bmelyk
Express the limit as a definite integral over the indicated interval [a,b]
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bmelyk
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\[\lim n->0 \sum_{i=1}^{n} [2(1+\frac{ 2i }{ n }-5]\frac{ 2 }{ n }\] over [-1,2]
bmelyk
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lim as n approaches 0 * if that doesnt look clear.
looking for a step by step explanation.
Jonask
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is it like this \[\huge \lim_{n \rightarrow 0} \sum_{i=1}^{n}[2(1+\frac{2i}{n})-5]\frac{2}{n}\]
bmelyk
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yes.
bmelyk
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@Jonask
Jonask
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we want to express the sum as\[\huge \sum_{i=1}^{n}f(c_i)\Delta x\]
where \[\Delta x=\frac{2}{n}\]
bmelyk
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okay.
Jonask
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we also wanna find c_i as the endpoint of each interval,
\[\huge c_i=\frac{2}{n}i\]this means that we have
\[\huge \sum_{i=1}^{n}[2(1+c_i)-5]\Delta x\]
Jonask
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hence you can steal your function easily from that
f(x)=
bmelyk
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ok
Jonask
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\[f(x)=2(1+x)-5=2x-3\]@shubhamsrg @ParthKohli
bmelyk
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and is that the answer?
Jonask
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\[\int\limits_{-1}^{2}2x-3\]
bmelyk
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so then i integrate normally?
Jonask
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yes but did you understand
bmelyk
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yes i do thank you
Jonask
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yw
bmelyk
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is the answer -5?
sirm3d
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\[\Large \Delta x=\frac{b-a}{n}=\frac{2-(-1)}{n}=\frac{3}{n}\]
bmelyk
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i subbed in the wrong number
Jonask
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\[x^2-3x {\huge|}_{-1}^{2}=4-6-1-3=-6\]
bmelyk
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so you're saying delta c is different than the other perosn?
bmelyk
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delta x
Jonask
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i was thinking 3/n but its not even in the equation
bmelyk
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im sorry i have to go to class but i will be back so please continue to explain further, thank you and chat soon
Jonask
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are you sure the class is {-1,2} not {0,2}
Jonask
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so we should write\[\frac{1}{3}\sum_{i=1}^{n}[2(3-2\frac{3i}{n})-5]2\frac{3}{n}\]
Jonask
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also 5(3)
\[f(x)=\frac{2}{3}[2(3-2x)-15]\]
Jonask
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@amistre64
sirm3d
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\[\left[2\left(1+2\frac{i}{n}\right)-5\right]\Delta x=\left[2\left(1+\frac{2}{3} \frac{3i}{n}\right)-5\right]\frac{2}{3}\frac{3}{n}\Delta x\]
\\[f(x)=\frac{2}3{}\left[2\left(1+\frac{2}{3}x\right)-5\right]\]
amistre64
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left side is stated as this correct? \[\lim_{n\to0}~\sum_a^b~f(a+i\frac{b-a}{n})\frac{b-a}{n}~;~i=0,1,2,...,b-\frac{b-a}{n}\]
amistre64
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gpt me i parts mixed around, and put abs in the wrong place, that summation
amistre64
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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
KatClaire
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@bmelyk @jonask it's [1,3], you typed it in from another question wrong
KatClaire
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@amistre64 @sirm3d ^
bmelyk
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yeah i fixed it.
Jonask
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so is it [-1,2] or [1,3]
sirm3d
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then it's just \[\Large \int_1^3 (2x-5)\;\mathrm dx\]
Jonask
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2x-3