Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

\[\lim n->0 \sum_{i=1}^{n} [2(1+\frac{ 2i }{ n }-5]\frac{ 2 }{ n }\] over [-1,2]
lim as n approaches 0 * if that doesnt look clear. looking for a step by step explanation.
is it like this \[\huge \lim_{n \rightarrow 0} \sum_{i=1}^{n}[2(1+\frac{2i}{n})-5]\frac{2}{n}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

we want to express the sum as\[\huge \sum_{i=1}^{n}f(c_i)\Delta x\] where \[\Delta x=\frac{2}{n}\]
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)=
\[f(x)=2(1+x)-5=2x-3\]@shubhamsrg @ParthKohli
and is that the answer?
so then i integrate normally?
yes but did you understand
yes i do thank you
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
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) \[f(x)=\frac{2}{3}[2(3-2x)-15]\]
\[\left[2\left(1+2\frac{i}{n}\right)-5\right]\Delta x=\left[2\left(1+\frac{2}{3} \frac{3i}{n}\right)-5\right]\frac{2}{3}\frac{3}{n}\Delta x\] \\[f(x)=\frac{2}3{}\left[2\left(1+\frac{2}{3}x\right)-5\right]\]
left side is stated as this correct? \[\lim_{n\to0}~\sum_a^b~f(a+i\frac{b-a}{n})\frac{b-a}{n}~;~i=0,1,2,...,b-\frac{b-a}{n}\]
gpt me i parts mixed around, and put abs in the wrong place, that summation
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
@bmelyk @jonask it's [1,3], you typed it in from another question wrong
yeah i fixed it.
so is it [-1,2] or [1,3]
then it's just \[\Large \int_1^3 (2x-5)\;\mathrm dx\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question