## bmelyk one year ago Express the limit as a definite integral over the indicated interval [a,b]

1. bmelyk

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

2. bmelyk

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

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

4. bmelyk

yes.

5. bmelyk

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

7. bmelyk

okay.

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

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

10. bmelyk

ok

$f(x)=2(1+x)-5=2x-3$@shubhamsrg @ParthKohli

12. bmelyk

$\int\limits_{-1}^{2}2x-3$

14. bmelyk

so then i integrate normally?

yes but did you understand

16. bmelyk

yes i do thank you

yw

18. bmelyk

19. sirm3d

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

20. bmelyk

i subbed in the wrong number

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

22. bmelyk

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

23. bmelyk

delta x

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

25. bmelyk

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

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

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

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

@amistre64

30. sirm3d

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

31. amistre64

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

32. amistre64

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

33. amistre64

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

34. KatClaire

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

35. KatClaire

@amistre64 @sirm3d ^

36. bmelyk

yeah i fixed it.

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

38. sirm3d

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