- #1
Artusartos
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I was trying to prove this, but something went wrong and I'm not sure what...
We know that the greatest value of sin(1/x) can be 1 and the lowest value can be -1...
We know that the upper darboux sum is...
[tex]U(f,P) = sup(f[t_{k-1} , t_t]) (t_k - t_{k-1}) = (1) (1-(-1)) = 2[/tex]
We also know that the lower darboux sum is...
[tex]L(f,P) = inf(f[t_{k-1} , t_t]) (t_k - t_{k-1}) = (-1) (1-(-1)) = -12[/tex]
So we know that...
[tex]U(f,P) - L(f,P) = 2 - (-2) = 4 > \epsilon[/tex]
Can anybody tell me where I went wrong?
Thanks in advance
We know that the greatest value of sin(1/x) can be 1 and the lowest value can be -1...
We know that the upper darboux sum is...
[tex]U(f,P) = sup(f[t_{k-1} , t_t]) (t_k - t_{k-1}) = (1) (1-(-1)) = 2[/tex]
We also know that the lower darboux sum is...
[tex]L(f,P) = inf(f[t_{k-1} , t_t]) (t_k - t_{k-1}) = (-1) (1-(-1)) = -12[/tex]
So we know that...
[tex]U(f,P) - L(f,P) = 2 - (-2) = 4 > \epsilon[/tex]
Can anybody tell me where I went wrong?
Thanks in advance