Integral of sin3x from -a to a: Is My Answer Right?

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SUMMARY

The integral of sin(3x) from -a to a is confirmed to be 0, as sine is an odd function. The calculation steps provided, which involve evaluating the integral as [-cos(3x)/3] from -a to a, are accurate. The final result of 0 is derived from the property that the integral of an odd function over a symmetric interval around the origin yields zero. Therefore, the conclusion that the integral equals 0 is correct.

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PrudensOptimus
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Take the integral of:

sin3x, with respect to x. [-a,a] - interval

I end up getting 0/3:

= [-cos(3x)/3]

= [-cos(3a) - (- cos (-3a))]/3

= [-cos (3a) + cos (3a)]/3 ---> cos(-3a) = cos 3a

= 0/3 = 0

I think I did something wrong, right?
 
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No you are correct , sine is an odd function and integrating it over any interval symmetric about the origin will give you 0.
 
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Yes, your answer is correct. The integral of sin3x from -a to a is indeed 0. Your steps are also correct, and you did not make any mistakes. Well done!
 

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