Question

Integrals help!

Answer 1

im posting the question right now

Answer 2

@Sheraz12345

Answer 3

@Bobmath help

Answer 4

@ganeshie8 help

Answer 5

(a) remember that change of x = (b-a) / n
which is (3-1)/n = 2/n

(b) the formula is a+ k( change of x) ---> plug in 2/n for change of x
which is 1 + ( (3k)/n)

Answer 6

ok

Answer 7

(c) Just multiply the part a and b
= (1 + (3k/n)) ( 2/n)
simplify if needed

Answer 8

ok im confused in part b,

Answer 9

so the answer is 1+ 3k/n?

Answer 10

or do i do i plug in the 2?

Answer 11

where the n is going?

Answer 12

\[1 + \frac{ 3k }{ n }\]

Answer 13

im getting it wrong

Answer 14

Oh sorry I meant 1 + 2k/n haha I got blind

Answer 15

2/n as the first answer c:

Answer 16

ohh ok haha

Answer 17

wouldnt that make "c" a different answer instead of (1 + (3k/n)) ( 2/n)?

Answer 18

(d) (1+ (2k/n)) (2/n). You can move the 2/n outside of \[\Sigma \]
so just (1+ (2k/n)
and c would be (1 + (2k/n)) (2/n)

Answer 19

it is wrong even when i simplify

Answer 20

not even that

Answer 21

Oh I see, asking for f(x) change of x
2 ( 1 + 2k/n) (2/n)

Answer 22

since the original equation is 2x change of x
Solve for
segma 2(1+ 2k/n) (2/n)
Move 2/n outside of segma, then simplify:
2/n segma : 2+ 4k/n ---> k = n(n+1) / 2
= 2n + ( (4/n) * (n(n+1)) /2))
simplify and find limit

Answer 23

i simplify 2n + ( (4/n) * (n(n+1)) /2))? right?

Answer 24

or move the 2 outside as well same ans:
would be:
4/n segma of 1+ 2k/n
= 4/n segma of n + (2/n)(n (n+1)/2)

Answer 25

i get 2 n+2 (n+1)? is that ok

Answer 26

what do you get?

Answer 27

for which answer?

Answer 28

for C

Answer 29

4/n + 8k /n^2

Answer 30

whats D and E?

Answer 31

Is C correct ?

Answer 32

yeah

Answer 33

\[\large \sum \limits_{k=1}^n \left(\dfrac{4}{n} + \dfrac{8k}{n^2}\right)\]

Answer 34

evaluate ^^

Answer 35

like previous posts take 4/n outside and solve for segma of ( 1+ 2k/n). that's a bit easier c:

Answer 36

ok im doing it give me a sec

Answer 37

\[\large \dfrac{4}{n}\sum \limits_{k=1}^n 1 + \dfrac{8}{n^2}\sum \limits_{k=1}^nk\]

Answer 38

im getting 8/n + 4

Answer 39

\[\sum_{k=1}^{n} 2 (1 + \frac{ 2k }{ n }) (2n)\]
\[\frac{ 4 }{ n } \sum_{k=1}^{n} 1 + \frac{ 2k }{ n }\]

\[\frac{ 4 }{ n } \sum_{k=1}^{n} 1 + \frac{ 2k }{ n } (\frac{ n(n+1) }{ 2 })\]

\[\lim_{n \rightarrow \infty} 4/n ( 1+n+ 1/2)\]

Solve for lim to find e

Answer 40

e = 4

Answer 41

correction lim 4/n ( 1+ 1+ n) = lim 4/n (2+ n) = lim 8n + 4

Answer 42

8/n* yes I think your answer is all perf

Answer 43

so what would part D be after getting this limit?

Answer 44

was 8/n +4

Answer 45

im getting that wrong lol

Answer 46

try 2+ n? since I simplify and remove the 4/n out of segma

Answer 47

nope

Answer 48

@ganeshie8 after i do what you told me i get 8/n + 4