anonymous
  • anonymous
can anyone define definite and indefinite integrals???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
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
anonymous
  • anonymous
ok...now, can you tone that down a bit?
amistre64
  • amistre64
not really, thats really the simplest explanation I got :)

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amistre64
  • amistre64
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
amistre64
  • amistre64
definite= whats the area of a sealed box? indefinite = whats the area of a box that has no top bottom or sides?
anonymous
  • anonymous
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?
amistre64
  • amistre64
i could, if I can ;)
anonymous
  • anonymous
\[\int\limits_{1}^{2}\sqrt{x-1}\times dx\]
anonymous
  • anonymous
oh, and how do we tell if a problem is definite or indefinite?
amistre64
  • amistre64
if it is definite that elongates "s" will have numbers around it; if its indefinite, then the "s" will be bare naked.
anonymous
  • anonymous
ok so the problem above is definite
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
ok. i think im getting it. so to solve the problem, what would you do from there?
amistre64
  • amistre64
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 :)
anonymous
  • anonymous
lol. they will have nothing to shatter because i have yet to build any confidence in this subject...
anonymous
  • anonymous
I think you're missing an x in this question..
amistre64
  • amistre64
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...
anonymous
  • anonymous
oh yes i am! sorry. its before \[\sqrt{x-1}\]
amistre64
  • amistre64
we will define u = x-1 ; therefore; du=1 we will also rewrite the expresion as an exponent to make life easier for us
amistre64
  • amistre64
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
amistre64
  • amistre64
2u^(3/2) 2(x-1)^(3/2) ------- = ------------- is our F(x) suited up function, 3 3
anonymous
  • anonymous
your already explaining it better than my teacher!:)
amistre64
  • amistre64
Now to determine the answer to our problem we do this: F(2) - F(1) and that equals the answer we seek. and thanx :)
amistre64
  • amistre64
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
amistre64
  • amistre64
didnt quite line up there did it :) 2/3 is the answer I get
anonymous
  • anonymous
thank you!!! this makes much more sense! are you a teacher?
amistre64
  • amistre64
no im not, im just old and smelly :)
anonymous
  • anonymous
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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