Question

Calc help!

Answer 1

so you are given a=2 and b=5
and f(x)=x^3
last question they told you the setup
\[A=\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(a+i \Delta x) \Delta x \text{ where} \Delta x=\frac{b-a}{n}\]

Answer 2

it is just plugging into that

Answer 3

So its f(2+3i/n)3/n?

Answer 4

yes and of course you would have that lim sum thingy in front
but I bet they want you to also give what f(2+3i/n) is
using f(x)=x^3

Answer 5

\[A=\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(2+i \frac{3}{n}) \frac{3}{n} \\ \text{ now use } f(x)=x^3 \text{ to evaluate } f(2+i \frac{3}{n})\]
just in case my math stuff was hard to read

Answer 6

(2+3i/n)^3(3/n)

Answer 7

\[A=\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(2+i \frac{3}{n}) \frac{3}{n} \\ \text{ now use } f(x)=x^3 \text{ to evaluate } f(2+i \frac{3}{n}) \\ \text{ so yes we have } \\ A=\lim_{n \rightarrow \infty} \sum_{i=1}^{n} (2+i \frac{3}{n})^3 \frac{3}{n}\]
now you could stop here if you see this as a choice
but if you don't see this as a choice that means they want us to expand (2+i3/n)^3

Answer 8

I see it as a choice

Answer 9

Thank you!!

Answer 10

k

Answer 11

though later when you are ask to evaluate these limit sum thingys
you will have to expand that awful cube thingy

Answer 12

\[(u+v)^3=u^3+3u^2v+3uv^2+v^3 \\ (2+i \frac{3}{n})^3 =(2)^3+3(2)^2(i \frac{3}{n})+3(2)(i \frac{3}{n})^2+(i \frac{3}{n})^3\]
and then use law of exponents a bit there
but anyways yeah

Answer 13

Thank you for the extra help! Ill remember this in the future!

Answer 14

a few more steps though to actually evaluate the sum part
and then the limit part