Can somone find the Riemann Sum of this intergral and also draw the volume of the solid:
$\displaystyle \int_0^1(-x+1)dx$
Thanx
Since, $\displaystyle f(x)=-x+1$ is countinous on $\displaystyle [0,1]$ its is integrable-meaning its Riemann Sum exists. Since the Riemann sum if its exists is independent of the partion of the interval (mathematical term is well-defined) we can use any one. The one I am going to use is the right-endpoint partion which is,
$\displaystyle \lim_{n\to\infty}\sum^n_{k=1}f(a+k\Delta x)\Delta x$
Where,
$\displaystyle a=0$ and $\displaystyle \Delta x=\frac{b-a}{n}=\frac{1-0}{n}$ and $\displaystyle f(x)=-x+1$ thus what we have is,
$\displaystyle \lim_{n\to\infty}\sum^n_{k=1}\left(1-\frac{k}{n}\right)\frac{1}{n}$=$\displaystyle \lim_{n\to\infty}\sum^n_{k=1}\frac{1}{n}-\frac{k}{n^2}$
Using the rule,
$\displaystyle \sum_{k=1}^n k=\frac{n(n+1)}{2}$
we have,
$\displaystyle \lim_{n\to\infty}1-\frac{n(n+1)}{2n^2}=1-1/2=1/2$
No i am looking for disks rotated by the x-axis.
And also Im planning to draw the figure of the disks on the poster board. Im planning to cut 15 circles too stack on top of each other and decrease the radius all the way close to zero. My graph is 10 cm apart from 0 to 1. So can some one list the radiuses of the 15 circles.
Hello,Originally Posted by Nimmy
If you want to draw 15 cylinders (a disk is a cylinder with a small height) it is advisible to make all cylinders with the same height.
With your problem you get a height of $\displaystyle \frac{1}{15}$.
You can calculate two different sets of cylinders: The smaller ones lay completely in a cone which is generated by the line $\displaystyle y=-x+1,\ x\in [0;1]$, the greater ones lay completely outside of this cone. I've calculated both sets of radii. See attachment.
(I assume, that you want to make a large drawing: So multiply the values of the radii by 10)
The complete diagram looks like the 2nd attachment.
Greetings
EB