I'm trying to find:
$$\int_0^{1} \sin 2\pi x \; dx =$$
My friend says the result should be zero. Is there any graphical way to prove it.
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1How have u achieved the result? – lab bhattacharjee Apr 25 '17 at 05:24
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The strangest result someone gets by hand. – Zaid Alyafeai Apr 25 '17 at 05:27
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1The result is exactly $0$ because $\int_0^1\sin{2\pi x}dx=(1/2\pi)(\cos{0}-\cos{2\pi})$. Looks like your result is a numerical approximation which is almost $0$ to the $10^{-5}$ – marwalix Apr 25 '17 at 05:28
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Just from heuristic reasoning, we see that the area under the curve from $0$ to $1/2$ is equal to, and opposite in sign from the area under the curve from $1/2$ to $1$. – Mark Viola Apr 25 '17 at 05:34
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Perhaps you could delete this post. – Mark Viola Apr 25 '17 at 05:35
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It says "Repeated deletion of answered questions can result in your account being blocked from asking. Are you sure you wish to delete?" What choices do I have? – Gin99 Apr 25 '17 at 05:38
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@Gin99, I doubt you can delete a question with an upvoted answer. – Zaid Alyafeai Apr 25 '17 at 06:39
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@Zaid Alyafeai, is it possible for you to close it? Please – Gin99 Apr 25 '17 at 06:42
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I voted to close. One more vote and it will be closed. – Zaid Alyafeai Apr 25 '17 at 06:46
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@Zaid Alyafeai. Thanks. At the others Please vote to close. Thanks – Gin99 Apr 25 '17 at 06:50
1 Answers
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Have you tried graphing the function?
Intuitively, what should the answer be? Because the area from $0$ to $0.5$ and the area from $0.5$ to $1$ effectively cancel each other out, the answer is $0$.
The difference between your answer and your friend's answer may be due to rounding errors.
As @marwalix pointed out, $\int_0^1 \sin(2\pi x)dx=(\frac{1}{2\pi})(\cos(0)−\cos(2\pi))$. Note that $(\cos(0)−\cos(2\pi)) = (1 - 1) = 0$. Then, since we are multiplying by $0$, the answer is also exactly $0$.
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@Gin99 I used Desmos, but any online graphing calculator will work. – DrowsyFugitive Apr 25 '17 at 07:41