Second Theorem

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We present the second fundamental theorem of calculus

We’ve seen the first fundamental theorem of calculus - that antiderivatives can be used to interpret the definite integral when the upper bound is a variable. But this doesn’t necessarily help us quickly compute definite integrals with constant bounds. This is where the second fundamental theorem of calculus comes in.

Video Lecture

Text with Additional Details

We’ve already established the needed groundwork for the second fundamental theorem, so we will start by stating it.

Second Fundamental Theorem of Calculus (FTC II): Let $F:\mathbb {R}\rightarrow \mathbb {R}$ be an antiderivative of an integrable function $f:\mathbb {R}\rightarrow \mathbb {R}$ , i.e. $F'(x) = f(x)$ . Fix $a,b\in \mathbb {R}$ , with $a < b$ . Then $\int \limits _a^b f(x)dx = F(b)-F(a)$

As usual, we translate this into more normal words. In essence, the second fundamental theorem of calculus asserts that, if you want to calculate a definite integral you follow these steps:

(a): Find an antiderivative of $f(x)$ (the thing being integrated) - we’ll denote it $F(x)$ .
(b): Evaluate that antiderivative at both bounds, i.e. $F(a)$ and $F(b)$ .
(c): Subtract the lower bound from the upper bound to find the definite integral, i.e. the definite integral is $F(b) - F(a)$ .
(d): ... profit?

This means that we no longer need to worry about taking a limit of a Riemann Sum, rather we can use the antiderivative knowledge from indefinite integrals, to solve our definite integrals!

It is also easy to overlook a rather intriguing fact from the above. Notice that, although there are infinitely many possible antiderivatives, we didn’t say we needed to find a specific anti-derivative, rather any antiderivative works. Importantly, this means we don’t have to worry about figuring out initial condition information or the “ $+C$ ” when using the second fundamental theorem of calculus. In reality the $+C$ part can be included, but you’d find that it ends up canceling itself out, which is why we don’t need to worry about it - it’s not that it isn’t there, just that it doesn’t end up making an impact.

From FTC I we established that there was a link between indefinite and definite integrals, at least when considering actual functions. But with FTC II we have that the relationship exists, even with constant bounds. In fact, FTC II gives us the key instrument to calculating definite integrals without needing to worry about complicated infinite sums; making computation of definite integrals just as “easy” as the indefinite integrals.