Question

Draw a picture that shows the area of int (x^2+x)dx from 1 to 3 using Riemann sums.

Answer 1

int: integral

Answer 2

sorry I am only learning integrals next week so I can't help u ;-(

Answer 3

YOu are on too late for calc questions

Answer 4

you are too far away from my lesson hahahah

Answer 5

ya I am just in calc 1

Answer 6

me too.

Answer 7

but i am almost at the end

Answer 8

i only have two weeks left of this course

Answer 9

ok

Answer 10

whtvr in any case I can't help u

Answer 11

\[f(x)=x^2+x\]\[\Delta x=\frac{3-1}{n}=\frac{2}{n},\quad f(1+k\Delta x)=(1+k\Delta x)^2+1+k\Delta x,\quad k=0\dots n-1\]\[f(1+k\Delta x)=(1+k\Delta x)^2+1+k\Delta x=2+3k\Delta x+k^2(\Delta x)^2\]\[S_n=\sum\limits_{k=0}^{n-1}f(1+k\Delta x)\Delta x=\Delta x\sum\limits_{k=0}^{n-1}\left(2+3k\Delta x+k^2(\Delta x)^2\right)=\]\[=\Delta x\left(2n+3\Delta x\sum\limits_{k=0}^{n-1}k+(\Delta x)^2\sum\limits_{k=0}^{n-1}k^2\right)=\]\[=\frac{2}{n}\left(2n+\frac{6}{n}\frac{(n-1)n}{2}+\frac{4}{n^2}\frac{(n-1)n(2n-1)}{6}\right)=\]\[=4+\frac{6(n-1)}{n}+\frac{4(n-1)(2n-1)}{3n^2}\]\[S=\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}\left(4+\frac{6(n-1)}{n}+\frac{4(n-1)(2n-1)}{3n^2}\right)=\]\[=4+6+\frac{8}{3}=\frac{38}{3}\]

Answer 12

where is the picture?

Answer 13

http://en.wikipedia.org/wiki/Riemann_sum :-)