Proof. Let be a partition of . Then is constant on the open subintervals of , so,
We compute,
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5 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 ?
No. He reordered the indexes. Observe that below the sigma he starts counting from k = 1, in this way talking about the subdivision $x_{k – 1} makes sense.
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![\[ \int_0^2 [t^2] \, dt = 5 - \sqrt{2} - \sqrt{3}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-196c27b47d66e6a14e1e9d7300138c76_l3.png "Rendered by QuickLaTeX.com")
![\[ \int_{-3}^3 [t^2] dt. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-93ae8936dcd88821c51250a993f3c681_l3.png "Rendered by QuickLaTeX.com")
be a partition of ![[0,2]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d2292f5d2ce11b1d657528e0cf2e91a6_l3.png "Rendered by QuickLaTeX.com")
. Then ![[t^2]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-6e5033486c298b86fdc770f39bd70920_l3.png "Rendered by QuickLaTeX.com")
is constant on the open subintervals of 
, so,
![\begin{align*} \int_0^2 [t^2] \, dt &= \sum_{k=1}^5 s_k \cdot (x_k - x_{k-1}) \\ &= (0 \cdot 1) + (1)(\sqrt{2}-1) + (2)(\sqrt{3}- \sqrt{2}) + (3)(2 - \sqrt{3}) \\ &= 5 - \sqrt{2} - \sqrt{3}. \qquad \blacksquare \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a8d4081366525e075f3006a81b802af7_l3.png "Rendered by QuickLaTeX.com")
![\begin{align*} \int_{-3}^3 [t^2] \, dt &= 2 \cdot \int_0^3 [t^2] \, dt \\ &= 2 \cdot \left( \int_0^2 [t^2] \, dt + \int_2^3 [t^2] \, dt \right) \\ &= 2 \cdot \left( 5 - \sqrt{2} - \sqrt{3} + 4(\sqrt{5}-2) + 5(\sqrt{6} - \sqrt{5}) + 6(\sqrt{7}-\sqrt{6}) \right.\\ & \qquad \quad \left.+ 7(\sqrt{8} - \sqrt{7}) + 8(3 - \sqrt{8}) \right) \\ &= 2 (21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7f5ee29c4c0bd44569fe2e49a4b26298_l3.png "Rendered by QuickLaTeX.com")
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 
?
No. He reordered the indexes. Observe that below the sigma he starts counting from k = 1, in this way talking about the subdivision $x_{k – 1} makes sense.