Proof. Let be a partition of . Then is constant on the open subintervals of , so,
We compute,
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4 comments
M.d.S says:
How come does the integral from -3 to 0 equals the one from 0 to 3?
The function from -1 to 0 equals to -1, while the function from 0 to 1 equals 0 != -(-1)
from -2 to -1 it is -2, while the positive part is 1 from 1 to 2
i guess i just didn’t understand
Anonymous says:
Actually the function from 0 to -1 is also equal to 0 ( 0 being the smallest integer between the interval). \int_{-3}^{3} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{-3}^{0} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{0}^{3} [(-t)^{2}] \ dx = 2\int_{0}^{3} [t^{2}] \ dx
Anonymous says:
Don’t forget about the exponent 2, the smallest integer between [(-1)^2,0] is 0. \int_{-3}^{3} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{-3}^{0} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{0}^{3} [(-t)^{2}] \ dx = 2\int_{0}^{3} [t^{2}] \ dx
Jamie Corkhill says:
Should the on your sigma be , and not , since the partition starts as ?
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How come does the integral from -3 to 0 equals the one from 0 to 3?
The function from -1 to 0 equals to -1, while the function from 0 to 1 equals 0 != -(-1)
from -2 to -1 it is -2, while the positive part is 1 from 1 to 2
i guess i just didn’t understand
Actually the function from 0 to -1 is also equal to 0 ( 0 being the smallest integer between the interval). \int_{-3}^{3} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{-3}^{0} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{0}^{3} [(-t)^{2}] \ dx = 2\int_{0}^{3} [t^{2}] \ dx
Don’t forget about the exponent 2, the smallest integer between [(-1)^2,0] is 0. \int_{-3}^{3} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{-3}^{0} [t^{2}] \ dx = \int_{0}^{3} [t^{2}] \ dx + \int_{0}^{3} [(-t)^{2}] \ dx = 2\int_{0}^{3} [t^{2}] \ dx
Should the on your sigma be , and not , since the partition starts as ?