Initially, I was attempting to find the function which expresses the area under enclosed between the function ##\arcsin(\sin(x))## and the ##x##-axis (so technically I am looking for ##\int_{0}^{x} \arcsin(\sin(t)) dt## specifically, but got caught up on finding the general antiderivative)...
Consider the following scenario:
Given that points ##M## and ##N## are the midpoints of their respective line segments, what would be the fastest way to determine what percentage of the squares total area is shaded purple?
I managed to determine that the purple shaded area is ##5\text{%}##...
Thank you for the reply, it helped a lot. I've amended the errors in the domain on the piecewise and on the integration bounds, so now I have:
$$f(x,y)=
\begin{cases}
1 & \text{if }\operatorname{arccoth}\left(\sec\left(x\right)+xy\right)>y \\
0 & \text{otherwise} \\...
The implicit curve in question is ##y=\operatorname{arccoth}\left(\sec\left(x\right)+xy\right)##; a portion of the equations graph can be seen below:
In particular, I'm interested in the area bound by the curve, the ##x##-axis and the ##y##-axis. As such, we can restrict the domain to ##[0...
Firstly, I'm aware that title doesn't really make sense but stick with me on this. I'm trying to find a way to define an operation which will "join" two numbers instead of adding them. So for example, ##12+34=1234##. Ideally, it would be great if it also had something similar to sigma notation...
Hi everyone! I want to apologise for bumping this old thread, but I was recently going through my old posts and found this again. After spending a couple of days having another go at it, I managed to find a solution on my own :) I've decided to share how I went about it here in case anyone else...
I've always been taught that the indefinite integral of ##\frac{1}{x}## is ##\ln(|x|)##. Extending this to definite integrals, particularly over limits involving negative values, should work just like any other integral:
$$\int_{-1}^{1} \frac {1} {x} dx = \ln(|-1|) - \ln(|1|) = \ln(1) - \ln(1)...
Hi, thank you for the reply. However, I'm still really struggling to make much progress with this problem (my fault entirely).
Solving for the zero discriminant gives us ##r_s = y_s##. While this initially seemed quite useful I've really struggled with making any more progress towards...
Summary:: Calculate the percentage of area remaining when a quarter-cirlce is deprived of 1 large circles and 2 smaller circles.
Hi,
I'm not sure if this is the right subforum for this question but it seemed to be the one that fit the best. Please consider the following diagram:
Before...
Thanks for the reply. I did the testing regarding this part using Desmos, which said that the integral should be converging to ##1## for the constants all equalling ##1##.
In fact, Desmos indicates that the integral converges so long as all the constants are are greater than ##0##:
One of the maths groups I'm apart of on Facebook posts (usually) daily maths challenges. Typically they act as small brain teaser for when I wake up and I can solve them without much trouble. However, today's was more challenging:
(Note: blue indicates a variable and red indicates a constant)...
Sorry for not being clear in the OP. You're right, I do already have the graph of ##y=x \text{ csch}(x+y)## and I'm just trying to find the area enclosed by the cruve and the ##x##-axis
Edit: Do you think it would be possible to use WolframAlpha's nest function to create and integrate a heavily...
The title and summary pretty much say it all. I was wondering if it's possible to accurately determine the area enclosed by the curve ## y=x \text{ csch}(x+y)## and the ##x##-axis?
I first tried solving for ##y## and then ##x##, however it doesn't appear possible to solve for either variable. I...
Hi all; I have a very basic understanding of sequences and series and recently encountered a sequence which really has me confused: $$(\frac{1}{5}+(\frac{1}{5}+(\frac{1}{5}+(...)^2 )^2)^2)^2$$ What type of sequence would you call this? I couldn't even google it because I couldn't work out how to...