Question

f(x)=4-(12/(x^2))Calculate the lower and upper sum estimates of the area under the curve over the interval [2,5] using subintervals of width ½.

Answer 1

what have you tried ?

Answer 2

I found delta x

Answer 3

I am not sure how to approach the problem though.

Answer 4

basically you need to estimate the area under given curve using rectangles

for left sum you need to find area of below rectangles
http://gyazo.com/7ae848ca5057eb28c8933bf534c1c481

Answer 5

I know that you use six rectangles. Every video and website says you have to use Riemann sum, but that is in the next lesson.

Answer 6

Do you have to use a calculator or is there and equation I can use?

Answer 7

you may use calculator
notice that you just need to find the area of those 6 rectangles and add up

Answer 8

what did you get for \(\Delta x\) ?

Answer 9

I don't now if deltax is 6 or 3/n

Answer 10

.5

Answer 11

hey we're given \(\Delta x = \frac{1}{2}\)

Answer 12

So delta x is the width?

Answer 13

yes, then total number of rectangles = \(\large \dfrac{5-2}{\frac{1}{2}} = ?\)

Answer 14

See i tried doing this problem a harder way, that's why i am confused.

Answer 15

its okay :) can we say the area using left sum is below :\[\Delta x \left[ f(2) + f(2+\Delta x) + f(2 + 2\Delta x) + \cdots + f(2+5\Delta x)\right]\]

?

Answer 16

yes?

Answer 17

replacing \(\Delta x\) by \(\frac{1}{2}\) gives

\[\frac{1}{2} \left[ f(2) + f(2+\frac{1}{2} ) + f(2 + 2\cdot \frac{1}{2} ) + \cdots + f(2+5\cdot \frac{1}{2} )\right]\]

Answer 18

see if you can evaluate ^

Answer 19

See i know that for the left sum you start at 2 and end with 4.5 and the right you use 3 and end with 5

Answer 20

fcor the right sum you start with 2.5

Answer 21

Ya sorry i meant 2.5

Answer 22

yes

Answer 23

since its intervals of .5

Answer 24

do we just plug those numbers into the original equation now?

Answer 25

yep

Answer 26

Alright I get thanks

Answer 27

I have no idea what is was doing, i was using

Answer 28

you're welcome :) just ask if something doesnt make sense still
right sum will give you overestimate :
http://gyazo.com/86dafae0a550f6d06fbd86f606ecc9e2

Answer 29

\[\sum_{i=1}^{n}\]

Answer 30

When do you use what i just posted?

Answer 31

both are same

Answer 32

I get the concept of how to do it its just the technical stuff that confuses me. So n is .5 then right?

Answer 33

Left sum :
\[\begin{align} &\Delta x \left[ f(2) + f(2+\Delta x) + f(2 + 2\Delta x) + \cdots + f(2+5\Delta x)\right] \\~\\&= \sum\limits_{i=1}^6 f(2 + (i-1)\Delta x)\times \Delta x\\~\\&=\sum\limits_{i=1}^6 f(2 + (i-1)\frac{1}{2})\times \frac{1}{2}\\~\\
\end{align}\]

Answer 34

n = number of rectangles = (5-2)/0.5 = 6

Answer 35

That's what exactly what I was doing, so this is the same thing as we just did?

Answer 36

Yep

Answer 37

This method is just so much more confusing.Any way thanks for your help i really appreciate it.

Answer 38

keep in mind you're just multiplying `width` and the `height` for area of each rectangle :
\[\begin{align}& f(2 + (i-1)\Delta x)\times \Delta x\\~\\
\end{align}\]

then adding them up :
\[\begin{align}& \sum\limits_{i=1}^{6} f(2 + (i-1)\Delta x)\times \Delta x\\~\\
\end{align}\]

Answer 39

`riemann sum` is just a fancy name for above method ^

Answer 40

I now know that, thanks.Also would the right sum be 3i/n +2

Answer 41

That's why i am confused, Riemann sum is the next lesson. Any way thanks again.

Answer 42

right sum would be
\[\begin{align}& \sum\limits_{i=1}^{6} f(2 + i\Delta x)\times \Delta x\\~\\
\end{align}\]

how did u get 3i/n+2 ?

Answer 43

Well my OS is broken, but