Question

help with calc!?!?!

Answer 1

@aaronq are you good in calc?? (:

Answer 2

Check the image that you uploaded.

Answer 3

hahah sorry!!!

Answer 4

@jim_thompson5910

Answer 5

Hint: \[\Large \int_{a}^{b}f(x)dx = \lim_{n \to \infty}\left[\sum_{k=1}^{n}f(x_{k})\Delta x\right]\]

Answer 6

\[\Large \int_{a}^{b}f(x)dx = \lim_{n \to \infty}\left[\sum_{k=1}^{n}f(x_{k})\Delta x\right]\]

where

\[\Large x_{k} = a + k*\Delta x\]

\[\Large \Delta x = \frac{b-a}{n}\]

Answer 7

yeah i have that formula infront of me in my book but im not getting the write answer (its practice question) i think im just applying it wrong, ill try again

Answer 8

were you able to determine 'a' and 'b' ? or delta x?

Answer 9

is delta x, 3/n??

Answer 10

yep, \[\Large \Delta x = \frac{3}{n}\]

Answer 11

yay, at least i got that right haha

Answer 12

what are 'a' and 'b' ?

Answer 13

im not sure, is a 1?

Answer 14

what is the function f(x) in this case?

Answer 15

-1+k 3/n

Answer 16

a=-1 b = 2

Answer 17

@jim_thompson5910

Answer 18

good

Answer 19

how about f(x)?

Answer 20

f(x)= -1+k 3/n

Answer 21

not quite

Answer 22

:/

Answer 23

am i close?

Answer 24

not really

Answer 25

then im not quite sure..

Answer 26

you know what delta x is

so why not remove it from the sigma and see what you can find (or see what's left over)

Answer 27

-1+k?

Answer 28

compare this
\[\Large \lim_{n \to \infty}\left[\sum_{k=1}^{n}\left[\left(-1+k\frac{3}{n}\right)^4 \frac{3}{n}\right]\right]\]
to this
\[\Large \lim_{n \to \infty}\left[\sum_{k=1}^{n}f(x_{k})\Delta x\right]\]

Answer 29

Cross off the delta x terms
\[\Large \lim_{n \to \infty}\left[\sum_{k=1}^{n}\left[\left(-1+k\frac{3}{n}\right)^4 \cancel{\frac{3}{n}}\right]\right]\]
\[\Large \lim_{n \to \infty}\left[\sum_{k=1}^{n}f(x_{k})\cancel{\Delta x}\right]\]

Answer 30

ohhh! (-1+k 3/n)^4

Answer 31

@jim_thompson5910

Answer 32

closer

Answer 33

still no

Answer 34

the whole equation??

Answer 35

it's actually simply f(x) = x^4

Answer 36

the x, aka \(\Large x_k\), gets replaced with -1+k*(3/n)

Answer 37

put this all together to say

\[\Large \lim_{n \to \infty}\left[\sum_{k=1}^{n}\left[\left(-1+k\frac{3}{n}\right)^4 \frac{3}{n}\right]\right] = \int_{-1}^{2}x^4dx\]

Answer 38

oh wow i shouldve known that

Answer 39

i look too much into it hahah

Answer 40

is that a definite integral? thats it?

Answer 41

@jim_thompson5910

Answer 42

yes

Answer 43

thank you!!

Answer 44

np