All Questions
8
questions
1
vote
0
answers
57
views
What is the equation and area under curve for Covid load dynamics?
Covid virions on infection, replicate exponentially and once the body's defense system starts attacking it then it also seems to decrease exponentially.
Source
The time period when the PCR test is ...
- 111
2
votes
2
answers
2k
views
Line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions.
There is a line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. Find m.
My solution:
When I compute my answer, I get $1-\frac{...
- 1,927
0
votes
3
answers
84
views
examples of cases showing that knowing the area under a curve really matters ( at the elementary level)
It is often said that integral calculus offers a means to solve the area problem. My question, simply aims at understanding what is the interest of this area problem ( at the most basic level).
...
user655689
3
votes
5
answers
110
views
Finding the area bounded by $y = 2 {x} - {x}^2 $ and straight line $ y = - {x}$
$$
y =\ 2\ {x} - {x}^2
$$
$$
y =\ -{x}
$$
According to me , the area
$$
\int_{0}^{2}{2x\ -\ { x} ^2}\, dx \ + \int_{2}^{3}{\ {x} ^2\ -\ 2{x} }\, dx \\
$$
Which gives the area $ \frac{8}{3}$
But ...
user589548
2
votes
2
answers
282
views
Find the area bounded by $x=-y^2$ and $y=x+2$.
Question
Find the area bounded by $x=-y^2$ and $y=x+2$.
My Attempt
I know it is a very simple question to ask on MSE, but I don't know why I get stuck.
If you trace the graph, then the point of ...
- 772
0
votes
2
answers
689
views
Double Integral Application with Disks
I have to find the average value of a function $f(x, y) = x + y + x^2 + y^2$
over the disk $0 \le x^2 + y^2 \le 4$, and I am certain this is a double integral problem but I am unsure of the limits ...
- 575
0
votes
1
answer
375
views
Area under a basketball shot
The other day, someone asked me how to find the area under a basketball shot. It looked something like this:
How would I go about doing this?
user319034
2
votes
1
answer
4k
views
Area bounded by$ y^2=x^2(1-x^2)$
Find the area bounded by $y^2=x^2(1-x^2)$?
I think in this way as the graph lies between -1 to 1 the area is 4 times of $\int x \sqrt{1-x^2} dx$ limits from 0 to 1. Am I correct?
- 1,144