Area Under A Curve
What formula's so I need/use for this worksheet ?

- anonymous

Area Under A Curve
What formula's so I need/use for this worksheet ?

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- schrodinger

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- anonymous

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- anonymous

which problem exactly?

- anonymous

all of them ,
i just want to know which formula's i need to use so that i can solve them all .

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## More answers

- anonymous

first of all find the critical points i.e. dy/dx=0 then proceed further when the slope is +ve or -ve between those intervals

- anonymous

r u from india?

- anonymous

no

- anonymous

ok...any other questions?

- anonymous

why would you need critical points for this problem...? it's one of those rectangle approximating things that everyone hates lol

- amistre64

everyone hates ? :)

- anonymous

I thought critical points could help in drawing the graph...could be wrong..

- anonymous

who needs to approximate when you can just integrate :P i always just do it and forget it

- amistre64

id say integrate really :)

- amistre64

why learn to walk if you are running marathons?

- anonymous

wait...so what am i supposed to do :O

- amistre64

you are supposed to learn that doing it the long and tedious way is proof that the shortway is better....

- anonymous

im not running marathons im barely even walking lol

- amistre64

when you have a function that is difficult to integrate, you can approximate the area under its curve thru numerical stuff like the trapaziod rule or simpsons rule and other time consuming methodss

- amistre64

but these? really? lol

- anonymous

what do you mean but these?
i dont understand anything yo u just said /:

- amistre64

[S] -x^2 +4 dx [0,2]
F(x) = -x^3/3 + 4x ....at 0 its useless so solve for 2
-8/3 + 12 = -8 +36//3 = 28/3

- amistre64

and I mess up doing simple multiplication lol....4(2) = 8 ;)

- amistre64

-8/3 +8 = -8 +24//3 = 16/3....maybe :)

- anonymous

integrating isnt the answer she wants though

- amistre64

its the answer, just not the "way" :)

- anonymous

you wont get teh same answer by approximating

- amistre64

you will if you take the reaimann sums and sll that fun stuff...right? if I recall correclt?

- amistre64

with inscribed and ourscribed rectangles you get the average and yadayada...

- anonymous

yeah but reimann sum is taking an infinite number of rectangles, this is just like a few boxes

- amistre64

then lets do boxes ;) we have to do it inscribed and outscribed.. but Im pretty sure you end up with the average being the limit of the 2..

- amistre64

mean value theorum or some such....

- anonymous

y'all are confusing mee /:

- amistre64

y = -x^2 + 4 find your f(x) values for your little rectangles.... do left right sides

- amistre64

y = -.5^2 + 4
y = -1^2 + 4
y = -1.5^2 + 4
y = -2^2 + 4

- amistre64

the area of each "rectangle" is gonna be .5 * its y value

- amistre64

then add the areas up

- amistre64

I can draw a picture if you like :)

- anonymous

yes please

- amistre64

like this

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- amistre64

each place a corner touches the curve; we get a value for y right?

- anonymous

\: im still confused .
i guess i got lost along y'alls "discussion" ..

- amistre64

do you see the rectangles I drew? they have a width and a height right?

- anonymous

yes

- amistre64

the right edge of each rectangle hits the curve...and stops because we drew if INside the curve.... inscribed

- amistre64

we are told that we do this every .5 units..... so the width of each rectangle is .5 makes sense?

- anonymous

yes

- amistre64

good, then all we need to figure out the area of any given rectangle is ...we already now its width (.5) is going to be its height...right?

- anonymous

yep

- amistre64

so at the first interval where x = .5; we go up until we hit the curve of y = -x+4
sine the value of "y" is our height we need to use the equation to figure out the height at x = .5
y = -.5^2 + 4
y = -2.5 +4
y = 1.5 is this right?

- amistre64

like this

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- amistre64

and of course I forgot how to square decimals...... -(.5*.5) = -.25
4-.25 = 3.75 thats better

- amistre64

revised and corrected version lol

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- amistre64

at our next step, .5 + .5 = 1. we stick x = 1 into our equation to get the height for the second rectangle; width = .5 and height =-x^2 +4.
h = -(1^2) + 4
h = -1 + 4
h = 3
Area2 = .5 * 3 = 1.5

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- amistre64

this would be easier if I didnt keep getting typos..... this is right.

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- amistre64

at our next step, .5 + .5 +.5 = 1.5 we stick x = 1.5 into our equation to get the height for the third rectangle; width = .5 and height =-x^2 +4.
h = -(1.5^2) + 4
h = -2.25 + 4
h = 1.75
Area3 = .5 *1.75 = .875

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- amistre64

at our next step, .5 +.5 +.5+.5 = 2 we stick x = 2 into our equation to get the height for the third rectangle; width = .5 and height =-x^2 +4.
h = -(2^2) + 4
h = -4 + 4
h = 0
Area4 = .5 *0 = 0 Do we really need to include this one? lets assume no :)

- amistre64

Now we add up all the areas of our rectangles:
A1 + A2 + A3 + A4 = answer
1.875 + 1.5 + .875 + 0 = answer
1.875 + 1.5 + .875 + 0 = 4.25 this is the smallest value for the area under the curve that is reasonably close to the actual area under the curve.
Any of this make sense?

- amistre64

if any of this doesnt make sense.....let me know.

- anonymous

im sorryy if i'm making you impatient but my mind is VERY blank right now /:

- amistre64

im not impatient dear :) just wanting to help you see what you really already know ;)

- amistre64

tell me what you are having trouble with...and I can explain it better for you.

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