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anonymous

  • 5 years ago

integrals...area under the curve?

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  1. anonymous
    • 5 years ago
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    Gowlet

  2. amistre64
    • 5 years ago
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    yeah, what about em?

  3. anonymous
    • 5 years ago
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    how do we do them...without any rule?

  4. amistre64
    • 5 years ago
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    well if you wanna live by no rule, then it doesnt matter HOW you do them ;) I gotta ask, what do you mean?

  5. anonymous
    • 5 years ago
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    we have a curve with a grid...we need to count the squares and find the area under the curve

  6. anonymous
    • 5 years ago
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    is there any more accurate way to do this than to simply count the squares?

  7. amistre64
    • 5 years ago
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    then you just gotta do your best guess... the area of a "trapazoid" is: (base)([height rightside + height leftside] /2) right?

  8. anonymous
    • 5 years ago
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    oh ok thanks i have another question

  9. anonymous
    • 5 years ago
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    how would i find the average value of f(x) over an interval?

  10. amistre64
    • 5 years ago
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  11. amistre64
    • 5 years ago
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    the average of any 2 numbers is add them together and divide by 2 5+8 = 13 13/2 = 6.5 is the average between them

  12. amistre64
    • 5 years ago
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    how big is the interval?

  13. amistre64
    • 5 years ago
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    the average f(x) is the number of each partition added together; then divide by the number of partitions used...

  14. amistre64
    • 5 years ago
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    the more subdivision of your interval, the more accurate your average f(x) will be

  15. anonymous
    • 5 years ago
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    f(x)=sqrt(25-x^2)

  16. amistre64
    • 5 years ago
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    what the integral does is divides the interval into an infinite number of peices then adds them all up to get an exact value ;)

  17. anonymous
    • 5 years ago
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    the interval is 1<=x<=5

  18. amistre64
    • 5 years ago
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    do you want integration? or the slower way like trapaziodal rule?

  19. anonymous
    • 5 years ago
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    probably integration

  20. amistre64
    • 5 years ago
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    yeah...probably :)

  21. amistre64
    • 5 years ago
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    so we have to turn this function into a higher order...... if you dont know the techniques than it aint gonna make alot of sense to you but here they are: sqrt(25 - x^2) is a disguised "cos". x = 5 sin(t) so that x^2 = 25sin^2(t) sqrt(25 - 25sin^2(t)) sqrt(25(1-sin^2(t))) = 5cos and then we can integrate:

  22. amistre64
    • 5 years ago
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    [S] 5cos(t) dt -> 5sin(t) +C but since we got and interval we can forgo the +C part.

  23. amistre64
    • 5 years ago
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    and thats the place I get lost at in doing this lol What I need is someone smarter than me to come along and tell me what or why I cant do it like this ;)

  24. amistre64
    • 5 years ago
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    thats my mistake....sqrt(1-sin^2) = cos^2...not cos; got it:)

  25. amistre64
    • 5 years ago
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    ....nah, I was right to begin with....if you got a show you can watch, now would be the time to do it ;)

  26. amistre64
    • 5 years ago
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    This is what I get for the area 978.338522 and Im like 67% sure im right lol

  27. amistre64
    • 5 years ago
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    yep, 98% sure 978.338522

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