kamaruddin
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- Oct 30, 2005
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I have solved this question by myself.
Thanks anyway.
Thanks anyway.
I have solved this question by myself.
- Thread starter kamaruddin
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
You REALLY need to get the idea of the first one. How would you do it with a single integral?
\(\displaystyle \int_{0}^{2}{2*(2-x)} dx = 4\)
or
\(\displaystyle \int_{0}^{4}{(1/2)*(4-y)} dy = 4\)
Compare to the double integral version:
\(\displaystyle \int_{0}^{2}{\int_{0}^{2*(2-x)}{1} dy} dx = 4\)
or
\(\displaystyle \int_{0}^{4}{\int_{0}^{(1/2)(4-y)}{1} dx} dy = 4\)
\(\displaystyle \int_{0}^{2}{2*(2-x)} dx = 4\)
or
\(\displaystyle \int_{0}^{4}{(1/2)*(4-y)} dy = 4\)
Compare to the double integral version:
\(\displaystyle \int_{0}^{2}{\int_{0}^{2*(2-x)}{1} dy} dx = 4\)
or
\(\displaystyle \int_{0}^{4}{\int_{0}^{(1/2)(4-y)}{1} dx} dy = 4\)