4:44 AM
Now closely observe it - do you see that you can sum up the areas of those teeny-tiny rectangular strips you can cut out from the figure?? The infinitesimal rectangles have a very unobservable sort of breadth and hence their area will be so small. But still, adding them up yields the total area
Now I have said it in the case where you wanted to measure the area of an amoeboid.
The same case comes into play when you integrate functions - you're basically finding the area under the graph
Now let's think it all with respect to the sum at hand
The sum we needed to approximate was : $\sum\limits^{1000000}_{i=4} \frac{1}{_3\sqrt{i}}$
Observe that each of the $i$'s are separated by 1 unit of length
Also, the length of each strip we take will be $\frac{1}{_3\sqrt{3}}$
So when we plot the values of the length on the y axis and the length on the x axis, we'll be getting a series of points
So our problem will be shrunk to finding the sum of the areas of each of the rectangle of dimensions $\frac{1}{_3\sqrt{i}} \text{units length} \times 1 \text{unit breadth}$ for each $i$
This problem can be approached by approximation, basically, since we can't keep finding the reciprocals of the cube roots of each $i$
Thus we consider the function $f(x) = \frac{1}{_3\sqrt{x}}$, continuous in the domain of interest (here, $\mathbb{R}^+$), to approximate the area covered
"Also, the length of each strip we take will be $\frac{1}{_3\sqrt{3}}$" - a mistake, it's supposed to be $\frac{1}{_3\sqrt{i}}$
"Thus we consider the function f(x)=13xβ, continuous in the domain of interest (here, R+), to approximate the area covered" - then we integrate the function within the limits $4$ to $1000000$, to get an approximate area.
Now the function we took and the sum we needed to calculate coincide at points corresponding to each $i$
@Arjun Hello!!! How's my content delivery in the above topic??
Would just like to know if I'd ever be a professor anywhere π
So basically, our continuous function grazes over each of the points $(i, f(i))$, so we have got the closest possible approximation, save for the fractional areas we accidentally added up. Now you just take the ceiling of the number to get the upper bound for the set that the number lies in... and voila! We have solved it!!
I pay my credits to Manu sir at Brilliant Study Centre's math faculty for having given us this superbly precise idea on integration
He's a vessel of we kids' respect and liking at Brilliant due to his young and cool behaviour :D He's so cool, tbh
Ahhh... I have a deadly test tomorrow at Brilliant
It's about conic sections...
Lost my vote in being a professor then π
Anyway, don't call that chapter a damn easy one
I have been reading through Resonance Eduventures' material on Coordinate Geometry... I must say 'Gosh, I wanna have a whole tank of water to drink!"
You can expect such hard questions from it
Plus I didn't study the chapter on time, so I've been working to get it done right
Now only the concepts of hyperbolae remain to be grazed through
@RajdeepSindhu Oh btw did you guys watch Minnal Murali (or Thunder Murali (I type it so since I don't know what the Hindi translation of the movie's name would be like), the new south-Indian superhero film with Tovino Thomas as the main actor in it)??? I did... it was AWESOME!!
@Anonymous I think using the wavy curve method should work... we were taught about it at Brilliant... I almost forgot it due to lack of usage... I need to get it back all right
@RajdeepSindhu 45 Mbps bro
