- #1
archaic
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- 214
There's one odd way to think about integration when it comes to interpreting it as a sum. suppose for a second that ##x## is in meters, you could think of distance as an infinite number ##n## of "points" in space, ##n→∞##, then in this case ##f(x)Δx## would mean that you now have ##nf(x)##.
Now, as usual, you think of ##Δx## as vanishing to ##0##, and then think of ##f(x)dx[meters]## as just, nearly, one point times ##f(x)##, and so summing all these would get you the sum of all ##f(x)##s, in meters that is, so if the integral equals ##5.36##, you consider it as the number of points in ##5.36## meters, which is infinite of course, but it's a way of "summing" infinity.
If you think of it this way, it kind of also seems directly deducible that the average of ##f(x)## is ##\int_a^b\frac{1}{b-a}f(x)dx##, you're diving by the "number of points" in ##b−a## meters, which corresponds to the number of ##f(x)##s. Again, infinite of course, but we're just manipulating infinities using names.
How bizarre is this, and what remarks could you make for such a way of thinking about integrals? Where can this pose problems?
Now, as usual, you think of ##Δx## as vanishing to ##0##, and then think of ##f(x)dx[meters]## as just, nearly, one point times ##f(x)##, and so summing all these would get you the sum of all ##f(x)##s, in meters that is, so if the integral equals ##5.36##, you consider it as the number of points in ##5.36## meters, which is infinite of course, but it's a way of "summing" infinity.
If you think of it this way, it kind of also seems directly deducible that the average of ##f(x)## is ##\int_a^b\frac{1}{b-a}f(x)dx##, you're diving by the "number of points" in ##b−a## meters, which corresponds to the number of ##f(x)##s. Again, infinite of course, but we're just manipulating infinities using names.
How bizarre is this, and what remarks could you make for such a way of thinking about integrals? Where can this pose problems?