The properties associated with the summation process are given in the following rule.
Rule: properties of sigma notation
Let
${a}_{1},{a}_{2}\text{,\u2026,}\phantom{\rule{0.2em}{0ex}}{a}_{n}$ and
${b}_{1},{b}_{2}\text{,\u2026,}\phantom{\rule{0.2em}{0ex}}{b}_{n}$ represent two sequences of terms and let
c be a constant. The following properties hold for all positive integers
n and for integers
m , with
$1\le m\le n.$
A few more formulas for frequently found functions simplify the summation process further. These are shown in the next rule, for
sums and powers of integers , and we use them in the next set of examples.
Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let
$f\left(x\right)$ be a continuous, nonnegative function defined on the closed interval
$\left[a,b\right].$ We want to approximate the area
A bounded by
$f\left(x\right)$ above, the
x -axis below, the line
$x=a$ on the left, and the line
$x=b$ on the right (
[link] ).
How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin by dividing the interval
$\left[a,b\right]$ into
n subintervals of equal width,
$\frac{b-a}{n}.$ We do this by selecting equally spaced points
${x}_{0},{x}_{1},{x}_{2}\text{,\u2026,}\phantom{\rule{0.2em}{0ex}}{x}_{n}$ with
${x}_{0}=a,{x}_{n}=b,$ and
${x}_{i}-{x}_{i-1}=\frac{b-a}{n}$
for
$i=1,2,3\text{,\u2026,}\phantom{\rule{0.2em}{0ex}}n.$
We denote the width of each subinterval with the notation Δ
x , so
$\text{\Delta}x=\frac{b-a}{n}$ and
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base.
a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec
Pls help me solve
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation
your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why
*I want to know the meaning of those symbols in sets*
e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis
*in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions
what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question
*in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have:
n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation
40 = 24 + 32 - n(M and P) + 4
Now solve for the unknown using algebra:
40 = 24 + 32+ 4 - n(M and P)
40 = 60 - n(M and P)
Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P)
Solution:
n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form:
Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.