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# Derivative and integral: Fundamental Theorem of Calculus

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

What does Riemann integral have to do with antiderivatives?

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

1. then antiderivative is $F(x)$, the position function;
2. 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.