pens 😄👏👏..
🤦🙎
I really appreciate this explanation.
Short but clear explanation! Well done! 🤗
Creative Math with Ching-Hui
I love this proof. My question however is why x = e^t and not any other value.
It actually does work for any exponential! For example, x = 2^t, dx=ln2(2^t) which would give ln2(2^t)/2^t dt
which gives ln(2)dt = ln(2)t = ln(2)*lnx/ln(2)
ln(2)'s cancel out to get lnx + C. I guess e^t is the easiest because no extra algebra but not necessary.
@@jessetrevena4338why does it have to be expontential? Why not anything else
@@ezxd5192 because e^x is the only function with its derivative equal to the initial function
@@ezxd5192Because it's the only method that works
@@ezxd5192the form e^t can express any value because t isn’t limited
Great explanation!!
Love love those shorts the original blackpenredpen I wasn't able to watch all the videos were 1 h
You can do it in the reverse direction by applying the inverse derivative theorem. To prove that:
Let f^-1(x) = g(x)
x = f(f^-1(x)) = f(g(x))
1 = f’(g(x))*g’(x)
g’(x) = 1/f’(g(x))
Now, for the real thing:
Let g(x)=lnx. Then, f(x)=e^x and f’(x)=e^x.
g’(x) = 1/e^(lnx)
g’(x) = 1/x
Therefore, the derivative of ln(x) is 1/x.
You ROCK dude!
I was looking for this explanation, thanks
Isn't that the same as implicit differentiation of lnx
I have no Idea at all but still watched till the end
Your are genius sir
Nice explanation
beautiful thank you ... your the best.
You could do it by Applying limit n -> -1 (integral(x^n))
Exponential substitution
Thanks from India
Smooth.
How do you know that you have to substitute e^t though
This is not a "find the answer", but a proof that it is true. The difference is significant, but not emphasized in the presentation.
You guess, making use of the fact that d/dx e^x = e^x.
Or more reasonably, e^x crops up in lots of places in mathematic and thus it's a fair bet that substituting it for something will help.
Thanks 👍
Finally learned why it was ln x 😅
Ur genius.
Its because diffrentiation of lnx is 1/x
Please make video on [Forcing By Part Integration].....
哈哈，可以直接求解1/udu的积分，就是lnx+c
sorry, why x = e^t? 🙏
Amazing
Can someone answer me why is "dx = e^t dt", not "dx = de^t" ?
It is equal, we have dx = d(e^t), but we also have : if a function f has a derivative f', then d(f(t)) = f'(t)dt (by definition df(t)/dt = f'(t)). And the derivative of the exponential is itself, so d(e^t) = e^t dt.
All of this is not very rigorous but its fine as long as you know what you are doing, i suggest you check the rigorous formula for a change of variables in an integral.
@@romaing.1510 oh I see, thank you
Couldnt we substitute x with something like a^t instead, giving the integral of log_a(x)?
This equation is very, very, very easy
How do you know the derivative of e^x without knowing derivative of lnx
If I substitute x with 2^t won't I get logbase2?
Nice proof. I have a question though. In this method, we have set x = e^t, meaning x will never be a negative quantity. Then why is a |x| necessary?
Because 1/x has values for x
Hey can you please tell integration of 1/ x-y it will be really helpful to me
Thank you
Love you ❤️
Its just ln(x-y)+C.Let x-y be u,du will simply be 1 dx,so du=dx,so we get integral of 1/u du witch is ln|u|,with is ln(x-y)+C
It's funny, at my calculs lessons we defined ln(x) to be the definite integral from 1 to x of 1/x, and then exp(x) to be its inverse. After this, we defined general exponentiation (what a^b means when b belongs to the real numbers ; a^b = exp[b*ln(a)]) and then saw that exp(x) = e^x.
I am curious, how do you define general exponentiatiation if ln(x) is not defined to be an antiderivative of 1/x ? :p
There are many ways, I'll write two :
1) e^x=sum of x^k/k! where k goes from 0 to infinity
2) e^x is the only function verifying f'=f and f(0)=1
@@SimsHacks Ah yeah, I was too tunnel-visionned haha. Thanks :D
Exponention can be defined as the limit of sequences corresponding to those real numbers exponentiated
You can also define exp(x) = lim (1+x/n)^n, n→∞. But I'd also go with the (formal) power series definition. It gives f'=f and relations to the trig functions etc etc "for free".
Evaluate ∫(1/dx)=? , Sir plzzz help me
Why are you allowed to say x = e^t ?
Whitecat are you?
It's a substitution :)
Because you are just defining a new variable, t equal to ln(x)
if exists an t such as e^t = x then u can say that freely, if u wanted to say that x = cos(t) u could do that (but obviously, x sjould be between -1 and 1)
Thanks for putting x=e^t as I have suggested earlier. DrRahul Rohtak Haryana India
May I know why you should only keep e^t but not 10^t? I mean literally, the problem would be much simpler i.e, logx +c instead of "ln(x) +c" ??!
Could you please explain to me sir? I am so confused
@@vishwakmusic9314 e^t is the only function whose derivative is itself (d/dx e^t = e^t). We can't bend the laws of mathematics, we have to use e^t because only it has the properties we need.
@@cppghost Ok..... Thank you veeeeerry much!!! I am actually a Biology student. So, I don't have a depth knowledge in maths. But, in our country(India), before joining a Medical University, you must study physics and chemistry! For that I have to learn maths also! They won't teach us but there is differentiation, integration, Matrices, Sets, Progressions (including all three types). I personally have no hate for maths and I loved it since childhood but, I never thought I had to learn it in depth till I reached Work, Energy and Power chapter and Rotatory motion in physics!!!!!!!! I am so stressed out learning math. Any tips?
why dx = e^t * dt?
Nice 👍
2/x
Leave integration, I m just looking at his marker and expression 😄😄
For x
I had traveled from planet to planet, searching for an answer. And now after all these millennia, I have found the answer
what about negative x?
I think we would have to say x = -e^t but it will end up to be the same result
the condition says its only valid when x > 0
@@idontknow1630 wtf I can't read thanks
@@acetonewong4608 For any real value of x, other than the problem point of x=0, the integral of 1/x is ln(|x|) + C.
This comes from the fact that logarithms are defined for complex numbers, where they are equal to the log of the magnitude for the real part, plus i*(the angle + any arbitrary integer multiple of 2*pi). Simply let your constant of integration cancel out the imaginary part, and you see that it is consistent with ln(|x|) + C.
If x
nice
Can blackpenredpen be an alien?🤔
Why the C? Just because I'd like to see.
t does not cover all x>0 domain and therefore your proof is incomplete!
Which bit of x>0 does it miss?
And is .......
log|x|+c
how do you integrate it if x>0 is not given
In real numbers, the integral of 1/x dx, is ln(|x|) + C, where the || brackets indicate absolute value.
In complex numbers, the integral of 1/z dz is the complex log of z, plus C. Complex log of z is ln(|z|) + i*(angle(z) + 2*pi*k), where k is any integer.
Since the angle of all negative real numbers is pi, this means for negative real values of z, it reduces to ln(|z|) + i*(2*k + 1)*pi.
We can let the constant of integration equal anything we want, including complex numbers. So we can set its imaginary part equal to -i*(2*k+1)*pi, and cancel this part out. This shows how it is consistent with ln(|x|) + C, as you learn in an introductory calculus class.
oh thanks@@carultch
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Happy Krishna
Hermoso
Asians.....Thanks for the derivation of this direct formula😂
Does the antiqutient rule where 1 becomes x+c really simplify to ln|x|+c 🤯🤣😂😭
It's not that 1 becomes x + c rather that d/dx(x + c) = d/dx x + d/dx c = 1 + 0 = 1.
d/dx (ln |x| + c) = d/dx ln |x| + d/dx c
= 1/x + 0
= 1/x
The + constanst is because shifting the final curve up or down the y-axis (+ constant) makes no difference to the slope of the curve at any point (dy/dx). So given the slope, you cannot know which of the infinite curves which are all parallel to each other, just shifted up/down the y-axis, was the original. Thus you have to add a constant (C) without knowing its exact value.
where are the numbers man... :[
What is c
Any constant value you like as when you differentiate ln x + c, the c becomes zero.
Hacks
Why dx=etdt but not dx=etd tho
dx/dt=d(e^t)
dx=e^tdt
It's the same but this form is easier for doing integration.
This guy lol
Noice 😅
What language is he speaking?
*There there*
If I assume x = 10^t
dx = 10^t dt
So,integral (1/10^t) 10^t dt
Integral (1) dt
=t
=log x
😅😅😅
I don't know why we assume x=e^t,
But don't do what I've done ya😂
e^t has the unique property that its derivative is e^t dt. the derivative of 10^t is not 10^t dt.
@@datknightguy6474 ya I know,I just wanna make a confusion
Whatever
:D
100th comment😎
It's *wrong* if you add +c in a line when the preceding line does not have it but the indefinite integral already disappeared. Then it's better not to put +c. Anyway +c doesn't make any sense if you don't make precise what is c. So it's better to omit it on the rhs. It should be integrated in the l.h.s., i.e. the integral sign: Just define \int to be equal to -c+\int.
But why did you make x = et?….you are just forcing it….make x be soemthing else e
Eska ans to log t +c hay ....ya kya likha hay esne😂
nice