Solve Integral: 50 + 14sin(πt/12) | 0 to 12

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Homework Help Overview

The discussion revolves around solving the integral of the function 50 + 14sin(πt/12) over the interval from 0 to 12. The subject area is calculus, specifically focusing on integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore various methods for solving the integral, including potential use of trigonometric substitution and integration by parts. There is discussion about the appropriateness of these methods, with some suggesting a simpler substitution might suffice. Questions arise regarding the correct substitution and the handling of limits of integration.

Discussion Status

Participants have provided differing perspectives on the best approach to take, with some suggesting straightforward substitutions while others discuss the integral's components. There is acknowledgment of multiple interpretations, and guidance has been offered regarding the integration process without reaching a definitive consensus.

Contextual Notes

Some participants note the importance of changing limits when applying substitutions, and there is mention of constants that need to be factored out during integration. The original poster expresses uncertainty about the methods to use, indicating a need for clarification on the integral's setup.

ziddy83
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integrals, oh yeah...

hey what's up,
Ok so...i am having a little bit of a problem on solving the following integral...

[tex]\int_{0}^{12} 50 + 14 sin\frac{\pi t}{12} dt[/tex]

would i use...trig substitution or by parts?... :rolleyes: yeah...i need some help..thanks.
 
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No,need for part integration,a simple,obvious substitution would do it.Pay attention to the change of limits (of integration).

Daniel.
 
wait maybe i don't see it...would i let u= sin pi t/12? and then get du...?
 
Last edited:
The [tex]\int (a + b) = \int a + \int b[/tex]

So the only thing that seems tricky is the second part.

Let [tex]u = \frac{\pi{t}}{12} , du = \frac{\pi}{12}[/tex]

Set [tex]\frac{\pi}{12} = 14dx[/tex] and put the new integral in the form of

[tex]C*\int sin(u)du[/tex]
 
You don't need any substitution. You have the following:

The second part is:

[tex]14 \int \sin (kt) dt[/tex]

where [tex]k=\frac{\pi}{12}[/tex]

The anti-derivative of sin kt is [tex]\frac{-\cos kt}{k}[/tex]
 
Or you could do it like that. Good point. :rolleyes: :-p
 
great...thanks a lot guys...i got the right answer for the problem. i just used the substitution of [tex]u = \frac{\pi{t}}{12}[/tex] and brought out all of the constants. Thanks again.
 

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