Schrodinger's Dog
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Ok I'm sure everyone is bored to death by this by now, but I'm finding it quite interesting and quite informative for someone who's forget a lot about integration, so with that in mind it might be of use to someone, here's how he justifies the a in the integral and calling it "definite".
I can't see a problem with this, but then I'm seriously not that good, or that au fait atm with it all, although it's coming back gradually.
(2)... The point about the constants is an important one. If, for example, you have I = e^{ax}dx, then you can effectively get rid of the constant, a, from being inside the integral, by simply substituting for, say, t = ax. Differentiating this, we get dt = a dx. Rearranging, therefore dx = dt/a. Substitute in the integral, for ax, and dx, and we now get,
I = (e^{t}dt)/a .We divide I by the factor a, but a has now come outside the integral of (e^{t}dt) .
(3)... We can do a similar substitution for pretty well any integral.So in general, if I = f(ax)dx, where f is some function, then I=(f(t)dt)/a .
I can't see a problem with this, but then I'm seriously not that good, or that au fait atm with it all, although it's coming back gradually.