<|im_start|>system
You are a helpful AI assistant.<|im_end|>
<|im_start|>user
Prove all parts of the fundamental theorem of calculus.<|im_end|>
<|im_start|>assistant
The Fundamental Theorem of Calculus is a pair of theorems that establishes a fundamental relationship between differentiation and integration. It consists of two parts, the first and the second:

**Part 1 (FTC I):**

If \(f(x)\) is continuous on the closed interval \([a, b]\), and \(F(x)\) is any antiderivative of \(f(x)\), then:

\[ \int_a^b f(x) dx = F(b) - F(a) \]

In other words, the definite integral of a continuous function \(f(x)\) over the interval \([a, b]\) is equal to the difference of the antiderivative \(F(x)\) evaluated at the endpoints \(b\) and \(a\).

**Part 2 (FTC II):**

If \(f(x)\) is continuous on an open interval containing \(a\), and \(F(x)\) is any antiderivative of \(f(x)\), then:

\[ \frac{d}{dx} \int_a^x f(t) dt = f(x) \]

In other words, the derivative of the integral with a variable upper limit (from \(a\) to \(x\)) is equal to the original function \(f(x)\).

**Proof of Part 1 (FTC I):**

We need to show that \(\int_a^b f(x) dx = F(b) - F(a)\), where \(F(x)\) is an antiderivative of \(f(x)\).

By the Fundamental Theorem of Calculus, we know that \(F'(x) = f(x)\). Therefore, we can use the Second Part of the Fundamental Theorem of Calculus to evaluate the integral:

\[
\begin{align*}
\int_a^b f(x) dx & = \lim_{h \to 0} \int_a^{a+h} f(x) dx \\
& = \lim_{h \to 0} [F(a+h) - F(a)] \\
& = F(b) - F(a) \text{ (since } F \text{ is continuous)}
\end{align*}
\]

This proves Part 1 of the Fundamental Theorem of Calculus.

**Proof of Part 2 (FTC II):**

We need to show that \(\frac{d}{dx} \int_a^x f(t) dt = f(x)\), where \(F(x)\) is an antiderivative of \(f(x\)).

Let \(G(x) = \int_a^x f(t) dt\). By the definition of the integral, \(G(x)\) is the area under the curve of \(f(t)\) from \(a\) to \(x\).

We want to find the derivative of \(G(x\)). Using the definition of the derivative, we have:

\[
\begin{align*}
G'(x) & = \lim_{h \to 0} \frac{G(x+h) - G(x)}{h} \\
& = \lim_{h \to 0} \frac{1}{h}\left(\int_a^{x+h} f(t) dt - \int_a^x f(t) dt\right)
\end{align*}
\]

Now, we can use the properties of integrals and the fact that \(F'(x) = f(x)\) (since \(F(x)\) is an antiderivative of \(f(x\)) to simplify the above expression:

\[
\begin{align*}
G'(x) & = \lim_{h \to 0} \frac{1}{h}\left(\int_x^{x+h} f(t) dt\right) \\
& = \lim_{h \to 0} \frac{1}{h}\left[F(x+h) - F(x)\right] \text{ (by FTC I)} \\
& = \lim_{h \to 0} \frac{F(x+h) - F(x)}{h} \\
& = F'(x) \\
& = f(x)
\end{align*}
\]

This proves Part 2 of the Fundamental Theorem of Calculus.<|im_end|>
<|im_start|>user
Very well explained, thank you.<|im_end|>
<|im_start|>assistant
You're welcome! I'm glad you found the explanation helpful. If you have any more questions or need further clarification on any topic, please don't hesitate to ask.<|im_end|>
<|im_start|>user
I'm Glad. Great learning<|im_end|>
<|im_start|>assistant
I'm pleased that you found my proof useful and educational. Do you have any other questions? I'm also here to chat with you on any topic you like, just let me know!<|im_end|>