Question

Use Riemann sums and a limit to compute the exact area under the curve.

Answer 1

y = x^2 + 2 on [0, 1]

Answer 2

how much time you got?

Answer 3

a lot why?

Answer 4

ok lets do it without the \(2\) because \[\int_0^1x^2+2dx=\int_0^1x^2dx+\int_0^12dx=\int_0^1x^2dx+2\]

Answer 5

now you have to break up the unit interval in to \(n\) parts, each of which will be of length \(\frac{1}{n}\)

Answer 6

if you use left hand endpoints you can write
\[x_0=0,x_1=\frac{1}{n},x_2=\frac{2}{n},...\] so in general
\[x_k=\frac{k}{n}\]

Answer 7

then you have
\[f(x_k)=\left(\frac{k}{m}\right)^2=\frac{k^2}{n^2}\]

Answer 8

you want \[\sum_{k=0}^nf(x_k)\Delta x\]
\[=\sum_{k=0}^n\frac{k^2}{n^2}\times \frac{1}{n}\]

Answer 9

as far as the summation is concerned, \(n\) is a constant
so you can rewrite this as
\[\frac{1}{n^3}\sum_{k=0}^nk^2\]

Answer 10

then use the well know formula for summing squares, get
\[\frac{1}{n^3}\times \frac{n(n+1)(2n+1)}{6}\]

Answer 11

take the limit as \(n\to \infty\) and get \(\frac{1}{3}\)

Answer 12

okay