I created this deck to support my study of Calculus 1 while attending university in Italy (Computer Science). I aimed to include all essential information in a clear and well-formatted way, making studying as effective as possible.
The deck is entirely text-based (I did not add thousands of images or videos), so its size is just a few kilobytes. However, to enhance it with graphical visualizations, I integrated several interactive HTML embeds from Geogebra, selecting the most useful ones available online, along with explanatory video clips from YouTube for convenient reference.
Some flashcards might be quite long, but this was a necessary compromise between the number of cards and the content per card. Personally, these cards were crucial for me to pass Calculus 1, and I hope they will be just as helpful for anyone using them.
🇮🇹 Utenti italiani: La versione originale di questo mazzo è in italiano e l'ho pubblicata separatamente. La puoi trovare su AnkiWeb cercando "Analisi 1 - Matematica" o cliccando sul seguente link diretto al deck in italiano.
⚠ Note: Sample cards might not display correctly on the Anki website, so I recommend downloading the deck and testing it directly in the Anki application.
If you find this deck useful, leave a rating here! I spent several days creating it, and if you feel generous and want to support my work, encouraging me to create more high-quality decks, you can freely support me here.
| Fronte | How do you calculate the integral of a function on an interval decomposed into multiple subintervals? |
| Retro | If \( f \) is integrable on \( [a, b] \) and \( c \in (a, b) \), then \( f \) is integrable on \( [a, c] \) and on \( [c, b] \), and \[ \int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \] |
| Fronte | What is the modulus of a complex number? |
| Retro | Given a complex number \( z = a + ib \), its modulus, denoted by \( |z| \), is defined as \( |z| = \sqrt{a^2 + b^2} \). Graphically, it corresponds to the distance from the origin to the point \(z = (a, b) = a + ib\) on the Gaussian plane, with real component \(a\) (abscissa) and imaginary component \(b\) (ordinate). |
| Fronte | What does Rolle's theorem state? [+ Proof] |
| Retro | Let \( f : [a, b] \to \mathbb{R} \) be a function such that:- \(f\) is continuous on \( [a, b] \); - \(f\) is differentiable on \( (a, b) \);- \( f(a) = f(b) \).Then there exists at least one point \( c \in (a, b) \) such that \( f'(c) = 0 \).[Proof]Since \( f\) is continuous on a closed and bounded interval \([a,b]\) by hypothesis, we can apply the Weierstrass theorem, so there exist \(x_m,x_M \in [a, b]\) minimum and maximum points for \(f\), that is, such that:\[\textcolor{gray}{m =} f(x_m) \leq f(x) \leq f(x_M) \textcolor{gray}{=M} \quad \forall x \in [a, b]\]If \(x_m = a\) and \(x_M = b\) (or vice versa), then \(f(x_m) = m = M = f(x_M)\) (from the hypothesis \( f(a) = f(b) \)) so the function \(f\) is constant and therefore its derivative \(f'(c) = 0\). Otherwise, at least one of the local extremum points \(x_m\) and \(x_M\) is inside the interval \((a, b)\), so (since \(f\) is differentiable by hypothesis), by the Fermat's theorem at that point the derivative is \(0\), which is what we wanted to prove.\(\square\)In summary: the Weierstrass theorem ensures the presence of an absolute maximum and minimum in the interval \([a, b]\) and by Fermat's theorem the first derivative at those points is zero. https://www.geogebra.org/m/ZFR2ERC3 |