Question

Calculus question

Answer 1

your asked to add the area of rectangle to approximate the integration with .... its a reimann sum thing

Answer 2

yes.

Answer 3

so the width of each rectangle is: the interval divided into 3 parts: 6/3 = 2

the heights are given as left, right, or mid which are simply stated in the table

Answer 4

so I find the area using each one?

Answer 5

2h1+2h2+2h3

2(h1+h2+h3) for the required heights

Answer 6

yes, or simplify it as i did

Answer 7

the last part is asking if the approx is more, less or about equal to the actual value

Answer 8

wait where in the chart is the left right and mid

Answer 9

you do realize that the interval from 0 to 6 is broken along the 2s correct?

| | | |
0 1 2 3 4 5 6

so the lefts are at 0,2,4
the rights are at 2,4,6
and the mids (centers) are at 1,3,5

Answer 10

so which of the values do I put to solve ?

Answer 11

the ones in the table that coorespond to the left, rights and mids accordingly

Answer 12

ive already given you a shorcut for it; and even indicated what parts to use .... not too sure what your confusion is at the moment

Answer 13

area = 2(h1+h2+h3)

such that the h's relate to the left, right, or mid value from the table ....

you have to work it 3 different times according to the instructions.

Answer 14

are the areas 12,24 and 18 @amistre64

Answer 15

@Hero @Luigi0210 @mathslover @nincompoop @iPwnBunnies

Answer 16

Well, if you use left end points

Rectangular's height = left end point * base

Answer 17

im confused aren't there 3 left endpoints?

Answer 18

Hmm wait, lemme understand the quest. give me 1 minute! :)

Answer 19

ook no problem

Answer 20

And btw, if I'm allowed to intrude, one of your areas will be negative,

Answer 21

Negative? Please explain Luigi.

Answer 22

Area can never be negative

Answer 23

Wait,

Answer 24

I'm thinking of something else aren't I?

Answer 25

Probably

Answer 26

If you follow what @amistre64 said :

\(\color{blue}{\text{Originally Posted by}}\) @amistre64
you do realize that the interval from 0 to 6 is broken along the 2s correct?

| | | |
0 1 2 3 4 5 6

so the lefts are at 0,2,4
the rights are at 2,4,6
and the mids (centers) are at 1,3,5
\(\color{blue}{\text{End of Quote}}\)

Then , the areas come out to be 12,24, 18

Answer 27

a) total sum of the left end points, x- 0 to 5, left end points - (-6) to 50

b) total sum of right end points, x - 1 to 6, right end points - 0 to 80

c) sum of the mid points, x- 0 to 6, all of f(x), and midpoints

Answer 28

So, the lefts are at (0,2,4) that is for f(x) it is (-6,8,30)

2(-6 + 8 + 30) = 2(32) = 64

For rights at (2,4,6) that is for f(x) it is (8,30,80)

2(8+ 30 + 80) = 2(118) = 236

For the mids at (1,3,5) that is for f(x) it is (0,18,50)

2(0 + 18 + 50) = 136

Answer 29

For the last part -> (as f(x) is an increasing function -> given in the quest.)

Left end points will be the smallest (least positive) values in the interval, or the area estimated will be smaller than the actual area under the curve. (under-estimate the integral)

Answer 30

Right end points will be the largest values in the interval or the area will be over-estimate for the integral.

Answer 31

the mid points give the value which is average of the 2 values above. hence, it will give exact value for the integral if f(x) is a straight line OR is made up of linear segments and all the turning points are at the intervals used.

Answer 32

I'm not 100% sure whether I am correct or not. I request @iambatman and @Luigi0210 to verify my above points!

Answer 33

Looks right. Btw, that -6 is what I meant by "negative area".. i probably have the wrong term. :l

http://prntscr.com/3lprm8

Answer 34

Oh! Okay :)