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anonymous

  • 5 years ago

can anyone define definite and indefinite integrals???

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  1. amistre64
    • 5 years ago
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    definite integrals are integrals that have bounds associated with them so that the answers you receive have a defininte value. indefinite integrals are boundless and therefore have no definite value unless given an initial condition to anchor it to the graph

  2. anonymous
    • 5 years ago
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    ok...now, can you tone that down a bit?

  3. amistre64
    • 5 years ago
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    not really, thats really the simplest explanation I got :)

  4. amistre64
    • 5 years ago
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    definite means it has a specific and determinable value because it fits inside a box. indefinite means that it cannot be measured; so we attach a generic "+C" to it until we can define it better

  5. amistre64
    • 5 years ago
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    definite= whats the area of a sealed box? indefinite = whats the area of a box that has no top bottom or sides?

  6. anonymous
    • 5 years ago
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    ok. i understand that now thank you. but i really do not understand how it works. if i gave a problem could you explain it a little?

  7. amistre64
    • 5 years ago
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    i could, if I can ;)

  8. anonymous
    • 5 years ago
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    \[\int\limits_{1}^{2}\sqrt{x-1}\times dx\]

  9. anonymous
    • 5 years ago
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    oh, and how do we tell if a problem is definite or indefinite?

  10. amistre64
    • 5 years ago
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    if it is definite that elongates "s" will have numbers around it; if its indefinite, then the "s" will be bare naked.

  11. anonymous
    • 5 years ago
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    ok so the problem above is definite

  12. amistre64
    • 5 years ago
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    it is..... it is bounded by the x axis, the x = 1 line the x = 2 line and the y = sqrt(x-1) line; so its completely boxed in :)

  13. anonymous
    • 5 years ago
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    ok. i think im getting it. so to solve the problem, what would you do from there?

  14. amistre64
    • 5 years ago
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    you would have to integrate that sqrt(x-1) part; which in this cqse is quite simple....they give you easy ones to begin with to build your confidence so later down the road they can shatter it to pieces :)

  15. anonymous
    • 5 years ago
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    lol. they will have nothing to shatter because i have yet to build any confidence in this subject...

  16. anonymous
    • 5 years ago
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    I think you're missing an x in this question..

  17. amistre64
    • 5 years ago
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    if youve done derivatives, then integrating is suiting it back up. Derivatives dress down a function; and integrals suit it back up.... thats about it...

  18. anonymous
    • 5 years ago
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    oh yes i am! sorry. its before \[\sqrt{x-1}\]

  19. amistre64
    • 5 years ago
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    we will define u = x-1 ; therefore; du=1 we will also rewrite the expresion as an exponent to make life easier for us

  20. amistre64
    • 5 years ago
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    du = dx (x-1)^(1/2) dx -> u^(1/2) du add 1 to the exponent , and divide by the exponent+1 u^(1/2 + 2/2) ------------ 1/2 + 2/2

  21. amistre64
    • 5 years ago
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    2u^(3/2) 2(x-1)^(3/2) ------- = ------------- is our F(x) suited up function, 3 3

  22. anonymous
    • 5 years ago
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    your already explaining it better than my teacher!:)

  23. amistre64
    • 5 years ago
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    Now to determine the answer to our problem we do this: F(2) - F(1) and that equals the answer we seek. and thanx :)

  24. amistre64
    • 5 years ago
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    2(2-1)^(3/2) - 2(1-1)^(3/2) ------------------------- 3 2(1)^(3/2) - 2(0)^(3/2) ------------------------- 3 2(1) - 2(0) 2-0 2 ---------- = ----- = --- 3 3 3

  25. amistre64
    • 5 years ago
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    didnt quite line up there did it :) 2/3 is the answer I get

  26. anonymous
    • 5 years ago
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    thank you!!! this makes much more sense! are you a teacher?

  27. amistre64
    • 5 years ago
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    no im not, im just old and smelly :)

  28. anonymous
    • 5 years ago
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    hehe :D i am sure your not. but really, i thought i was a lost cause...but now there is hope! maybe..i need to try one on my own..

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