Question

Can somone please help me with this problem
Estimate
using left hand sums 3 subdivisions

Answer 1

Approximate the area between the curve of y = x 2 and the x - axis from x = 0 to x = 1 :


Figure %: Area bounded by a curve


The method to be used in this section to solve this problem is Riemann sums, which involves subdividing the region into rectangles of equal width and adding up the areas of all of the rectangles to approximate the area of the region.

Let's first use three subdivisions to approximate this area:


Figure %: Three subdivision of the region
Each of the regions has a width of 1/3 the interval from x = 0 to x = 1 which is 1/3. These will form the bases of the rectangles, but what should be chosen for the height? One possibility would be to use the value of the function at the left endpoint of each subdivision as the height. This is called a left-hand approximation.


Figure %: Left-hand approximation of area using three subdivisions
The left-hand approximation for the area is as follows:

(0) + + =

Another possibility is to use the value of the function at the right endpoint of each subdivision as the height. This is called a right-hand approximation.


Figure %: Right-hand approximation of area using three subdivisions

The right-hand approximation for the area is as follows:

+ + (1) =

Finally, a third possibility is to use the value of the function at the midpoint of each of the subdivisions as the height. This is called a midpoint approximation.


Figure %: Right-hand approximation of area using three subdivisions
The midpoint approximation for the area is as follows:

() + () + () =

Each of these three methods seems reasonable, yet each generated a different result for the area. It seems as though no matter which point we choose for the height of the box, we will never get the exact area, since there is always a margin of error present.

However, it also seems intuitive that if 10 subdivisions or 100 subdivisions were used instead of only three, the error would decrease. This is indeed true. As the number of subdivisions increases, the accuracy of the approximation goes up regardless of what endpoint is being used as the height of the rectangles. This is a very important observation, but discussing it further requires the establishment of some notation first:

Let's generalize the procedure of doing left-hand approximations to find the area under the graph of f on the closed interval [a, b] using n subdivisions. With equal subdivisions, each rectangle has a width of Δx = .

If we let x 0 = a and x n = b , then x 1 = x 0 + Δx , x 2 = x 0 +2Δx , etc:


Figure %: Generalized subdivision method
Each of the subdivisions uses the left endpoint for the height, so the height of the first subdivision is f (x 0) , the height of the second subdivision is f (x 1) , etc. Thus, L n , the "nth left-hand approximation" is equal to

L n = f (x 0)Δx + f (x 1)Δx + ...f (x n-1)Δx

Defining x k as x 0 + kΔx allows us to collapse the left-hand approximation into:

L n = f (x k)Δx

Similarly, the right-hand approximation could be written as

R n = f (x k)Δx

Notice how this formula is just slightly different from the one for the left-hand approximation. It was conjectured previously that using more subdivisions would improve the accuracy of the approximations. To make this approximation exact, one would need an infinite number of subdivisions, and limits allow us to get close to this: if A is the exact area of a region bound by a function f , the x -axis, and the lines x = a and x = b , then

A = L n andA = R n

This means that

A = f(xk)Δx

and

A = nf(xk)Δx

Answer 2

\[\int\limits_{2}^{8}\frac{ x+6 }{ x }dx\]

Answer 3

this is the problem that goes after the word estimate

Answer 4

can you help me out

Answer 5

I still dont get it can you walk me through the steps please

Answer 6

This will show you how

Answer 7

I don't unsderstand still Im sorry but I dont

Answer 8

ok I have two answers can you tell me which one is correct

Answer 9

I got\[-\frac{ 9 }{ 4 }\]

Answer 10

and I got 14.3178

Answer 11

I wanna say the decimal one. But fractios are also a big part of it. If you divide the -9/4 it gives you -2.25..... I would say the (-9/4) but just in case I am wrong I would keep your work for your other one and try to see the difference that you got

Answer 12

What is the answer that you got cause I don't even know if I did it right can you please do it to see what you get

Answer 13

can you tell me if my answers are correct or not I don't know if I did it right

Answer 14

show me your work on here and I will help you from thier

Answer 15

@electrokid what answer do you get

Answer 16

\[
h={8-2\over3}=2\\

x_0=2+0=2\quad f(x_0)={2+6\over2}=4\\
x_1=2+2=4\quad f(x_1)={4+6\over4}={5\over2}\\
x_2=2+4=6\quad f(x_2)={6+6\over6}=2\\
\;\\
A\approx h[f(x_0)+f(x_1)+f(x_2)]\\
A\approx2\left(4+{5\over2}+2\right)\\
A\approx 8+5+12=\mathbf{25}
\]

Answer 17

We were both wrong....

Answer 18

notice that since the integrand is a decreasing function, the left-hand approximation should give an over-estimate of the area

Answer 19

ok I understand now thanks so much

Answer 20

yw

Answer 21

I messed up with the addition!!!
\[A\approx8+5+4=\mathbf{17}\]

Answer 22

Hey, did I not get "2" medals here??

Answer 23

Yea you did. I gave you one and she gave you one

Answer 24

and now I see only one!!

Answer 25

I'm not forsure what happend