anonymous
  • anonymous
Area Under A Curve What formula's so I need/use for this worksheet ?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
which problem exactly?
anonymous
  • 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
  • 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
  • anonymous
r u from india?
anonymous
  • anonymous
no
anonymous
  • anonymous
ok...any other questions?
anonymous
  • anonymous
why would you need critical points for this problem...? it's one of those rectangle approximating things that everyone hates lol
amistre64
  • amistre64
everyone hates ? :)
anonymous
  • anonymous
I thought critical points could help in drawing the graph...could be wrong..
anonymous
  • anonymous
who needs to approximate when you can just integrate :P i always just do it and forget it
amistre64
  • amistre64
id say integrate really :)
amistre64
  • amistre64
why learn to walk if you are running marathons?
anonymous
  • anonymous
wait...so what am i supposed to do :O
amistre64
  • amistre64
you are supposed to learn that doing it the long and tedious way is proof that the shortway is better....
anonymous
  • anonymous
im not running marathons im barely even walking lol
amistre64
  • 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
  • amistre64
but these? really? lol
anonymous
  • anonymous
what do you mean but these? i dont understand anything yo u just said /:
amistre64
  • 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
  • amistre64
and I mess up doing simple multiplication lol....4(2) = 8 ;)
amistre64
  • amistre64
-8/3 +8 = -8 +24//3 = 16/3....maybe :)
anonymous
  • anonymous
integrating isnt the answer she wants though
amistre64
  • amistre64
its the answer, just not the "way" :)
anonymous
  • anonymous
you wont get teh same answer by approximating
amistre64
  • amistre64
you will if you take the reaimann sums and sll that fun stuff...right? if I recall correclt?
amistre64
  • amistre64
with inscribed and ourscribed rectangles you get the average and yadayada...
anonymous
  • anonymous
yeah but reimann sum is taking an infinite number of rectangles, this is just like a few boxes
amistre64
  • 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
  • amistre64
mean value theorum or some such....
anonymous
  • anonymous
y'all are confusing mee /:
amistre64
  • amistre64
y = -x^2 + 4 find your f(x) values for your little rectangles.... do left right sides
amistre64
  • amistre64
y = -.5^2 + 4 y = -1^2 + 4 y = -1.5^2 + 4 y = -2^2 + 4
amistre64
  • amistre64
the area of each "rectangle" is gonna be .5 * its y value
amistre64
  • amistre64
then add the areas up
amistre64
  • amistre64
I can draw a picture if you like :)
anonymous
  • anonymous
yes please
amistre64
  • amistre64
like this
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amistre64
  • amistre64
each place a corner touches the curve; we get a value for y right?
anonymous
  • anonymous
\: im still confused . i guess i got lost along y'alls "discussion" ..
amistre64
  • amistre64
do you see the rectangles I drew? they have a width and a height right?
anonymous
  • anonymous
yes
amistre64
  • amistre64
the right edge of each rectangle hits the curve...and stops because we drew if INside the curve.... inscribed
amistre64
  • amistre64
we are told that we do this every .5 units..... so the width of each rectangle is .5 makes sense?
anonymous
  • anonymous
yes
amistre64
  • 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
  • anonymous
yep
amistre64
  • 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
  • amistre64
like this
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amistre64
  • amistre64
and of course I forgot how to square decimals...... -(.5*.5) = -.25 4-.25 = 3.75 thats better
amistre64
  • amistre64
revised and corrected version lol
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amistre64
  • 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
  • amistre64
this would be easier if I didnt keep getting typos..... this is right.
1 Attachment
amistre64
  • 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
  • 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
  • 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
  • amistre64
if any of this doesnt make sense.....let me know.
anonymous
  • anonymous
im sorryy if i'm making you impatient but my mind is VERY blank right now /:
amistre64
  • amistre64
im not impatient dear :) just wanting to help you see what you really already know ;)
amistre64
  • amistre64
tell me what you are having trouble with...and I can explain it better for you.

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