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The Timpson Lyceum · Real Analysis (2020)

Lecture 22 Part 2 (Real Analysis 2020)

Overview

This final part of the course collects the corollaries that follow from the Second Fundamental Theorem of Calculus (FTC). Having established that differentiating an integral with a variable upper limit returns the integrand, we extend this in several directions: integrating toward a lower variable limit, adopting conventions for reversed and degenerate bounds, splitting an integral additively across an arbitrary triple of points, freeing the base point from the endpoint of the interval, and finally combining FTC with the chain rule to differentiate integrals whose upper limit is itself a function. Together these results turn the FTC into a flexible toolkit for differentiating integral-defined functions.

Key Concepts

Second FTC (starting point). For ff continuous on [a,b][a,b] and a≤x≤ba \le x \le b, ddx∫axf(t) dt=f(x).\frac{d}{dx}\int_a^x f(t)\,dt = f(x). The derivative "cancels" the integral, returning the integrand evaluated at the upper limit.

Corollary — variable lower limit. If instead the variable sits at the lower bound, the sign flips: ddx∫xbf(t) dt=−f(x).\frac{d}{dx}\int_x^b f(t)\,dt = -f(x). This is proved by splitting ∫abf=∫axf+∫xbf\int_a^b f = \int_a^x f + \int_x^b f. The left side is a constant in xx, so differentiating gives 0=f(x)+ddx∫xbf0 = f(x) + \frac{d}{dx}\int_x^b f, hence the derivative equals −f(x)-f(x).

Definition — reversed and equal bounds. For an integrable ff and points c<dc < d, define

These are conventions (definitions), not theorems, but they make the next result hold for points in any order.

Additivity over three points. For an integrable ff and any three points x1,x2,x3x_1, x_2, x_3 (in any order), ∫x1x3f=∫x1x2f+∫x2x3f.\int_{x_1}^{x_3} f = \int_{x_1}^{x_2} f + \int_{x_2}^{x_3} f. The point x2x_2 need not lie between x1x_1 and x3x_3. There are only six orderings of three points; each reduces to the standard split once the reversed-bound convention is applied to put the limits in increasing order.

Corollary — FTC with an arbitrary base point. Let ff be continuous on an open interval (a,b)(a,b) and let x0x_0 be any point of it. Then for all x∈(a,b)x \in (a,b), ddx∫x0xf(t) dt=f(x).\frac{d}{dx}\int_{x_0}^{x} f(t)\,dt = f(x). Unlike the original statement, the base point x0x_0 is any interior point, not a fixed endpoint aa. The proof writes ∫x0xf=∫x0af+∫axf\int_{x_0}^x f = \int_{x_0}^a f + \int_a^x f (legitimate even if a∉(a,b)a \notin (a,b), since the value of a definite integral does not depend on whether an endpoint lies in the open interval). Differentiating, the first term is constant (derivative 00) and the second gives f(x)f(x).

Corollary — chain-rule (Leibniz) form. Let ff be continuous on (a,b)(a,b) and let h:(c,d)→Rh:(c,d)\to\mathbb{R} be differentiable with image contained in (a,b)(a,b), so the composition is defined. Fix x0∈(a,b)x_0 \in (a,b). Then ddx∫x0h(x)f(t) dt=f(h(x)) h′(x).\frac{d}{dx}\int_{x_0}^{h(x)} f(t)\,dt = f\bigl(h(x)\bigr)\,h'(x). This is the chain rule applied to the integral-defined function. Define F(x)=∫x0xf(t) dtF(x) = \int_{x_0}^{x} f(t)\,dt, so the quantity above is G(x)=F(h(x))G(x) = F(h(x)). By the chain rule G′(x)=F′(h(x)) h′(x)G'(x) = F'(h(x))\,h'(x), and by the FTC F′=fF' = f, giving G′(x)=f(h(x)) h′(x)G'(x) = f(h(x))\,h'(x).

Worked Example

Differentiate an integral whose upper limit is a function — the Leibniz form combining the FTC with the chain rule.

ddx∫x0h(x)f(t) dt=F′(h(x)) h′(x),where F(x)=∫x0xf(t) dt=f(h(x)) h′(x)(since F′=f by the FTC).\begin{aligned} \frac{d}{dx}\int_{x_0}^{h(x)} f(t)\,dt &= F'\bigl(h(x)\bigr)\,h'(x), \quad \text{where } F(x)=\int_{x_0}^{x} f(t)\,dt \\ &= f\bigl(h(x)\bigr)\,h'(x) \quad (\text{since } F' = f \text{ by the FTC}). \end{aligned}

Concretely, for ddx∫0x2cos⁡t dt\dfrac{d}{dx}\displaystyle\int_{0}^{x^2} \cos t\,dt we take f(t)=cos⁡tf(t)=\cos t and h(x)=x2h(x)=x^2:

ddx∫0x2cos⁡t dt=f(h(x)) h′(x)=cos⁡ ⁣(x2)⋅2x=2xcos⁡ ⁣(x2).\begin{aligned} \frac{d}{dx}\int_{0}^{x^2} \cos t\,dt &= f\bigl(h(x)\bigr)\,h'(x) \\ &= \cos\!\left(x^2\right)\cdot 2x = 2x\cos\!\left(x^2\right). \end{aligned}

Summary