Couple of integrals questions (and with limits)

In summary, for (a) the integral of e^(x^2) is not possible to solve, for (b) the absolute value sign only applies to the x value and does not change the integral, and for (c) to express the integral of x/(x+1) as a limit, divide the interval into sub-intervals and use the midpoint of each interval as c_{i}=\frac{x_{i+1}+x_{i}}{2}.
  • #1
dnt
238
0

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 don't 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 can't 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!
 
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  • #2
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
a) thanks

b) i don't see why that should change anything. do i need to break it into two separate 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
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]
 

1. How do you solve integrals with limits?

To solve integrals with limits, you first need to find the antiderivative of the integrand. Then, plug in the upper and lower limits into the antiderivative and subtract the resulting values to find the definite integral.

2. What is the purpose of using limits in integrals?

Limits in integrals help define the range or interval over which the integration is performed. This allows for more accurate and specific calculations, especially when dealing with functions that have discontinuities or infinite bounds.

3. What is the difference between definite and indefinite integrals?

A definite integral has specific limits or bounds, while an indefinite integral does not. This means that a definite integral will result in a single numerical value, while an indefinite integral will result in a function (with a constant of integration) that can be used to find the value of the integral at any point within its domain.

4. Can limits in integrals be negative or non-numeric?

Yes, limits in integrals can be negative or non-numeric, as long as they are within the domain of the function being integrated. The limits can also be variables or expressions.

5. How can I check if my solution to an integral with limits is correct?

You can check your solution by taking the derivative of the antiderivative you found. If the resulting function matches the original integrand, then your solution is correct. You can also use online calculators or graphing tools to visually confirm your solution.

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