Is My Reimann Sum Solution Correct for 6/(1+2x) from 0 to 2 with 4 Subintervals?

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In summary, the conversation discussed approximating the value of an integral using 4 subintervals of equal width and left endpoints. The integral was simplified to a sum and then solved using the given values, resulting in a final answer of 25/4. The individual also asked for confirmation on their work and expressed gratitude for the help received on previous tests.
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Qube
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Homework Statement



Approximate the value of the integral of 6/(1+2x) with respect to x from 0 to 2. Use 4 subintervals of equal width and use the left endpoints.

Homework Equations



delta x = (b-a)/N

The Attempt at a Solution



The integral is the sum of 6/(1+i) from i = 0 to i = N-1 or 3 all multiplied by delta x, or 1/2.

This yields:

(1/2)(6 + 3 + 2 + (6/4))

= 3 + 3/2 + 1 + 3/4 = 4 + 1.5 + 0.75 = 5.5 + 0.75 = 6.25 = 25/4.

1) This is correct without the N term in the sum, right? I'm wondering because usually I have to take the limit as N approaches infinity but the N doesn't exist here, since N has already been defined.

2) Also is my work correct in general? I'm still getting a hang of this Reimann sum notation with the indices and n subintervals and the x star notation. I'll have to learn Latex another day!
 
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Qube said:

Homework Statement



Approximate the value of the integral of 6/(1+2x) with respect to x from 0 to 2. Use 4 subintervals of equal width and use the left endpoints.

Homework Equations



delta x = (b-a)/N

The Attempt at a Solution



The integral is the sum of 6/(1+i) from i = 0 to i = N-1 or 3 all multiplied by delta x, or 1/2.

This yields:

(1/2)(6 + 3 + 2 + (6/4))

= 3 + 3/2 + 1 + 3/4 = 4 + 1.5 + 0.75 = 5.5 + 0.75 = 6.25 = 25/4.

1) This is correct without the N term in the sum, right? I'm wondering because usually I have to take the limit as N approaches infinity but the N doesn't exist here, since N has already been defined.

2) Also is my work correct in general? I'm still getting a hang of this Reimann sum notation with the indices and n subintervals and the x star notation. I'll have to learn Latex another day!

Yes, it's fine. And your work is fine in general. I don't know why you are asking all of these questions. As far as Riemann sums go, you can do a left sum, in which case i=N doesn't appear or a right sum in which case i=0 doesn't appear. These finite sums are just approximations to the true integral which you get by taking N->infinity. In which case it shouldn't matter whether it's left or right or something else.
 
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Thanks again! Well, you guys helped me get a perfect score on my last chem test when the average was a 68% and the historic average was in the 50s. I got help here for free while my friends paid for help and still did dismally, or completely failed. I just want to make sure I'm right because I'd like to have a repeat performance on my calculus test! On the last calc test I actually scored over a 100 while the average was in the 70s, thanks again to PF.

Physics Forum is a grade saver! Thanks again to everyone who helped me and keeps helping me :)!
 
  • #4
Qube said:
Thanks again! Well, you guys helped me get a perfect score on my last chem test when the average was a 68% and the historic average was in the 50s. I got help here for free while my friends paid for help and still did dismally, or completely failed. I just want to make sure I'm right because I'd like to have a repeat performance on my calculus test! On the last calc test I actually scored over a 100 while the average was in the 70s, thanks again to PF.

Physics Forum is a grade saver! Thanks again to everyone who helped me and keeps helping me :)!

Scoring over 100% is certainly worth a mention. At this point you should be getting a little more confident. Congratulations!
 

FAQ: Is My Reimann Sum Solution Correct for 6/(1+2x) from 0 to 2 with 4 Subintervals?

What is a Riemann Sum with No N term?

A Riemann Sum with No N term is a mathematical concept used to approximate the area under a curve by dividing it into smaller rectangles. Unlike a normal Riemann Sum, which uses a specific number of rectangles (N) to approximate the area, a Riemann Sum with No N term does not have a set number of rectangles.

How is a Riemann Sum with No N term calculated?

A Riemann Sum with No N term is calculated by taking the limit of the sum of the areas of the rectangles as the width of the rectangles approaches zero. This is represented by the integral symbol (∫) in calculus notation.

What is the purpose of using a Riemann Sum with No N term?

The purpose of using a Riemann Sum with No N term is to get a more accurate approximation of the area under a curve. By allowing the number of rectangles to approach infinity, the rectangles become infinitely small and the sum of their areas becomes a more precise approximation of the actual area under the curve.

What are the limitations of using a Riemann Sum with No N term?

One limitation of using a Riemann Sum with No N term is that it can be computationally intensive. As the number of rectangles approaches infinity, the calculation becomes more complex and time-consuming. Additionally, the accuracy of the approximation depends on the shape of the curve and the choice of intervals used to divide the curve.

What are some practical applications of a Riemann Sum with No N term?

Riemann Sums with No N term have various applications in science and engineering, such as calculating the volume of irregularly shaped objects, estimating the amount of chemical reactions, and predicting the displacement of a moving object. They are also used in financial modeling to estimate the value of complex investments and in statistics to approximate probabilities.

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