Question

Evaluate the integral by computing the limit of Riemann sums.?
2
∫ x^2dx
0

Answer 1

the Riemann sum can be written$$\sum_{k=0}^n\left(\frac{k}n\right)^2\cdot\frac2n$$do you agree?

Answer 2

oops that should be \((2k/n)^2\) in the sum

Answer 3

since $$\sum_{k=0}^nf(x_k)\Delta{x}$$where$$f(x)=x^2\\\Delta x=\dfrac{2-0}n=\dfrac2n\\x_k=0+k\cdot\Delta{x}=\frac{2k}n$$

Answer 4

so$$\sum_{k=0}^n\left(\frac{2k}n\right)^2\frac2n=\frac2n\sum_{k=0}^n\frac4{n^2}k^2=\frac8{n^3}\sum_{k=0}^nk^2$$

Answer 5

recall$$\sum_{k=0}^nk^2=\frac{n(n+1)(2n+1)}6$$

Answer 6

so we have $$\frac8{n^3}\cdot\frac{n(n+1)(2n+1)}6=\frac43\cdot\frac{2n^2+3n+1}{n^2}=\frac43\cdot\left(2+\frac3n+\frac1{n^2}\right)$$

Answer 7

now take the limit as \(n\to\infty\):$$\lim_{n\to\infty}\frac43\cdot\left(2+\frac3n+\frac1{n^2}\right)=\frac43\cdot2=\frac83$$

Answer 8

okay

Answer 9

isn't it i=1 not k=0?

Answer 10

I use \(k=0\) and it makes no difference

Answer 11

okay

Answer 12

thanks

Answer 13

NP remember to reward the best answer with a medal

Answer 14

I don't think I'd be able to do this any better. I haven't done it in a while, so I can't completely confirm it. But if the \(\sum k^2\) part was handled correctly, it seems to be good!

Answer 15

use the fundamental theorem to verify:$$\int_0^2 x^2\ dx=\left(\frac13 x^3\right)_0^2=\frac132^3-\frac130^3=\frac83$$