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Fundamental Theorem of Calculus

Part One

g'(x) & =\lim_{ h \to 0 } \frac{g(x+h)-g(x)}{h} \
& =\lim_{ h \to 0 } \frac{1}{h}\left( \int_{a}^{x+h} f(t) , dt -\int_{a}^{x} f(t) , dt \right) \
& =\lim_{ h \to 0 } \frac{1}{h}\left( \int_{x}^{x+h} f(t) , dt \right) \
& =\lim_{ h \to 0 } f(c)\text{ for }c \in [x,x+h]\quad \text{ by def. of average } \
& =f(x)\quad \text{ since } c\to x \text{ as } h\to 0
\end

−Nowconsider$H(x)=∫x2excos⁡(t2)dt$−Find$H′(x)$:let$u=x2$,$v=ex$−

\begin{align}
H & =\int_{u}^{v} \cos(t^{2}) , dt \
& =\int_{u}^{a} \cos(t^{2}) , dt +\int_{a}^{v} \cos(t^{2}) , dt \
& =-\int_{a}^{u} \cos(t^{2}) , dt +\int_{a}^{v} \cos(t^{2}) , dt \
H'(x) & =-f(u) \frac{ \mathrm{d}u }{ \mathrm{d}x }+f(v)\frac{ \mathrm{d}v }{ \mathrm{d}x } \
& =e^{x}\cos(e^{2x})-2x\cos(x^{4})
\end

- This result leads to the extended version: - If $H(x)=\int_{g(x)}^{h(x)} f(t) \, dt$, then $H'(x)=f(h(x))h'(x)-f(g(x))g'(x)$ - If you think about it, this just came from splitting up the integral, using the simple version, and chain rule ## Part 2 - If $f$ is cts on $[a,b]$, then $\int_{a}^{b} f(x) \, dx=F(b)-F(a)$ - Where $F'(x)=f(x)$ - $F(x)$ is the anti-Derivative - Do not forget the C ($F+C$ is also an antiderivative) - For FTC2, we choose any one $F$, and the $C$s will cancel out, but you must choose a single $F$ - $F(x)\big|_{a}^{b}=F(b)-F(a)$ > [!NOTE]- Proof: > Split $[a,b]$ into $n$ intervals $x_{0},x_{1},x_{2},\dots,x_{n}$ > Derivative#Mean Value Theorem > Begin with the right side >

\begin{align}
F(b)-F(a) & =F(x_{n})-F(x_{0}) \
& =(F(x_{1})-F(x_{0}))+(F(x_{2})-F(x_{1}))+\dots+(F(x_{n})-F(x_{n-1})) \
\text{ define } t_{i} & \in (x_{i-1},x_{i}) \
F(x_{i-1})-F(x_{i}) & =F'(t_{i})(x_{i}-x_{i-1}) \text{ by MVT } \
& =f(t_{i})\Delta x_{i} \
F(b)-F(a) & =f(t_{1})\Delta x_{1}+f(t_{2})\Delta x_{2}+\dots+f(t_{n})\Delta x_{n} \
& =\sum_{i=1}^{n}f(t_{i})\Delta x_{i}
\end

Whichisjusta[[F0−Glossary/ConcreteConcepts/RiemannSum∥RiemannSum]]