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# Derivative and integral: Fundamental Theorem of Calculus

### From Mathematics Is A Science

After Integral: properties, we consider the most important one.

What does Riemann integral have to do with *antiderivatives*?

We already know answer, for motion.

If $y = f(x)$ is the velocity, $x$ time,

- then antiderivative is $F(x)$, the position function;
- then integral is $\int\limits_{a}^{b} f(x)\, dx$, the displacement during $[a,b]$.

$$\underbrace{F(b)}_{\text{Current position}} - \underbrace{F(a)}_{\text{initial position}}$$

These aren't *equal* to each other. After all the former is a function and the latter is a number. But if we know the position, we can certainly compute the displacement.

What about vice versa? We would have to compute the displacement for all $x$'s... How?

It's simple. The position is
$$F(x) = \int\limits_{a}^{\overbrace{x}^{\text{varies}}} F(x) \, dx.$$
In other words, this is our Riemann integral (area under the graph) but with a *variable upper limit*. It's illustrated below: $x$ runs from $a$ to $b$.

**General Setting:** $f$ is just a function. Then
$$\begin{aligned}
g(x) &= \int_{a}^{x} f(x) \, dx \text{ is the area function} \\
&= \text{ the area under the graph of } f \text{ from } a \text{ to } x
\end{aligned}
$$

**Example:** Consider
$$f(x) = 2x.$$

Then
$$\begin{aligned}
g(x) &= \int_{0}^{x} 2x\, dx \\
&= \text{area of triangle} \\
&= \frac{1}{2} \text{width}\cdot\text{height} \\
&= \frac{1}{2} x \cdot 2x \\
&= x^{2}
\end{aligned}
$$
What's the relationship between $f(x) = 2x$ and $g(x) = x^{2}$?
$$\left(x^{2}\right)^{\prime} = 2x, $$
so $g$ is an *antiderivative* of $f$!

This idea can be used for all functions.

**The Fundamental Theorem of Calculus (Part 1)**

Given a continuous function, $f$ on $[a,b]$, then $$g(x) = \int_{a}^{x} f(x) \, dx $$ is an antiderivative of $f$.

**Proof:** We need to show: $g^{\prime} = f$.

To do that, we have to go all the way back to the definition of the derivative (derivative as a limit). $$\begin{aligned} g^{\prime}(x) &= \lim_{h \to 0}\frac{1}{h} \left( g(x+h) - g(x) \right) \\ &= \lim_{h \to 0} \frac{1}{h} \left( \int_{a}^{x+h} f(x)\, dx + \int_{a}^{x} f(x)\, dx\right) \\ &= \lim_{h \to 0} \frac{1}{h} \int_{x}^{x+h} f(x) \, dx \\ \end{aligned} $$ We used the additivity of integral and simplified.

What happens to this limit?

Let's try to make $h$ small. The rectangle will be thinner and thinner and its top edge will look more and more horizontal. Then its height is "almost" $$\frac{\int_{x}^{x+h} f(x)\, dx }{h}. $$

Why? Because $$\frac{\text{Area of Rectangle}}{\text{Width}} = \text{ Height}.$$ $\blacksquare$

Observation: Displacement with a given velocity $f$ over $[a,b]$ is
$$\int_{a}^{b} f(x)\, dx $$
the area under the graph.

Find the position function first: $F(x)$, then $$\text{displacement } = \underbrace{F(b)}_{\text{final}} - \underbrace{F(a)}_{\text{initial}}.$$ Note: $F$ may vary by a constant.

Then it follows,

**The Fundamental Theorem of Calculus (Part 2)**

Given a continuous function, $y=f(x)$ on $[a,b]$, then
$$\int_{a}^{b} f(x)\, dx = F(b) - F(a) $$
where $F$ is *any* antiderivative of $f$.

**Note:** If $G(x)$ is another antiderivative of $f$, then
$$G(x) = F(x) + C.$$
If we substitute:
$$\begin{aligned}
G(b) - G(a) &= \left( F(b) + C \right) + \left( F(a) + C \right) \\
&= F(b) - F(a)
\end{aligned}
$$
That's good for computations...

**Example:** Compute the area under $y=x^{2}$ from 0 to 1.

We need the antiderivative of $f(x) = x^{2}$. Use the Power Formula: $$F(x) = \frac{x^{3}}{3} + C$$ Then $$ \begin{aligned} \int_{0}^{1} x^{2}\, dx &= F(1) - F(0) \\ &= \frac{1^{3}}{3} - \frac{0^{3}}{3} = \frac{1}{3}. \end{aligned} $$ Easy in comparison to the Riemann sum approximation.

**Notation:** This is how we can record the above:
$$\int_{0}^{1} x^{2}\, dx = \left.\frac{x^{3}}{3}\right|_{0}^{1}.$$
Or, generally:
$$
\begin{aligned}
\int_{a}^{b} f(x) \, dx &= \left. F(x) \right|_{a}^{b}\\
&= F(b) - F(a).
\end{aligned}
$$
As you can see, the limits of integration just jump over and are kept for reference.

Now, the Fundamental Theorem of Calculus is a connection between these two:

Riemann Integral | and | Antiderivatives |
---|---|---|

Defined via Riemann sums (and limit) | Defined via derivatives | |

Can always be computed for continuous $f$. | Not always easily computable. Ex: $\int e^{x^{2}}dx$ is a new function. |

**Example:**
$$\begin{aligned}
\int_{-1}^{1} e^{x} \, dx &= \left. e^{x} \right|_{-1}^{1} \\
&= e^{1} - e^{-1} = e - \frac{1}{e}
\end{aligned}
$$
That's what we have found:

**Example:**

$$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \cos x \, dx &= \left.\sin x \right|_{0}^{\frac{\pi}{2}} \\ &= \sin \frac{\pi}{2} - \sin 0 \\ &= 1 - 0 = 1. \end{aligned} $$

**Example:** What about the area of a circle?

$$\text{Area } = \int_{-1}^{1} \sqrt{1 - x^{2}} \, dx = ?$$ What's antiderivative? We don't know, yet.

**Terminology:** Don't confuse these two.

1. $\int_{a}^{b} f(x) dx=$

- Riemann Integral,
- area under graph,
*definite integral*;

These are numbers.

2. $\int f(x) dx=$

*indefinite integral*,- antiderivative(s).

These are functions.