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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which problem exactly?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0all of them , i just want to know which formula's i need to use so that i can solve them all .

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok...any other questions?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why would you need critical points for this problem...? it's one of those rectangle approximating things that everyone hates lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought critical points could help in drawing the graph...could be wrong..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0who needs to approximate when you can just integrate :P i always just do it and forget it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0id say integrate really :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0why learn to walk if you are running marathons?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait...so what am i supposed to do :O

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you are supposed to learn that doing it the long and tedious way is proof that the shortway is better....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not running marathons im barely even walking lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0but these? really? lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean but these? i dont understand anything yo u just said /:

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0[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
 5 years ago
Best ResponseYou've already chosen the best response.0and I mess up doing simple multiplication lol....4(2) = 8 ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.08/3 +8 = 8 +24//3 = 16/3....maybe :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0integrating isnt the answer she wants though

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its the answer, just not the "way" :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you wont get teh same answer by approximating

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you will if you take the reaimann sums and sll that fun stuff...right? if I recall correclt?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0with inscribed and ourscribed rectangles you get the average and yadayada...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah but reimann sum is taking an infinite number of rectangles, this is just like a few boxes

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then 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
 5 years ago
Best ResponseYou've already chosen the best response.0mean value theorum or some such....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y'all are confusing mee /:

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = x^2 + 4 find your f(x) values for your little rectangles.... do left right sides

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y = .5^2 + 4 y = 1^2 + 4 y = 1.5^2 + 4 y = 2^2 + 4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the area of each "rectangle" is gonna be .5 * its y value

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0then add the areas up

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I can draw a picture if you like :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0each place a corner touches the curve; we get a value for y right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\: im still confused . i guess i got lost along y'alls "discussion" ..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do you see the rectangles I drew? they have a width and a height right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the right edge of each rectangle hits the curve...and stops because we drew if INside the curve.... inscribed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we are told that we do this every .5 units..... so the width of each rectangle is .5 makes sense?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0good, 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0and of course I forgot how to square decimals...... (.5*.5) = .25 4.25 = 3.75 thats better

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0revised and corrected version lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this would be easier if I didnt keep getting typos..... this is right.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0at 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
 5 years ago
Best ResponseYou've already chosen the best response.0Now 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
 5 years ago
Best ResponseYou've already chosen the best response.0if any of this doesnt make sense.....let me know.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im sorryy if i'm making you impatient but my mind is VERY blank right now /:

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0im not impatient dear :) just wanting to help you see what you really already know ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0tell me what you are having trouble with...and I can explain it better for you.
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