The general answer. after inegration, would be x/2. It is pretty visual. You get half of the difference between the boundaries.

I absolutely love your new intro! Thanks for a lot of great content.

General solutions are: floor(x)/2 + ({x}^2)/2 for the fractional part and (floor(x)*(floor(x) - 1))/2 + floor(x)*(x - floor(x)) for the floor function

It help me so much the graph thank you!

One thing I pulled from this: When dealing with a negative number inside a floor/ceiling function, you flip the function direction (floor->ceil, ceil->floor) and pull the negative out front. Is that a safe statement?

In synthesis, thats what we call a 'ramp' wave! a saw wave would be 1 - {x}

Suppose we want to integrate {x} from arbitrary a to b. We want to split this into three integrals: from a to ceil(a), from ceil(a) to floor(b), and from floor(b) to b, since this makes the process much easier. The first Integral is simply [ceil(a) - a]*[1 - {a}]/2 = [ceil(a) - {a}ceil(a) - a + a{a}]/2 since this is just the area of a trapezoid. The second integral is simply [floor(b) - ceil(a)]/2 since it is half the area of the rectangle of height one, and the base is the difference of the boundaries. The third integral is {b}^2/2. Adding it all gives [{b}^2 + floor(b) - {a}ceil(a) - a(1 - {a})]/2. Equivalently, we can evaluate the first two integrals using the floor function only. The first may go from floor(a) to a, but this we will subtract from the total. This will give {a}^2/2. The second integral will then run from floor(a) to floor(b), and it will equal to [floor(b) - floor(a)]/2. So the total integral from a to b of {x} is simply [{b}^2 + floor(b) - {a}^2 - floor(a)]/2. If we ever want to integrate {f(x)} from a to b instead, then we let u = f(x) so that dx = u’*(f^-1)’(u) du, with the lower bound u = f(a) and the upper bound u = f(b). Then you can perform a very similar decomposition of the integrals into three parts, but the middle integral with the second approach will be much harder to do now that we have extra-factors in the integrand. So instead we split this middle integrals, the first of the many going from floor(f(a)) to floor(f(a)) + 1, the second going from floor(f(a)) + 1 to floor(f(a)) + 2, so on and so forth until the integral that goes floor(f(b)) - 1 to floor(f(b)). And then you replace {u} with u - floor(f(a)) - n, and this you perform the integration. Or, one can also recur to simply doing {f(x)} = f(x) - floor[f(x)], but only if it is simpler. Now suppose we want to integrate floor(x) from arbitrary a to b. Then we can split this into separate integrals as well. We subtract the integral from floor(a) to a from the total. This integral is simply equal to {a}floor(a). Then from floor(a) to floor(b). This is equal to Sum n = floor(a) to n = floor(b) - 1 of f(n)=n. This is tricky to evaluate since floor(a) and floor(b) can be negative or have different signs. Let me simply call this summation I(floor(b) - 1, floor(a)). Then the integral from floor(b) to b simply evaluates to {b}floor(b). So the total integral is {b}floor(b) - {a}floor(a) + I(floor(b) - 1, floor(a)), without loss of generality. As for the summation part, it can be evaluated to T(|floor(b) - 1|) - T(|floor(a)|), where T(n) = n(n + 1)/2. Hence I(floor(b) - 1, floor(a) = [floor(b) - 1]^2/2 + |floor(b) - 1|/2 + floor(a)^2/2 - |floor(a)|/2. Thus the integral from a to b of floor(x) is {b}floor(b) - {a}floor(a) + [floor(b) - 1]^2/2 + |floor(b) - 1|/2 + floor(a)^2/2 - |floor(a)|/2. To integrate some floor[f(x)] instead, repeat the same procedure outlined as for {f(x)}, but rather than replacing {u} by u - floor[f(a)] - n, simply replace it for floor[f(a)] - n, which will be a constant. To integrate ceil(x), notice that ceil(x) = -floor(-x). Then proceed to do a substitution and win. Oh, by the way, the antiderivative of floor(x) = x*floor(x) + c[floor(x)] + C, where c[floor(x)] is some constant for every integer.

Imagine him as your maths teacher 😍

Great. Thank u.

Man i love calculus now that you teach me😍

Thanks for evaluating the integral by graphically. Otherwise it may be easily evaluate. {X} is a periodic function with the period =1.So, the integral {X}dx, between 0 to 4=4*integral {X}dx, between 0 to 1=4*integral x dx, between 0 to 1=4*(x^2), between 0 to 1.=4*(1/2)=2. [ integral f(X)dx, from 0 to nT=n*integral f(X)dx, from 0 to T, where T is the period of the function f(X)]

you are amazing man!!!

I like your video, understood everything Also I liked your "well well"

This is similar to the question I asked in your last video: Is the fractional part of 0.999...=0.999... or is it 0? Because 0.999...=1

thanks a lot man

I haven't watched the video yet. If I'm wrong, oh well. Anyway, wouldn't you just integrate the function normally then subtract off the integral of the floor of the function? Not sure how to type calculus. Hope this works for you.. "S" is integral sign (a,b) would indicate integral bounds. So: S(0,4) frac(x) dx = S(0,4) x dx - S(0,4) floor(x) dx

Sir It is pretty simple by using graphical approach......

The integral of the floor function is just the sum notation of x shifted by 1. So from ... X(x-1)/2= x^2-x/2 . so the fractional part is x^2/2 -(x^2/2-x/2)=x/2... It just make sens

Know i saw the previous video so i wander how a function with both would look like

Please, how to calculate function erf (x)?

I think more generally it would be useful to split the integral, using the definition {x} = x - [x] (just pretend those are the brackets for the floor function) Becuase I think the floor function is easier to handle on its own.

Will you please make a video on iit jee Giv and fractional . Function questions

Very impressive but can you do the indefinite integral of floor(x) or {x}?

So, we can define a number by n={n}+[n] Like we can define it by n=sgn(n)*|n|

Will you pls help me regarding this problem? f (x) = {x}, 0

How would one handle integrating the fractional part of an arbitrary function of x?

Excelente

thanks for ur help 😍😍😍

Could you pls make a video solving the equation; „1/x=ln(x)“

Sir why you consider the perpendicular of the triangles as 1 as this is not exact 1.

Love your video

2:36 i watched this one first... BUT i did understood that x=[x]+{x} means a number equal to it's digits on the left side of the dot as a whole number (my keyboard don't have the right simbols) and all the fractional part in other words it's just to split the number by the dot

How does this look for the integral. Seems to work for any a and b (|x|/x)(floor(|x|)/2 + (1/2)({|x|})({x}+|floor(x)|-floor(|x|)))

integration of [x]dx with limits. Where [•] represent greatest integer function. Please please please

Can you plz do this for me. Find the value of sin16degree.

is 0.9 the height for the triangles? cause the domain never goes to 1

Great👍❤️

Can anyone solve these integration of fractional function part of x^2 from 0 to 2.

Sir ..can solve please integral 0 to 100 fractional part of root x

Why isn't this fractional part just a piecewise function? f(x) = {x} where f(x) = x-floor(x) if >= 0; f(x) = x-ceiling(x) if < 0? Wouldn't that be a more accurate result?

5:52 But what does the graph look like for negative (-) values of x?

Do a video about Functional derivative

so, what is integral of fractional part of x^2 from 0 to 4

Hey nice explanation ।love from india 🥰🤩

Thankyou sir 😃😃😃

Definitely a floor function person.

I have a good question: integral(sqrt(tanx+ln(sin x))

Sir, please what's the integration of (e^y /y )dy

Well well is the new phrase isn’t it isn’t it

Plzzz upload integration over 0 to 1 of [ x ]

How to integrate [x]^[x]dx

But the height is not exact 1 because there is broken continuity. Isn’t it?

Bro can u help me to solve int of 1 to 4 {x}^[x]dx please

2.. let me see the answer

How do we integrate the sigma function?

Nice video

Interesting :) When I think of the fractional part of x, I first think of x modulo 1, as opposed to x minus floor(x). No difference in the positive domain, just for x < 0.

Interesting. Is there ceil function, which rounds number to min bigger integer?

Sir, can you make a video on a² + b² = a + ib Please sir. Finding values of a, b

Make a video on (i)! pleasd

Why not integrate ( pi/2 +atan( tan(pi*x+pi/2) ) )/pi ? ;)

I miss your old intro ;( #YAY

We can do it by graph easily

I prefer the definition frac(x) = x - truncate(x), where truncate(x) := sign(x)*floor(|x|), because for x = -1.7, for instance, truncate(x) = -1, so {x} = -0.7

good new intro!

in fact ∫{x}dx from 0 to k is k/2 for all k

Which language is this??

Now do integral from 0 to 2018 of {x}/x

before watching: Well I think it would be a periodic line that cuts off at every integer, much like [[x]] - x is. This means the area would be 4 x (l x w / 2) since they are triangular pieces. But length and width are both 1, therefore A = 2 (4 triangles make 2 squares) Ok let's see

Integrate y=x xor k please

0.999... = 1, {0.999...} = 0. A few guys already have answered this question.

Good, now do ceiling

One way to write the indefinite integral: ∫{x}dx=(x²+⌊x⌋²+⌊x⌋)/2-x⌊x⌋

i'd like to see the anti-derivative... please

{sqrt2} = 1?

0:07 isn't that doraemon intro song?

I’m mad that {-|x|} isn’t just -({|x|}) I’m sorry for the absolute values I’m just so paranoid that x could be negative AHH

your channel deserve millions views but the people prefer songs and something trivial than mathematics and this very sad🙄🙄