Question

Calculus Question!! Medal!!

Answer 1

\[\lim_{n \rightarrow \infty} \sum_{k=1}^{n} f(a+i \frac{b-a}{n}) \frac{b-a}{n}= \int\limits_a^b f(x) dx\]

Answer 2

compare the left hand side to your expression given
what is f(x)=?

Answer 3

uh f(x)= (5+ k 2/n)^10 2/n

Answer 4

what I was hoping you would give me f(x)
but you have given me f(5+k2/n) times 2/n

Answer 5

ok we will come back to that
can you compare what I wrote to what is given to find a ?

Answer 6

\[\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(a+i \frac{b-a}{n}) \frac{b-a}{n}= \int\limits\limits_a^b f(x) dx \\ \lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(a+i \frac{b-a}{n}) \frac{b-a}{n}=\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(5+i \frac{2}{n}) \frac{2}{n}\]
and by the way k was suppose to be i

Answer 7

or i was to be k either way

Answer 8

compare that last equality on both sides
what is b-a=?
what is a=?

Answer 9

a-b is 2 and a=5 ?

Answer 10

right so b=?

Answer 11

b-a is 2 (not a-b)

Answer 12

b-a=2
a=5
so
b=?


notice you have
b-a=2
and you know a=5
so you have
b-5=2
so b=?

Answer 13

7

Answer 14

right now you just need to figure out f(x)

Answer 15

can you tell me what
\[f(5+k \frac{2}{n})=?\]

Answer 16

notice 5+k*2/n is in ...

Answer 17

what is that inside of in your expression given
what is happening to the 5+k*2/n

Answer 18

it is being raised to the 10th power?

Answer 19

right

Answer 20

\[f(5+k \frac{2}{n})=(5+k \frac{2}{n})^{10} \]

Answer 21

\[f(x)=...\]

Answer 22

so f(x) is just (5+k 2/n) ^10 ?

Answer 23

no

Answer 24

just replace the 5+k2/n with x

Answer 25

\[f(x)=x^{10}\]

Answer 26

it is a machine
whatever you replace the old variable with that is what you do everywhere else

Answer 27

for example if I had
\[f(a+b)=(a+b+2)^n\]
then I could say \[f(x)=(x+2)^n\]
notice the old variable was a+b
and I wanted to know what f(x) was so I just replaced all the a+b's with x's

Answer 28

oohh okay got ya. So know since i know f(x)= x^10 and b=7 and a =5 do I just integrate x^10 ?

Answer 29

\[\lim_{n \rightarrow \infty} \sum_{k=1}^{n} f(a+k \frac{b-a}{n}) \frac{b-a}{n}=\lim_{n \rightarrow \infty} \sum_{k=1}^{n} (5+k \frac{2}{n})^{10} \frac{2}{n} \\ \implies a=5 \\ b-a=2 \\ \implies b=7 \\ f(a+k \frac{b-a}{n})=(5+k \frac{2}{n})^{10} \implies f(x)=x^{10} \text{ since } a+k \frac{b-a}{n}=5+k \frac{2}{n} \\ \text{ since we already said } a=5 \\ \text{ and } b-a=2\]

Answer 30

aren't you done with the question?

Answer 31

just write as a definite integral

Answer 32

it doesn't say evaluate

Answer 33

ohh right srry I forgot it didn't say evaluate.

Answer 34

you are just suppose to replace that thing earlier that I mentioned with three things

Answer 35

a,b, and f(x)
and you are done
replace them in
\[\int_a^b f(x) dx\]

Answer 36

okay just to be sure the final answer is \[\int\limits_{5}^{7} x^(10) dx\]

Answer 37

@freckles

Answer 38

a=5 and b=7 and f(x)=x^(10)

Answer 39

yeah isn't that was i wrote?

Answer 40

oh wait the x^10 look a little weird but that what i meant

Answer 41

so is my answer correct?

Answer 42

yes that is what we found above

Answer 43

okay thanks for all the help!

Answer 44

np