Question

Hi can someone help me with this question!?

Answer 1

@rational

Answer 2

\[\large \int\limits_a^bf(x)\,dx=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(a+\Delta x)*\Delta x\]

where \(\Delta x = \frac{b-a}{n}\)

Answer 3

how do you compute it? thats all i need! and finding the lim

Answer 4

In the current problem, we have \(a=0,~b=1\)
plug them in first

Answer 5

by any chnace can you post the solution to it?

Answer 6

not to that but to the problem?

Answer 7

yup thats what i got so far as well

Answer 8

is it i insetad of 1

Answer 9

exactly! there is a mistake in the formula, ima fix it

Answer 10

okay yeah! noo thats fine! we can just move on!

Answer 11

because i neeed to be done with this in ten minutes :O

Answer 12

\(\color{blue}{\Delta x=\frac{b-a}{n} = \frac{1-0}{n}=\frac{1}{n}}\)

so
\[\large{

\begin{align}\int\limits_a^bf(x)\,dx
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(a+i*\color{blue}{\Delta x})\color{blue}{\Delta x} \\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(0+i*\color{blue}{\frac{1}{n}})*\color{blue}{\frac{1}{n}}\\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(\color{blue}{\frac{i}{n}})*\color{blue}{\frac{1}{n}}\\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^n \left[2(\color{blue}{\frac{i}{n}})^2+3(\color{blue}{\frac{i}{n}})-5\right]*\color{blue}{\frac{1}{n}}\\~\\

\end{align}
}\]

Answer 13

see if it makes sense now

Answer 14

Yeah okay now thats what i really had lol

Answer 15

good, lets evaluate the sum first
forget about limit for time being

Answer 16

\(\color{blue}{\Delta x=\frac{b-a}{n} = \frac{1-0}{n}=\frac{1}{n}}\)

so
\[\large{

\begin{align}\int\limits_a^bf(x)\,dx
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(a+i*\color{blue}{\Delta x})\color{blue}{\Delta x} \\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(0+i*\color{blue}{\frac{1}{n}})*\color{blue}{\frac{1}{n}}\\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(\color{blue}{\frac{i}{n}})*\color{blue}{\frac{1}{n}}\\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^n \left[2(\color{blue}{\frac{i}{n}})^2+3(\color{blue}{\frac{i}{n}})-5\right]*\color{blue}{\frac{1}{n}}\\~\\
&=\color{gray}{\lim\limits_{n\to\infty}} \color{blue}{\frac{1}{n}}\sum\limits_{i=1}^n \left[\frac{4}{n^2}(\color{blue}{i^2})+\frac{3}{n}(\color{blue}{i})-5\right]\\~\\
&=\color{gray}{\lim\limits_{n\to\infty}} \color{blue}{\frac{1}{n}} \left[\frac{4}{n^2}\sum\limits_{i=1}^n(\color{blue}{i^2})+\frac{3}{n}\sum\limits_{i=1}^n(\color{blue}{i})-\sum\limits_{i=1}^n5\right]\\~\\

\end{align}
}\]

Answer 17

see if it still looks fine

Answer 18

yeah it does!

Answer 19

remember the formulas for sum of first n natural numbers ?
and the sum of squares of first n natural numbers ?

Answer 20

not really...

Answer 21

google

Answer 22

okay and then?

Answer 23

what do we do next

Answer 24

pls help me its due soon! i promise ill help u if u ever need help!!

Answer 25

google and see if u can get me the formulas for them

Answer 26

your professor expects you to know them

Answer 27

n(n+2)/ 2

Answer 28

Yes!

Answer 29

OH WAIT JK I DID know it i just didnt know what i was called

Answer 30

what about the formula for sum of squares ?

Answer 31

n(n+1)(2n+1)/6

Answer 32

good, simply plug them in

Answer 33

what do u mean?

Answer 34

\(\color{blue}{\Delta x=\frac{b-a}{n} = \frac{1-0}{n}=\frac{1}{n}}\)

so
\[\large{

\begin{align}\int\limits_a^bf(x)\,dx
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(a+i*\color{blue}{\Delta x})\color{blue}{\Delta x} \\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(0+i*\color{blue}{\frac{1}{n}})*\color{blue}{\frac{1}{n}}\\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^nf(\color{blue}{\frac{i}{n}})*\color{blue}{\frac{1}{n}}\\~\\
&=\lim\limits_{n\to\infty} \sum\limits_{i=1}^n \left[2(\color{blue}{\frac{i}{n}})^2+3(\color{blue}{\frac{i}{n}})-5\right]*\color{blue}{\frac{1}{n}}\\~\\
&=\color{gray}{\lim\limits_{n\to\infty}} \color{blue}{\frac{1}{n}}\sum\limits_{i=1}^n \left[\frac{4}{n^2}(\color{blue}{i^2})+\frac{3}{n}(\color{blue}{i})-5\right]\\~\\
&=\color{gray}{\lim\limits_{n\to\infty}} \color{blue}{\frac{1}{n}} \left[\frac{4}{n^2}\sum\limits_{i=1}^n(\color{blue}{i^2})+\frac{3}{n}\sum\limits_{i=1}^n(\color{blue}{i})-\sum\limits_{i=1}^n5\right]\\~\\
&=\color{gray}{\lim\limits_{n\to\infty}} \color{blue}{\frac{1}{n}} \left[\frac{4}{n^2}
\frac{n(n+1)(2n+1)}{6}+\frac{3}{n}
\frac{n(n+1)}{2}-5n\right]\\~\\

\end{align}
}\]

Answer 35

i just plugged in the formulas u gave
see if you're okay wid that

Answer 36

i am

Answer 37

\[\large{

\begin{align}
&=\color{gray}{\lim\limits_{n\to\infty}} \color{blue}{\frac{1}{n}} \left[\frac{4}{n^2}
\frac{n(n+1)(2n+1)}{6}+\frac{3}{n}
\frac{n(n+1)}{2}-5n\right]\\~\\
&=\color{gray}{\lim\limits_{n\to\infty}} \frac{-17n^2+15n+2}{6n^2}\\~\\
\end{align}
}\]

Answer 38

evaluate the limit next

Answer 39

so would it just be neg infinity?

Answer 40

Nope. the degrees of numerator and denominator are same'
so the limit will be simply the ratio of "leading coefficients"

Answer 41

-17/6

Answer 42

Yep!

Answer 43

thank you!!! do u mind helping me with one more questionn? its not as long

Answer 44

ill post it so u can get another medal

Answer 45

okay
i just want you see that we don't need to do all this work once we start using "fundamental theorem of calculus"

Answer 46

yeah i know how to do that! i just didnt know how to do it the long way!!

Answer 47

good :)

Answer 48

yhnk u for ur help!