![]() an indefinite integral, defined by the "has the correct derivative" test, and written $ \int f(t) dt$,.a definite integral, defined via limits of Reimann sums, and written $ \int_a^b f(t)dt $, and.If you now understand this, and in particular understand the difference between With just a bit of renaming and re-arranging. Which is the same as the equation from the 1st thm So if you look back at the equation in the 2nd thm, and imagine that you also know that for your $F$, $F(a) = 0$, notice that the equation ![]() In fact, it is the unique one that has the value $0$ at $a$. ![]() ![]() So the 2nd thm actually gives us a whole family of $F(x)$ functions that will work, and $g(x)$ is one of them. As to why it's pulled out separately, and isn't just a step in the Second Theorem, I'm not sure, but it's pretty established as a separate theorem by now, so it might just be historical tradition.)Īs to your question about the relation between the $g$ of the First Theorem and the $F$ of the Second Theorem: Note that $F$ isn't uniquely defined we know that we can add or subtract a constant to any existing $F$ that passes the 2nd thm's test, and get another function that also passes the same test. Well, it is used in proving the second theorem. (If the Second Theorem is the meat of the matter you might ask why we even bothered with the First Theorem. If we didn't have the Second Theorem, our calculus class might end right here, with your prof saying "Well, that's how an integral is defined, but there aren't many we can evaluate, and it's a pain to try to evaluate new ones, as we have to look for neat, tricky patterns in the Riemann sums each time." Instead we just throw out a guess, check that it has the right derivative, and we are done. This is what is going let us go forward and start actually evaluating integrals. This is amazing - no Riemann sums, no limits, just find an $F$ that passes the test, and you can evaluate definite integrals with two function evaluations and a subtraction. And at this point we're absolute whizzes at finding the derivatives of function. The Second Theorem tells us if we have such an $F$, then (and here's where the sun breaks through the clouds and a chorus of angels starts singing), we can evaluate integrals of $f$ by just evaluating $F$ at two points, and subtracting. It can be any wild-ass function, except it does have to pass a test: $F$'s derivative has to be equal to $f$, at each point in $$. How is $F$ defined in terms of $f$? It's not, at least not like $g$ was. Also notice that one of the things that's true about $g$, which appears to be to obvious to mention, is that $g(a) = 0$. But the First Theorem does give us some information about how $g$ behaves, and that's going to help us in proving the Second Theorem. $g$ is explicitly defined, but it's a real pain in the ass to evaluate $g$ at even one point - a Riemann sum and a limit each time. Okay, how do you find that? Well, you've got to construct a bunch of Riemann sums, and then prove that they converge to a limit as the mesh gets smaller, and then that limit is the value of $g$ at $7$. So the First Theorem defines a function $g(x)$ more-or less explicitly: What's, say, $g(7)$? Well, (assuming $7$ is between $a$ and $b$), it is $\int_a^7 f(t)dt $. I like to understand these theorems as kind of a 1-2 punch, where the first theorem sets things up, and the second theorem knocks them down (where "knocking things down" = "evaluating definite integrals".)
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