## Theorem Statement If $f$ is continuous on $[a, b]$ and $F(x) = \int_a^x f(t)\, dt$, then: $F'(x) = f(x) \quad \text{for all } x \in (a, b)$ Or in Leibniz notation: $\frac{d}{dx}\int_a^x f(t)\, dt = f(x)$ ### What It Means * Integration and differentiation are inverse operations. * Every continuous function has an antiderivative. * The rate of change of accumulated area equals the height of the function. --- ## Rigorous Proof ### Step 1: Set Up the Difference Quotient For $h > 0$: $\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\left[\int_a^{x+h} f(t)\, dt - \int_a^x f(t)\, dt\right]$ ### Step 2: Apply Interval Additivity $\int_a^{x+h} f(t)\, dt = \int_a^x f(t)\, dt + \int_x^{x+h} f(t)\, dt$ Therefore: $\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\, dt$ > **Note:** This is the average value of $f$ on $[x, x+h]$. ### Step 3: Use Extreme Value Theorem Since $f$ is continuous on $[x, x+h]$, it attains: * Minimum $m_h = f(c_h)$ at some $c_h \in [x, x+h]$ * Maximum $M_h = f(C_h)$ at some $C_h \in [x, x+h]$ Thus: $m_h \leq f(t) \leq M_h \quad \text{for all } t \in [x, x+h]$ ### Step 4: Bound the Integral $m_h \cdot h \leq \int_x^{x+h} f(t)\, dt \leq M_h \cdot h$ Dividing by $h$: $m_h \leq \frac{1}{h}\int_x^{x+h} f(t)\, dt \leq M_h$ ### Step 5: Apply Squeeze Theorem As $h \to 0$: * $c_h \to x$ and $C_h \to x$ (both trapped in $[x, x+h]$) * By continuity of $f$: $\lim_{h \to 0} m_h = \lim_{h \to 0} M_h = f(x)$ Therefore: $\lim_{h \to 0} \frac{F(x+h) - F(x)}{h} = f(x)$ $\boxed{F'(x) = f(x)}$ --- ## Why the Chain Rule Is Needed for $u(x)$ ### The Problem FTC2 requires the upper limit to be exactly $x$. For: $G(x) = \int_a^{x^2} f(t)\, dt$ The upper limit is $u(x) = x^2$, not $x$. ### Where the Proof Breaks Form the difference quotient: $\frac{G(x+h) - G(x)}{h} = \frac{1}{h}\int_{x^2}^{(x+h)^2} f(t)\, dt$ Critical observation: The interval width is: $(x+h)^2 - x^2 = 2xh + h^2 \neq h$ ### The Interval Width Problem | Feature | Naked $x$ | Composite $u(x) = x^2$ | | :--- | :--- | :--- | | **Width** | $h$ | $2xh + h^2$ | | **Ratio** $\left(\frac{\text{width}}{h}\right)$ | $1$ | $2x + h \to 2x$ | ### Extracting the Derivative Rewrite: $\frac{1}{h}\int_{x^2}^{(x+h)^2} f(t)\, dt = \frac{(x+h)^2 - x^2}{h} \cdot \frac{1}{(x+h)^2 - x^2}\int_{x^2}^{(x+h)^2} f(t)\, dt$ $= (2x + h) \cdot (\text{average value of } f \text{ on } [x^2, (x+h)^2])$ Taking the limit: * Average value $\to f(x^2)$ by continuity * $(2x + h) \to 2x = u'(x)$ Result: $\lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = f(x^2) \cdot 2x$ ### General Formula $\frac{d}{dx}\int_a^{u(x)} f(t)\, dt = f(u(x)) \cdot u'(x)$ --- ## Variable in Lower Bound: Where the Negative Sign Emerges ### The Setup Define: $G(x) = \int_x^a f(t)\, dt$ ### Forming the Difference Quotient $G(x+h) - G(x) = \int_{x+h}^a f(t)\, dt - \int_x^a f(t)\, dt$ ### The Critical Step: Interval Additivity Apply the defining property: $\int_x^a f(t)\, dt = \int_x^{x+h} f(t)\, dt + \int_{x+h}^a f(t)\, dt$ Solving for the difference: $\int_{x+h}^a f(t)\, dt - \int_x^a f(t)\, dt = -\int_x^{x+h} f(t)\, dt$ **The necessity:** The algebra forces the negative sign because we are removing area from the left side of the interval. ### Completing the Proof $\frac{G(x+h) - G(x)}{h} = -\frac{1}{h}\int_x^{x+h} f(t)\, dt$ Taking the limit: $\lim_{h \to 0} \frac{G(x+h) - G(x)}{h} = -\lim_{h \to 0}\frac{1}{h}\int_x^{x+h} f(t)\, dt = -f(x)$ $\boxed{\frac{d}{dx}\int_x^a f(t)\, dt = -f(x)}$ ### Where the Necessity Emerges | Step | Upper Bound Variable | Lower Bound Variable | | :--- | :--- | :--- | | **Difference** | $\int_x^{x+h} f(t)\, dt$ | $-\int_x^{x+h} f(t)\, dt$ | | **Divide by $h$** | $\frac{1}{h}\int_x^{x+h} f(t)\, dt$ | $-\frac{1}{h}\int_x^{x+h} f(t)\, dt$ | | **Limit as $h \to 0$** | $f(x)$ | $-f(x)$ | --- ## The Deep Reason: Orientation The negative sign is unavoidable due to the algebraic structure of oriented integrals: $\int_b^a f(t)\, dt = -\int_a^b f(t)\, dt$ This is required by: $\int_a^b f + \int_b^a f = \int_a^a f = 0$ When solving $\int_{x+h}^a f - \int_x^a f$, the algebra demands: $= \left(-\int_a^{x+h} f\right) - \left(-\int_a^x f\right) = -\int_a^{x+h} f + \int_a^x f = -\left(\int_a^{x+h} f - \int_a^x f\right) = -\int_x^{x+h} f$ --- ## General Case: Both Bounds Vary Define: $H(x) = \int_{v(x)}^{u(x)} f(t)\, dt$ Split using constant reference point $a$: $H(x) = \int_{v(x)}^a f(t)\, dt + \int_a^{u(x)} f(t)\, dt = -\int_a^{v(x)} f(t)\, dt + \int_a^{u(x)} f(t)\, dt$ Differentiate: $\frac{dH}{dx} = -f(v(x)) \cdot v'(x) + f(u(x)) \cdot u'(x)$ ### General Leibniz Rule: $\boxed{\frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\, dt = f(u(x)) \cdot u'(x) - f(v(x)) \cdot v'(x)}$ --- ## Summaries & Key Takeaways ### Integral Derivative Rules | Integral | Derivative | Intuition | | :--- | :--- | :--- | | $\int_a^x f(t)\, dt$ | $f(x)$ | Upper bound moves right, adding area | | $\int_x^a f(t)\, dt$ | $-f(x)$ | Lower bound moves right, removing area | | $\int_a^{u(x)} f(t)\, dt$ | $f(u(x)) \cdot u'(x)$ | Chain rule for upper bound | | $\int_{v(x)}^a f(t)\, dt$ | $-f(v(x)) \cdot v'(x)$ | Chain rule for lower bound | | $\int_{v(x)}^{u(x)} f(t)\, dt$ | $f(u)u' - f(v)v'$ | Net rate of change | ### Key Principles * The derivative equals (value at upper bound) × (speed of upper bound) minus (value at lower bound) × (speed of lower bound). * The negative sign for the lower bound is algebraically unavoidable—it emerges from the orientation property $\int_b^a = -\int_a^b$. ### Connection to First FTC | Feature | First FTC | Second FTC | | :--- | :--- | :--- | | **Statement** | $\int_a^b f(x)\, dx = F(b) - F(a)$ | $\frac{d}{dx}\int_a^x f(t)\, dt = f(x)$ | | **Role** | Practical: evaluate integrals | Theoretical: guarantees antiderivatives exist | | **Dependency** | Uses Second FTC to ensure $F$ exists | Independent foundation | > **Final Note:** The Second FTC provides the theoretical bedrock that makes the First FTC possible.