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Couple of integrals questions (and with limits)

  • Thread starter dnt
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dnt
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Homework Statement



a) integral of e^(x^2) (ie, integral of e to the x squared power)

b) integral from -1 to 6 of the square root of 3 plus the absolute power of x (sorry i dont know how to make it look nice on these forums...havent learned the code for it yet)

c) this one is slightly different: express the integral from 1 to 5 of x/(x+1) as a limit


Homework Equations



n/a

The Attempt at a Solution




(a) integral of e^(x^2) (ie, integral of e to the x squared power)

i know the integral of e^x is itself. but i cant figure out how to adapt to the fact that the x is squared. i tried substitution but that didnt help much.

(b) integral from -1 to 6 of the square root of 3 plus the absolute power of x

since the function is always positive, i feel i can just take the integral of the square root of 3 + x (without the absolute power) and plug us -1 and 6 just like normal. is that correct?

(c) express the integral from 1 to 5 of x/(x+1) as a limit

always had a little trouble with these...

i know it will be: the limit as n -> oo (infinity) of the sum from i=1 to n of f(ci)*xi

(hope you can read that :) )

the f(ci) represents the heights of the infinite rectangles and the delta xi represents the equal widths of all the rectangles (and you just sum them all up).

but i always get confused on what to plug into the function itself. in this case x/(x+1). what is ci?

and also if xi = (b-a)/n (in this case 4/n) how does the integral change if it was from 2 to 6 because xi would again be 4/n? or does it not matter? i would think it should be different since the end points of the function change but xi seems to be the same.

thanks for any help!
 

Answers and Replies

  • #2
arildno
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a) Forget it, you won't be able to do that one.
b) Nope; remember that the absolute value sign is only about the x
c) You multiply the length of a SMALL INTERVAL with the function value to a chosen point within that interval, and let the number of intervals go to infinity.
 
  • #3
dnt
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a) thanks

b) i dont see why that should change anything. do i need to break it into two seperate integrals (one from -1 to 0 and another from 0 to 6)?

c) still a little confused on how that relates to the function and ci - what do i plug into the function to work with the fact that it represents the height of the infinite rectangles (am i logically saying this correct)?
 
  • #4
arildno
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b) that would be best.

c) Divide your interval into (n-1) sub-intervals as follows:
[tex]1=x_{1}<x_{2}<\cdots<x_{n}=5[/tex]

The difference [itex]x_{i+1}-x_{i}[/itex] is the length of the i'th sub-interval.

As your [itex]c_{i}[/itex] value, take for example the midpoint in the i'th interval, [itex]c_{i}=\frac{x_{i+1}+x_{i}}{2}[/itex]
 

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