Question

Express the limit as a definite integral over the indicated interval [a,b]

Answer 1

\[\lim n->0 \sum_{i=1}^{n} [2(1+\frac{ 2i }{ n }-5]\frac{ 2 }{ n }\] over [-1,2]

Answer 2

lim as n approaches 0 * if that doesnt look clear.
looking for a step by step explanation.

Answer 3

is it like this \[\huge \lim_{n \rightarrow 0} \sum_{i=1}^{n}[2(1+\frac{2i}{n})-5]\frac{2}{n}\]

Answer 4

yes.

Answer 5

@Jonask

Answer 6

we want to express the sum as\[\huge \sum_{i=1}^{n}f(c_i)\Delta x\]
where \[\Delta x=\frac{2}{n}\]

Answer 7

okay.

Answer 8

we also wanna find c_i as the endpoint of each interval,
\[\huge c_i=\frac{2}{n}i\]this means that we have
\[\huge \sum_{i=1}^{n}[2(1+c_i)-5]\Delta x\]

Answer 9

hence you can steal your function easily from that
f(x)=

Answer 10

ok

Answer 11

\[f(x)=2(1+x)-5=2x-3\]@shubhamsrg @ParthKohli

Answer 12

and is that the answer?

Answer 13

\[\int\limits_{-1}^{2}2x-3\]

Answer 14

so then i integrate normally?

Answer 15

yes but did you understand

Answer 16

yes i do thank you

Answer 17

yw

Answer 18

is the answer -5?

Answer 19

\[\Large \Delta x=\frac{b-a}{n}=\frac{2-(-1)}{n}=\frac{3}{n}\]

Answer 20

i subbed in the wrong number

Answer 21

\[x^2-3x {\huge|}_{-1}^{2}=4-6-1-3=-6\]

Answer 22

so you're saying delta c is different than the other perosn?

Answer 23

delta x

Answer 24

i was thinking 3/n but its not even in the equation

Answer 25

im sorry i have to go to class but i will be back so please continue to explain further, thank you and chat soon

Answer 26

are you sure the class is {-1,2} not {0,2}

Answer 27

so we should write\[\frac{1}{3}\sum_{i=1}^{n}[2(3-2\frac{3i}{n})-5]2\frac{3}{n}\]

Answer 28

also 5(3)
\[f(x)=\frac{2}{3}[2(3-2x)-15]\]

Answer 29

@amistre64

Answer 30

\[\left[2\left(1+2\frac{i}{n}\right)-5\right]\Delta x=\left[2\left(1+\frac{2}{3} \frac{3i}{n}\right)-5\right]\frac{2}{3}\frac{3}{n}\Delta x\]

\\[f(x)=\frac{2}3{}\left[2\left(1+\frac{2}{3}x\right)-5\right]\]

Answer 31

left side is stated as this correct? \[\lim_{n\to0}~\sum_a^b~f(a+i\frac{b-a}{n})\frac{b-a}{n}~;~i=0,1,2,...,b-\frac{b-a}{n}\]

Answer 32

gpt me i parts mixed around, and put abs in the wrong place, that summation

Answer 33

hmm interval (-1,2) then b-a= 2--1 = 3 so there seems to be a little confusion unless we are spose to factor within the given summation fo accomodate that

Answer 34

@bmelyk @jonask it's [1,3], you typed it in from another question wrong

Answer 35

@amistre64 @sirm3d ^

Answer 36

yeah i fixed it.

Answer 37

so is it [-1,2] or [1,3]

Answer 38

then it's just \[\Large \int_1^3 (2x-5)\;\mathrm dx\]

Answer 39

2x-3