The FTC isn't what professors make it sound like
The first fundamental theorem of calculus is really just a shortcut that tells you when an integral is actually calculable without getting lost in Riemann sums. Most people learn it backwards — they see the limit-of-a-sum definition first, then later discover that evaluating integrals by hand is basically just finding antiderivatives. The theorem itself says something very specific and limited: if f is continuous on [a,b] and F is any antiderivative of f, then the definite integral from a to b of f(x)dx equals F(b) minus F(a). That's it. The proof takes about three lines if you already know the mean value theorem for integrals, but most textbooks spend six pages on it because they're trying to justify the entire enterprise of calculus courses. Here's where people go wrong. They treat the theorem as if it guarantees you can integrate anything. It doesn't. The continuity requirement is not optional, and the antiderivative must actually exist on the closed interval. I ran into this once when someone was trying to use the theorem to evaluate an integral of a function that had a jump discontinuity at an interior point. The function was piecewise — equal to x^2 on [0,1] and equal to x+2 on (1,3]. They plugged in the endpoints and subtracted, got some answer, and then spent two days debugging why their numerical integration routine disagreed. The function isn't continuous at x=1, so the FTC doesn't apply across the whole interval. The fix is splitting the integral at the discontinuity and applying the theorem separately on each subinterval, then summing the results. That's not mentioned prominently in any intro textbook I've ever seen. Another practical detail that gets glossed over: the antiderivative F only needs to exist, not be unique. Any constant works. People get confused by this and spend unnecessary time trying to find the "right" antiderivative when you can literally pick C=0 and move on. The constants cancel in the subtraction F(b)-F(a). There's no right choice.
The theorem also fails in cases where the function is integrable but has no elementary antiderivative. Functions like e^(-x^2), sin(x)/x, and the elliptic integrals are the classic examples. They have well-defined definite integrals. They're continuous everywhere. The FTC is still technically valid — an antiderivative exists, it's just not expressible in terms of elementary functions. So you can't use the theorem as a computational tool even though the theorem itself holds. This distinction matters more than you'd think when you're actually working with applied problems. Numerical quadrature methods like Gaussian quadrature or adaptive Simpson's rule become necessary, and those are completely different from the antiderivative approach the FTC describes. The version of the theorem that actually shows up in engineering work is sometimes called the second form, where the integral defines a function and the derivative of that function gives you back the original integrand. That's the one used in control theory and differential equations. Both forms are equivalent, but the version most students memorize — the evaluation formula — is the one with the strictest conditions and the most common points of failure. If your function isn't continuous, if your interval is infinite, or if you're dealing with a parameter-dependent integrand, you're outside the clean statement of the theorem and into a different set of tools entirely.
