Name $-$Surname:alfarooq alsamarray

Student ID:$190406181$

Rewrite your student ID number into boxes below:

Student ID: $$

abc is the last $3$ digit of student id number: $2^{**********}$ abc

$a$:$=$ the third digit from the right of your student ID number $($which is highlighted green$)$

$b$:$=$ the second digit from the right of your student ID number $($which is highlighted blue$)$

c :$=$ the last digit of your student ID number $($which is highlighted yellow$)$

For example, if your student id is $2405114786,$ then $a=7,b=8,c=6.$ Find $a,b,c$ for your student ID number:

$a=$

$b=$

$c=$

$q,$

$($These $3$ digits are special for you. If $a,b,c$ are not produced from your student ID number, then you can't get point!$)$

Now $f(x)=(a+1)x^2-bx+c.$ Write the $f(x)$ which is specified for you.

$f(x)=$Sketch the graph of $f(x).$We'll estimate the area A under the graph of $f(x)$ where $x$ is between $2$ and $8.$

$($The shape in the left is just for representation. Draw your own sketching in question $1)$

a$)$ The shape in the left and right are not same but it may give an idea about the area of the left one.

$A$~~$A_1$

$A_1$ is the area of the rectangle whose horizontal length is $8-2=6,$ and height $($vertical lenght$)$ equals to $f(2).$ Calculate $A_1$b$)$ The shape in the left and right still are not same but it may now give a better idea about the area of the left one.

$A$~~ubrace$(A_1+A_2$ubrace${)}_{R_2}$

Now $A_1$ is the area of the rectangle whose horizontal length is $5-2=3,$ and height $($vertical lenght$)$ equals to $f(2).A_2$ is the area of the rectangle whose horizontal length is $8-5=3,$ and height $($vertical lenght$)$ equals to $f(5).$ Calculate $A_1$ and $A_2.$ We can call the sum of $A_1$ and $A_2$ as Riemann sum with $2$ subinterval and notate it by $R_2.$ Calculate $R_2$ for better approximation of $A.$c$)$ Now we divide the shape into $3$ pieces and estimate its area by using $3$ rectangles. Lets calculate $R_3.$ Generally $R_3$ gives better approximation than $R_2.$

:$.$cdotsubrace$(A_1+A_2+A_i$ubrace${)}_{R_3}$

Hint: Now $A_1$ is the area of the rectangle whose horizontal length is $4-2=2,$ and height $($vertical lenght$)$ equals to $f(2).$

$A_2$ is the area of the rectangle whose horizontal length is $6-4=2,$ and height $($vertical lenght$)$ equals to $f(4).A_3$ is defined similarly.d$)$ Now we divide the shape into $6$ pieces and estimate its area by using $6$ rectangles. Lets calculate $R_6.$ Generally $R_6$ gives better approximation than $R_2$ and $R_3.$So far we have made better approximations in each step. We observed that as we increase the number of subintervals $($in other words as we divide the shapes to larger number of pieces$),$ the approximation becomes better. Now we no longer seek for "better approximations". Lets go for "the best one". For this purpose, a$)$ We find a formula for $R_n.($This time we divide the shape into $n$ pieces. $n$ is undetermined so it can be any positive integer. We appromiate each pieces to a rectangle and $R_n$ is the sum of areas of these $n$ rectangles.$)$

$A$~~$R_n={\displaystyle \underset{i=1}{\overset{n}{}}}{A}_i$

$={\displaystyle \underset{i=1}{\overset{n}{}}}\frac{b}{n}f(x_{i-1})$

$=\frac{6}{n}{\displaystyle \underset{i=0}{\overset{n-1}{}}}f(x_i)$b$)$ For the best approximation we should choose the largest possible $n.$ Howver there is no bound for choice of n $.$ Let's take limit of the $R_n$ as $n.$ This is not only the best appriximation but also exact result.

Hint: Before calculating the limit you should use.