## anonymous 5 years ago can anyone define definite and indefinite integrals???

1. 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

2. anonymous

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

3. amistre64

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

4. 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

5. amistre64

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

6. 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?

7. amistre64

i could, if I can ;)

8. anonymous

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

9. anonymous

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

10. amistre64

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

11. anonymous

ok so the problem above is definite

12. 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 :)

13. anonymous

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

14. 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 :)

15. anonymous

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

16. anonymous

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

17. 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...

18. anonymous

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

19. 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

20. 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

21. amistre64

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

22. anonymous

23. 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 :)

24. 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

25. amistre64

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

26. anonymous

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

27. amistre64

no im not, im just old and smelly :)

28. 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..