Another calc. question to check

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The discussion focuses on calculating the area bounded by the functions y=sin(x) and y=(x*pi)/12. The initial answer provided was A=1.969, which was later questioned due to potential errors in the integration limits and method. Participants clarified that the correct limits for integration should be from -2.446 to +2.446, and that the area calculation should consider both sides of the graph. It was noted that the area calculated for one side was 0.9845, which, when doubled, led to confusion about the final area. Ultimately, the correct area is expected to be less than 1, emphasizing the need for accurate integration techniques.
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Find the area bounded by y=sin(x) and y= x(pi)/12

the final answer i got was A= 1.969

Can anyone please confirm this?
 
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If that second function is y = x*pi/12 it is not correct
 
yeah..that's the second function..
well I am checking it over again, I am not sure how else i can do it
 
just to make sure, another way to write the second function is , (X/12) multiplied by pi
 
laker_gurl3 said:
just to make sure, another way to write the second function is , (X/12) multiplied by pi

That is correct. And you are trying to find the area bounded by the two functions. What integral are you doing, including the limits you are using?
 
i am using 2.446 and 0

2.446 2.446
| |
| sin (x) - | (x/12) multipled by pi
| |
0 0
 
Those are the correct limits. The correct answer is less than 1. Are you using a calculator function to integrate, or doing it by hand?
 
Well, the first part of the answer i got was .9845.. Then i multiplied by 2 to get the area of the other side... does that sound good?
 
laker_gurl3 said:
Well, the first part of the answer i got was .9845.. Then i multiplied by 2 to get the area of the other side... does that sound good?

OK sorry. I assumed you were talking about one side. My mistake.

You should have said your limts were from -2.446 to +2.446 with positive areas
 
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