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".) On the other hand, $g(x)=\int_a^x f(t)\ dt$ is a special antiderivative of $f$: it is the antiderivative of $f$ whose value at $a$ is $0$. That's why in the Fundamental Theorem of Calculus part 2, the choice of the antiderivative is irrelevant since every choice will lead to the same final result. In another words, if $F'(x)=G'(x)=f(x)$ for every $x$ in $I$, then there exists a constant $C$ such that $G(x)=F(x) C$ for every $x$ in $I$.įor example, $F(x)=\frac C$ for some constant $C$.Ī consequence of the theorem above is that, if $a$ and $b$ are numbers in $I$ and $F$ and $G$ are any antiderivative of $f$, then The antiderivatives of a function $f$ on an interval $I$ differ by a constant. Now, a key point about antiderivatives is that Note that $F(b) - F(a)$ is not two antiderivatives, but the difference of the values of one antiderivative at two numbers $a$ and $b$.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |