Expressing Integral as Reimann Sum

In summary: To find the limits of integration, you just have to analyze the interval you are given and see what values of x would make sense to use as your endpoints. In this case, the interval is [0,pi] so the limits of integration would be 0 and pi since those are the endpoints of the given interval.Also, just a side note, we typically use "i" for the index of summation and "n" for the number of subintervals, not vice versa as you have in your solution. It's not a big deal, just a standard convention.Hope that helps!As the summation goes from i = 0 to N, N is our upper limit. So
  • #1
Qube
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Homework Statement



Express the integral of (sinx + 1) dx over the interval [0,pi] with a Reimann Sum using 4 subintervals of equal width and letting x_i^* be the left endpoint of the subinterval [x_(i-1), x_i]

Homework Equations



Δx = [b-a] / n

The Attempt at a Solution



Δx = pi/4.

The Reimann sum is

(pi/4)Ʃ(1 + sin (pi/4)i) with i = 0 and the upper bound being N-1 or 3.

Is this correct?
 
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  • #2
Qube said:

Homework Statement



Express the integral of (sinx + 1) dx over the interval [0,pi] with a Reimann Sum using 4 subintervals of equal width and letting x_i^* be the left endpoint of the subinterval [x_(i-1), x_i]

Homework Equations



Δx = [b-a] / n

The Attempt at a Solution



Δx = pi/4.
Yes, that is correct.

The Reimann sum is

(pi/4)Ʃ(1 + sin (pi/4)i) with i = 0 and the upper bound being N-1 or 3.[/quote]
What you have written is not correct but it may be just bad notation. Dividing [0, pi] into four equal intervals and taking x to be the left endpoint, the values of x would be 0, pi/4, pi/2, and 3 pi/4 (which would be (pi/4)i for i equal to 0, 1, and 2) or [tex](\pi/4)\sum_{i= 0}^3 (1+ sin((pi/4)i)[/tex] which, personally, I would just write as (pi/4)((1)+ (2)+ (1)+ (-1))= 3pi/4.

Do you see the difference? You have "sin(pi/4)i" which, for i from 0 to 3 is 0(sin(pi/4))+ 1(sin(pi/4) 2(sin(pi/4))+ 3(sin(pi4). That is, you have the "i" outside the "sin(pi/4)"- sin(pi/4) times i rather than sine of "pi/4 times i".

Is this correct?[/QUOTE]
 
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  • #3
Qube said:

Homework Statement



Express the integral of (sinx + 1) dx over the interval [0,pi] with a Reimann Sum using 4 subintervals of equal width and letting x_i^* be the left endpoint of the subinterval [x_(i-1), x_i]

Homework Equations



Δx = [b-a] / n

The Attempt at a Solution



Δx = pi/4.
Yes, that is correct.

The Reimann sum is

(pi/4)Ʃ(1 + sin (pi/4)i) with i = 0 and the upper bound being N-1 or 3.
Is this correct?

What you have written is not correct but it may be just bad notation. Dividing [0, pi] into four equal intervals and taking x to be the left endpoint, the values of x would be 0, pi/4, pi/2, and 3 pi/4 (which would be (pi/4)i for i equal to 0, 1, and 2) or [tex](\pi/4)\sum_{i= 0}^3 (1+ sin((pi/4)i)[/tex] which, personally, I would just write as (pi/4)((1)+ (2)+ (1)+ (-1))= 3pi/4.

Do you see the difference? You have "sin(pi/4)i" which, for i from 0 to 3 is 0(sin(pi/4))+ 1(sin(pi/4) 2(sin(pi/4))+ 3(sin(pi4). That is, you have the "i" outside the "sin(pi/4)"- sin(pi/4) times i rather than sine of "pi/4 times i".
 
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  • #4
I see now. The i should be inside the sine function.

Am I correct that the i refers to the subintervals? I know that N refers to the number of sub-intervals; so i just refers to which sub interval we're talking about, correct? Like i = 0 refers to the left endpoint of the first sub-interval and so on to i = N referring to the right endpoint of the last sub-interval.

Something like this?

https://scontent-b-mia.xx.fbcdn.net/hphotos-frc3/v/1441369_10201128827724538_205517185_n.jpg?oh=0a1438fcfb7b95a530d0f60982eac16f&oe=52915B86
 
  • #5
What does [itex] f(x_i)\Delta x [/itex] give you?

You should also draw your rectangles better than that. :wink:
 
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  • #6
That's the area of the rectangle (height * width).
 
  • #7
Okay, so you know that [itex] f(x_i) [/itex] is your height, given a point [itex] x_i [/itex] and that [itex] \Delta x [/itex] gives you the width.

So yes, technically you can think of each index in the summation as a point on the x-axis and a height given by [itex] f(x_i) [/itex] that breaks the function up into subintervals.

You do know that picture is wrong right?
 
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  • #8
How is the picture wrong? Apart from the fact I didn't draw rectangles?

Also thanks for clarifying. I stands for index. I forgot that.
 
  • #9
Qube said:
Apart from the fact I didn't draw rectangles?

That’s what I meant. :-p


i is just a dummy variable, i, j, k, whichever you use they all stand for index. The value you start at is lower limit of summation, and n is the upper limit.
 
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  • #10
Awesome, makes sense! Thanks again!
 
  • #11
Can x_i^* also be expressed as i*Δx, where i is the index of summation? I find this form easier to work with.

Also, how do I determine the bounds of integration given a Reimann sum? I know that [b-a]/n = Δx but what's my a and b?
 
  • #12
Qube said:
Can x_i^* also be expressed as i*Δx, where i is the index of summation? I find this form easier to work with.

Also, how do I determine the bounds of integration given a Reimann sum? I know that [b-a]/n = Δx but what's my a and b?

The a and b are the limits in the integral you are approximating$$
\int_a^b f(x) ~dx$$
 
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  • #13
Qube said:
Can x_i^* also be expressed as i*Δx, where i is the index of summation? I find this form easier to work with.

Also, how do I determine the bounds of integration given a Reimann sum? I know that [b-a]/n = Δx but what's my a and b?

[itex] x_i^* [/itex] can be chosen arbitrary in the sub intervals. However, it's customary to choose the left end point, right end point or midpoint.

[itex] x_i^* = x_{i-1}= a + (i -1) \Delta x [/itex] for the left endpoint
[itex] x_i^* = x_i = a + i \Delta x [/itex] for the right.

Now if delta x is equal to say 1/n than i* delta x is correct for the right endpoint , is this true for the left end point?
 
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  • #14
Student100 said:
Now if delta x is equal to say 1/n than i* delta x is correct for the right endpoint , is this true for the left end point?

Seems like it should be as long as we start with the correct value of i (0).
 
  • #15
Qube said:
Seems like it should be as long as we start with the correct value of i (0).

Yep. Just make sure you start with the correct value.

A and b is the interval of your definite integral ...


[tex] \int_a^b f(x) dx = \lim_{max \Delta x_i \rightarrow 0 } \sum _{i= 1}^n f(x_i^*) \Delta x_i [/tex]

For your HW example you'd have [tex] \int_0^\pi (sin x + 1) dx [/tex]
 
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  • #16
Student100 said:
Yep. Just make sure you start with the correct value.

Definitely. I kept what you mentioned in mind today and I stumbled upon this problem (upper bound of integral is 5; got cut off, sorry).

http://i.minus.com/jbrTfeyOpuB4ky.png

i is not 0 in this case. Nor is it 1. We have to consider i within the context of each problem. We're talking about 4 subintervals of equal width here, and we're talking about midpoints here. Given the interval and upon calculating Δx, I knew that i should start at 1.5 and end at 4.5. That way we get 4 subintervals each of width 1.

Thanks so much :).
 
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Related to Expressing Integral as Reimann Sum

What is "Expressing Integral as Reimann Sum"?

"Expressing Integral as Reimann Sum" is a mathematical concept where an integral can be approximated by a sum of smaller parts, called Reimann sums. It is a way of breaking down a complex integral into simpler parts for easier computation.

Why is it important to express integrals as Reimann sums?

Expressing integrals as Reimann sums is important because it allows us to approximate the value of complex integrals without having to use advanced techniques such as anti-derivatives or the Fundamental Theorem of Calculus. This makes it a useful tool in many fields of science and engineering.

How do you express an integral as a Reimann sum?

To express an integral as a Reimann sum, the interval over which the integral is being evaluated is divided into smaller sub-intervals. The function being integrated is then evaluated at a specific point within each sub-interval and multiplied by the width of that sub-interval. The sum of these products gives an approximation of the integral.

What is the difference between left, right, and midpoint Reimann sums?

Left, right, and midpoint Reimann sums differ in the point at which the function is evaluated within each sub-interval. In a left Reimann sum, the function is evaluated at the left endpoint of each sub-interval. In a right Reimann sum, the function is evaluated at the right endpoint of each sub-interval. In a midpoint Reimann sum, the function is evaluated at the midpoint of each sub-interval.

How do you improve the accuracy of a Reimann sum approximation?

The accuracy of a Reimann sum approximation can be improved by increasing the number of sub-intervals, which decreases the width of each sub-interval. This gives a better approximation of the integral because the function is evaluated at more points within the interval. Alternatively, using higher-order Reimann sums (such as the trapezoidal rule or Simpson's rule) can also improve accuracy.

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