BEC'S methods questions

[bec]

alright so finding the area enclosed by y = 2x² and the x-axis from x=0 to x=3, what would i put in? i did this:

?

but that gave me 60 and the answer's 28...

[Mao]

alright so finding the area enclosed by y = 2x² and the x-axis from x=0 to x=3, what would i put in? i did this:

?

but that gave me 60 and the answer's 28...

i start at 1!!!
and your interval beginning, "a", should be 0, as you are measuring from x=0 to x=3.

Left ended:

Right ended:

Midpoint:

[bec]

...sorry, i don't understand. why 1? is it always one, even if the width of the segments are something else?

if i do this i get 58 - isn't the answer 28? or am i just doing something wrong?

[Mao]

i represent the numeral rectangle. when i=1, it represents the first rectange, i=2, the second, and so forth

the bounds here are meant for i, not the upper and lower bounds of x.

i.e. for this example, we are approximating with right hand rectangles [width of 1] between 0 and 3 for

that is,
Right ended:

hence, we will have a total of 3 rectangles, with the first rectangle beginning at the interval beginning, i.e. a=x=0
[yes I realised that I overlooked that mistake (there were two), "a" does not start at 1, a is the interval beginning, i is the rectangles. that's now fixed in my previous post]

hence, we have

[bec]

Left ended:

Right ended:

Midpoint:

So: using this formula, you sub in that
a= lower limit (eg if it says finding the area under a curve between x = 0 and x = 7, this is 0)
b = upper limit (so if Q was what i said above, this would be 7)
= width of rectangle
i = total number of rectangles =

is that right?

[Collin Li]

I would suggest getting a grasp of summation notation rather than trying to understand someone else's formula.

http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/summationdirectory/Summation.html

It's pretty simple, and it will become clear to you how to transform something like:

Example

Find the right-endpoint estimate for the curve from x=1 to x=11 with a step-size of 2.

Your relevant points are .

Your rectangles are of those heights, and are of width 2.

Hence, your sum is:

So we need to find a general way of expressing each term. We could try , but we have to remember that summation notation goes by increments of 1, so we would have to remedy this by doing something like . However, to get our desired , we would need a starting point of (so that we get ). It's not desirable to have a non-integer starting point, as summation notation wasn't designed for that.

So to get around that, we make the expression in your sum from to
(where corresponds to and corresponds to )

[bec]

hmm i think i understand, roughly...

how would i do this?

"The interval [0,4] is divided into n equal subintervals by the points x₀<x₁<...x_n-1<x_n<=4
Let = x_i - x _i^-1 for i = 1,2,...,n.

Then lim -->0  (x_i is equal to...

A)
B)
C) 0
D) 4
E) 8

[Mao]

Left ended:

Right ended:

Midpoint:

So: using this formula, you sub in that
a= lower limit (eg if it says finding the area under a curve between x = 0 and x = 7, this is 0)
b = upper limit (so if Q was what i said above, this would be 7)
= width of rectangle
i = total number of rectangles =

is that right?

yes

basically, it means:

sum of [area * height] for each rectangle at regular intervals.

[Mao]

hmm i think i understand, roughly...

how would i do this?

"The interval [0,4] is divided into n equal subintervals by the points x₀<x₁<...x_n-1<x_n<=4
Let = x_i - x _i^-1 for i = 1,2,...,n.

Then lim -->0  (x_i is equal to...

A)
B)
C) 0
D) 4
E) 8

you need to recognise that:



that is basically the definition of integrals, that as our approxmiation start involving psycho amounts of rectangles [], our approximation becomes more and more accurate, i.e. exact.

you should be able to figure this one out =]

[bec]

so the answer is 8?

how do we know (or do we know?) that f(x) = x? is that implied in the x_i?

[Mao]

so the answer is 8?

how do we know (or do we know?) that f(x) = x? is that implied in the x_i?

yeah, pretty much.
the question tell us that "The interval [0,4] is divided into n equal subintervals"
hence when we have infinite numbers of n, [that's really intuitive work, but you should see it =] ]

[bec]

hey,

Find the value of k such that the area bounded by the line with equation y = 2x, the x-axis and the line x = 4 is divided into two regions of equal area by the line with equation y = k.

i got , and that seems to work but i just wanted to confirm that i'm right in case there's something i'm missing (the text solutions say ...)

edit: the line x=4, not y=4

[Neobeo]

hey,

Find the value of k such that the area bounded by the line with equation y = 2x, the x-axis and the line y = 4 is divided into two regions of equal area by the line with equation y = k.

i got , and that seems to work but i just wanted to confirm that i'm right in case there's something i'm missing (the text solutions say ...)

The line y = 4 is parallel to the x-axis, so there is no region  :police:

That said, if the region is bounded by y-axis rather than x-axis, then .

[bec]

sorry, i meant x=4
thanks!

[bec]

you know when you forget how to do something you've been doing for years? i've forgotten how to convert expanded cubics into y = (x+3)^3+c form...
can anyone explain? haha you don't need to go into it just give me an indication and i'll remember...
thanks!