- anonymous

can anyone define definite and indefinite integrals???

- jamiebookeater

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- 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

ok...now, can you tone that down a bit?

- amistre64

not really, thats really the simplest explanation I got :)

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## More answers

- 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

definite= whats the area of a sealed box?
indefinite = whats the area of a box that has no top bottom or sides?

- 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

i could, if I can ;)

- anonymous

\[\int\limits_{1}^{2}\sqrt{x-1}\times dx\]

- anonymous

oh, and how do we tell if a problem is definite or indefinite?

- amistre64

if it is definite that elongates "s" will have numbers around it; if its indefinite, then the "s" will be bare naked.

- anonymous

ok so the problem above is definite

- 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

ok. i think im getting it. so to solve the problem, what would you do from there?

- 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

lol. they will have nothing to shatter because i have yet to build any confidence in this subject...

- anonymous

I think you're missing an x in this question..

- 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

oh yes i am! sorry. its before \[\sqrt{x-1}\]

- 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

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

2u^(3/2) 2(x-1)^(3/2)
------- = ------------- is our F(x) suited up function,
3 3

- anonymous

your already explaining it better than my teacher!:)

- 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

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

didnt quite line up there did it :)
2/3 is the answer I get

- anonymous

thank you!!! this makes much more sense! are you a teacher?

- amistre64

no im not, im just old and smelly :)

- 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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