Question

Verify if I worked this out correctly.

Answer 1

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

\[\lim_{n \rightarrow \infty }\sum_{i=1}^{n}((i/n)^2 +2)(1/n) \]

Answer 2

\[\lim_{n \rightarrow \infty } (1/n^3)((n(n+1)(2n+1)/(6) +2)\]

Answer 3

\[\lim_{n \rightarrow \infty }(2n^3/6n^3) +(3n^2/6n^3) +(n/6n^3) + 2\]

Answer 4

substituting infinity for x netted me 1/3 + 0 + 0 + 2 = 7/3 ?

Answer 5

original question should also state "use the limit process to find the area of the
region between the graph of the function and the -axis over the
given interval"

Answer 6

Do you mind if I write down a formula, just to make it easier to remember.

Answer 7

go right ahead

Answer 8

Assuming \( \Delta x = \frac{b-a}{n}\) $$ \lim_{n \to \infty } \sum_{i=1}^{n} f(a + i \Delta x) \Delta x = \int_{a}^{b}f(x)dx$$

Answer 9

Alright is that Riemann's Sum?

Answer 10

yes

Answer 11

Wow, now i actually see where these values are coming from. My teacher flew through this.

Answer 12

Define \( \large \Delta x = \frac{b-a}{n}\) \[
\lim_{n \to \infty } \sum_{i=1}^{n} f(a + i \Delta x) \cdot \Delta x = \int_{a}^{b}f(x)dx
\]
Plug in
\( a = 0 , b = 1 \)
\( \large \Delta x = \frac{1-0}{n} = \frac 1 n \)
\( f(x) = x^2 + 2 \)

\( \large \lim_{n \to \infty } \sum_{i=1}^{n} f(0 + i \cdot \frac 1 n ) \cdot \frac 1 n
\\ \large \lim_{n \to \infty } \sum_{i=1}^{n} f( \frac i n ) \cdot \frac 1 n
\\ \large \lim_{n \to \infty } \sum_{i=1}^{n} ( (\frac i n)^2 + 2 )\cdot \frac 1 n

\)

Answer 13

Thank you for this explanation. I seriously appreciate it. Its like the fog has been lifted!!

Answer 14

Your welcome :)
The rest of your work looks correct.

Answer 15

alright thanks again!!