Week 13 - Fundamental Theorem of Calculus #
The Fundamental Theorem of Calculus is appropriately named because it establishes a connection between the two branches of calculus: differential calculus and integral calculus.
Theorem #
Suppose $f: [a, b] \to \mathbb{R}$ is continuous. Let $F$ be the accumulation function, given by $$F(x) = \int_a^x f(t) dt$$. Then $F$ is continuous on $[a, b]$, differentiable on (a, b), and $F’(x) = f(x)$.
- To prove: $$F’(x) = \frac{F(x+h) - F(x)}{h} \approx \frac{f(x) \cdot h}{h} = f(x)$$
Suppose$f: [a, b] \to \mathbb{R}$ is continuous, and $F$ is an antidirevative of $f$. Then $$\int_a^b f(x) dx = F(b) - F(a)$$
- To prove:
- We say $G(x)$ is the antidirevative of $f$. Then, $G’(x) = f(x)$, and also, we know $f(x) = F’(x)$.
- where $F(x) = \int_a^x f(t) dt$
- In the chapter of antiderivative we have proved that, the antiderivative of $f$ should be some function plus a constant, then we can get $F(x) = G(x) + C$
- $F(x)$ and $G(x)$ are both the antidirevative functions of $f$.
- Another fact, we know, is: $F(a) = \int_a^a f(t) dt = 0 = G(a) + C$, so $C = -G(a)$.
- Then, we get $F(x) = G(x) - G(a)$, so $\int_a^x f(t) dt = G(x) - G(a)$
- Let x = b, we get: $\int_a^b f(t) dt = G(b) - G(a) = (F(b) - C) - (F(a) - C) = F(b) - F(a)$
- We say $G(x)$ is the antidirevative of $f$. Then, $G’(x) = f(x)$, and also, we know $f(x) = F’(x)$.
- To prove:
Examples #
$\int_0^\pi \sin{x} dx$:
- $$\begin{aligned} F(x) &= \int_0^\pi \sin{x} dx \\ F’(x) &= f(x) = \sin{x} \\ F(x) &= -\cos{x} + C\\ \int_0^\pi \sin{x} dx &= F(\pi) - F(0) \\ &= -\cos{\pi} - (-\cos{0}) \\ &= -(-1) - (-1) = 2 \end{aligned}$$
$\int_0^1 x^4$:
- $$\begin{aligned} F(x) &= \int_0^1 x^4 \\ F’(x) &= f(x) = x^4 \\ F(x) &= \frac{x^5}{5} + C \\ \int_0^1 x^4 &= \frac{1^5}{5} - \frac{0^5}{5} \\ &= \frac{1}{5} \end{aligned}$$
The area between $\sqrt{x}$ and $x^2$:
$$\int_0^1 \sqrt{x} - x^2 dx = [\frac{x^{3/2} }{3/2} - \frac{x^3}{3}]_0^1 = (\frac{2}{3} - \frac{1}{3}) - 0 = \frac{1}{3}$$
$\int_0^t\sqrt{1-x^2}dx$
Method 1 (use geometrical calculation):
- Break the area to a circular sector and a triangle.
- Then, we get:
- the triangle = $\frac{t \cdot \sqrt{1-t^2} }{2}$
- the circular sector = $\frac{\arcsin{t} }{2}$
- So $$\int_0^t\sqrt{1-x^2}dx = \frac{t \cdot \sqrt{1-t^2} }{2} + \frac{\arcsin{t} }{2}$$
- The area of the circular is $\frac{\theta}{2\pi} \cdot \pi r^2$, where $r = 1$, so the area is $\frac{\theta}{2} = \frac{\arcsin{t} }{2}$
Method 2 (Use calculus):
- Let’s say $F(x) = \frac{x \cdot \sqrt{1-x^2} }{2} + \frac{\arcsin{x} }{2}$. Then we just need to prove $F’(x) = \sqrt{1-x^2}$
- // skip
The Fundamental Theorem in Physics #
- In physics, $v(t)$ = velocity at time $t$, the displacement can written by $\int_a^b v(t) dt$ (Distance from time t=0 to t=b).
- Use Riemann sum, we can get:
- $\text{from } t = 0 \text{ to } t = h$, $D_1 = h \cdot v(0)$
- $\text{from } t = h \text{ to } t = 2h$, $D_2 = D_1 + h \cdot v(h)$
- $\text{from } t = 2h \text{ to } t = 3h$, $D_3 = D_2 + h \cdot v(2h)$
- keep add this down to $t=b$, and as h goes to zero, the Riemann sum will compute $\int_a^b v(t) dt$
- Summarize this, the accumulation function of velocity is displacement, and the derivative of the displacement is velocity.
d/da integral f(x) dx from x = a to x =b #
- $\frac{d}{da} \int_a^bf(x)dx$
- We know that $\frac{d}{db} \int_a^bf(x)dx = f(b)$
- The rate of the change of the accumulation function is the functions value
- Imagine we are calculating the area from a to b. When calculating $\frac{d}{db}$, we want to know how does that integral change when wiggling b.
- Compare to $\frac{d}{da}$, we want to know how does the integral change when wiggling a.
- Let’s set the x changes h, So the integral’s change should be $\frac{\int_a^bf(a+h)dx - \int_a^bf(a)dx}{h} \approx \frac{-hf(a)}{h} = -f(a)$