Confusing Integration Question?

In summary, the question is asking if I(y)=-tan-1y+C, where C is a constant. When you integrate ∂/∂y=-1/(1+y2), you get I(y)=-tan-1y+C.
  • #1
unscientific
1,734
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Confusing Integration Question??

Homework Statement



The question is as attached.

The Attempt at a Solution



I am very confused by the question. Firstly is I(y) a function like (tan-1y) or an operator like ∫y?

Then they say I(y) - I(∞) = (∏/2) - tan-1y so I suppose that I(y) is basically just -tan-1y.I(y) - I(∞) = (∏/2) - tan-1y

lim(y→0) [ I(y) - I(∞) ] = (∏/2) - 0 = (∏/2)

But returning to the question, as you let y→0, all you get is

y(0) = ∫ sinx /x dx = tan-10 = 0

But that doesn't look right at all.I hope someone can clear up all the terminology they are using...
 

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  • #2


unscientific said:

Homework Statement



The question is as attached.



The Attempt at a Solution



I am very confused by the question. Firstly is I(y) a function like (tan-1y) or an operator like ∫y?

Then they say I(y) - I(∞) = (∏/2) - tan-1y so I suppose that I(y) is basically just -tan-1y.


I(y) - I(∞) = (∏/2) - tan-1y

lim(y→0) [ I(y) - I(∞) ] = (∏/2) - 0 = (∏/2)

But returning to the question, as you let y→0, all you get is

y(0) = ∫ sinx /x dx = tan-10 = 0

But that doesn't look right at all.


I hope someone can clear up all the terminology they are using...

The terminology is very clear: I(y) is a function of y, given by a formula in the picture.

RGV
 
  • #3


But when they take the limit of y->0, all you get is I(0) = ∫ sinx /x dx = tan-10 = 0When you take the limit of y-> ∞, all you get is I(∞) = 0 (If you substitute ∞ into the question)
 
Last edited:
  • #4


unscientific said:
Then they say I(y) - I(∞) = (∏/2) - tan-1y so I suppose that I(y) is basically just -tan-1y.

They say very plainly that [itex]I(\infty) = 0[/itex], so [itex]I(y) = \frac {\pi} {2} - tan^{-1}(y)[/itex], including [itex]y = 0[/itex].
 
  • #5


voko said:
They say very plainly that [itex]I(\infty) = 0[/itex], so [itex]I(y) = \frac {\pi} {2} - tan^{-1}(y)[/itex], including [itex]y = 0[/itex].

Oh I think I got it now! Cause when you integrate ∂/∂y = - 1/(1+y2) you get

I(y) = -tan-1y + C

Based on the first equation, we know y(∞) but not y(0).

I(∞) = 0, so you solve C = ∏/2...


Everything makes sense now, thank you!
 

Related to Confusing Integration Question?

1. What is integration and why is it important in science?

Integration is the process of combining different pieces of information or data to form a unified understanding or conclusion. In science, integration is important because it allows researchers to make connections and draw conclusions from various experiments, observations, and theories.

2. How do you know which pieces of information to integrate?

The process of integration involves critically evaluating and selecting information that is relevant, reliable, and credible. Scientists use their knowledge and expertise in a particular field to determine which information is most important and applicable to their research.

3. Can integration lead to biased conclusions?

Yes, integration can lead to biased conclusions if the selected information is incomplete or if there are underlying biases in the interpretation of the data. It is important for scientists to remain objective and consider all available information when integrating data.

4. How can integration be used to solve complex problems?

Integration can be a powerful tool for solving complex problems as it allows scientists to combine different perspectives and approaches to find a comprehensive solution. By integrating data from multiple sources, scientists can gain a better understanding of complex systems or phenomena.

5. What are some challenges of integrating data from different fields of study?

One challenge of integrating data from different fields of study is the potential for conflicting or contradictory information. It can also be difficult to combine data from different disciplines that use different terminology or methods. Collaboration and communication between scientists from different fields can help overcome these challenges.

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