- #1

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I solved the integral and got [x+sin(x) -1]

and that´s the farthest that I got. I would appreciate the help.

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- #1

- 6

- 0

I solved the integral and got [x+sin(x) -1]

and that´s the farthest that I got. I would appreciate the help.

- #2

fresh_42

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Can you explain, what ##sen## means and where the ##1## in your antiderivative comes from?Hello good evening to all, I was studying here and got stuck with this.

View attachment 107864

I solved the integral and got [x+sin(x) -1]

and that´s the farthest that I got. I would appreciate the help.

- #3

Mark44

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I believe it is another abbreviation for sine.Can you explain, what sen means

- #4

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Ok, the antiderivative should be

[ t + sin(t) ] from 0 to x.

Then I get [ x + sin(x) - 0 - sin(0) ] and Yep, I found my first mistake. sin(0) = 0

Btw I would like a solution without using L' Hopital rule, if there is one of course.

- #5

fresh_42

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Are you allowed to use ##\lim_{x \rightarrow 0}\frac{\sin (x)}{x}=1##? If so, you could cancel the quotient by an ##x##.

Ok, the antiderivative should be

[ t + sin(t) ] from 0 to x.

Then I get [ x + sin(x) - 0 - sin(0) ] and Yep, I found my first mistake. sin(0) = 0

Btw I would like a solution without using L' Hopital rule, if there is one of course.

- #6

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Yep, I can use sinx/x limit but it still leave me with 0 in the denominator

- #7

fresh_42

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And a ##2## in the nominator, which tells you?Yep, I can use sinx/x limit but it still leave me with 0 in the denominator

Edit: Or more directly: What happens, if you divide a real number by a number that gets closer and closer to zero?

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- #8

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it goes to + infinite, but Wolfram gives me as answer +inf and - inf.

- #9

fresh_42

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You'd probably ignored the absolute value of your denominator. How do you manage to get a negative number?it goes to + infinite, but Wolfram gives me as answer +inf and - inf.

- #10

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- #11

fresh_42

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##\sin x## and ##x## are pretty much the same in a small neighborhood around zero, aren't they? This means ##\frac{\sin x}{x} \, \approx \,1##.

If ##x## is negative, then ##\sin x## is as well.

- #12

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I've never seen it this way, thank you very much!

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