Question

how to find integral(1/x^2) dx by using riemann sums

Answer 1

that is going to be hard to do a Riemann sum with no limits of integration

Answer 2

sorry, limits are from 2 to 3\[\int\limits_{2}^{3}\frac{ 1 }{ x^{2} } dx\]

Answer 3

you can use the mean value theorem to help do the sum

let \[f(x)=-\frac{1}{x}\]

then \[f'(x)=\frac{1}{x^2}\]

Answer 4

then how can i use riemann sums there?

Answer 5

\[\int\limits_{2}^{3}\frac{1}{x^2}dx=\sum_{i=1}^{n}\frac{1}{(x_i^*)^2}\Delta x\]


by the MVT
\[\frac{-1}{2+(i+1)/n}-\frac{-1}{2+i/n}=\frac{1}{(x_i^*)^2}\frac{1}{n}\]

then

\[=\sum_{i=1}^{n}\frac{1}{(x_i^*)^2}\Delta x=\sum_{i=1}^{n}\left[\frac{-1}{2+(i+1)/n}-\frac{-1}{2+i/n}\right]\]

Answer 6

left off my limit on the first equation

Answer 7

\[\sum_{i=1}^{n}\left[\frac{-1}{2+(i+1)/n}-\frac{-1}{2+i/n}\right]=\frac{1}{3(3n+1)}-\frac{1}{2(2n+1)}+\frac{1}{6}\to \frac{1}{6}\]

as \(n\to\infty\)

Answer 8

cool, thanks a lot