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

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

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

yes.

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

okay.

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

ok

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

and is that the answer?

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

so then i integrate normally?

yes but did you understand

yes i do thank you

yw

is the answer -5?

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

i subbed in the wrong number

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

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

delta x

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

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

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

yeah i fixed it.

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

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

2x-3