Get Ready: Module 5 — Differential Equations, Series & BC Extensions
This is the capstone module, so it leans on the whole calculus year at once. Before we build Taylor series and solve differential equations, we'll re-pour the four slabs that hold them up — integration, sequences & series, derivatives, and the exponential/log functions. Clearly, one step at a time, with nothing assumed. Go at your own pace.
Module 5 takes integration and turns it into a way to undo a rate of change (differential equations), and takes the sequences & series you met in Pre-Calculus and lets calculus add up infinitely many terms. If antiderivatives, the difference between a sequence and a series, or \(\frac{d}{dx}e^x\) feel fuzzy right now, that's completely okay — we'll firm them up together below, then you'll be ready for the lab.
Skills to build first
Four short skills, one per prerequisite. Read the idea, follow the worked example, then try one yourself — the answer is one click away whenever you want it.
Integration & antiderivatives (Module 4)
What it is: integrating is un-differentiating. The antiderivative of \(f\) is a function whose derivative is \(f\) — and because the derivative of a constant is \(0\), there's always a \(+C\): \(\int x^n\,dx = \frac{x^{n+1}}{n+1}+C\) (for \(n\neq -1\)). Why you need it: solving a differential equation is integrating both sides, and the \(+C\) is what an initial condition later pins down. Forgetting the \(+C\) is the single most common Module 5 mistake.
- Integrate term by term. Raise each power by one and divide: \(\int 6x^2\,dx = 6\cdot\frac{x^3}{3} = 2x^3\).
- The constant term \(4\) integrates to \(4x\) (think of it as \(4x^0\)).
- Add the constant of integration. Always finish with \(+C\): the answer is \(2x^3 + 4x + C\).
- Check by differentiating. \(\frac{d}{dx}(2x^3+4x+C) = 6x^2 + 4\). ✓
The check — differentiate your answer and see the original back — is your safety net on every integral.
Try it 1 — Find \(\int (3x^2 - 2x)\,dx\).
\(\int 3x^2\,dx = x^3\) and \(\int -2x\,dx = -x^2\). Answer: \(x^3 - x^2 + C\). Check: derivative is \(3x^2 - 2x\). ✓
Try it 2 — Find \(\int \cos x\,dx\) and \(\int e^x\,dx\).
\(\int \cos x\,dx = \sin x + C\) (since \(\frac{d}{dx}\sin x = \cos x\)); \(\int e^x\,dx = e^x + C\) (it's its own antiderivative).
Sequences & series (Pre-Calculus Module 5)
What it is: a sequence is an ordered list of terms \(a_1, a_2, a_3, \dots\); a series is what you get when you add them, written with \(\sum\). A geometric series \(a + ar + ar^2 + \cdots\) has a finite sum \(\frac{a}{1-r}\) exactly when \(|r| < 1\). Why you need it: a Taylor series is an infinite series, and the first convergence test you'll use compares against the geometric series. Knowing a sequence (the terms) from its series (the sum) prevents the most common confusion in this module.
- Identify the pieces. First term \(a = \left(\tfrac12\right)^0 = 1\); common ratio \(r = \tfrac12\).
- Check convergence. \(|r| = \tfrac12 < 1\), so the infinite sum exists.
- Apply the formula. \(\dfrac{a}{1-r} = \dfrac{1}{1 - \frac12} = \dfrac{1}{\frac12} = 2\).
- Sanity check the partial sums. \(1, 1.5, 1.75, 1.875, \dots\) — creeping up toward \(2\). ✓
Contrast: the sequence \(\left(\tfrac12\right)^n\) goes \(1, \tfrac12, \tfrac14, \dots \to 0\). The terms shrink to \(0\), but their sum is \(2\), not \(0\). Sequence and series are different questions.
Try it 1 — Does \(\sum_{n=0}^{\infty} 3\left(\tfrac{1}{4}\right)^n\) converge? If so, to what?
\(a = 3,\ r = \tfrac14\). Since \(|r| < 1\) it converges to \(\dfrac{3}{1 - \frac14} = \dfrac{3}{\frac34} = 4\).
Try it 2 — The sequence \(a_n = \tfrac{1}{n}\) tends to \(0\). Does that mean the series \(\sum \tfrac1n\) converges?
No. The terms going to \(0\) is necessary but not sufficient. The harmonic series \(\sum \tfrac1n\) actually diverges — this is the classic trap Module 5 will make explicit.
Derivatives — the power, product, and chain rules
What it is: the derivative \(f'(x)\) is the instantaneous rate of change — the slope of the curve. The power rule \(\frac{d}{dx}x^n = nx^{n-1}\) and the chain rule (differentiate the outside, times the derivative of the inside) cover most cases. Why you need it: a Taylor series is made of derivatives — its \(n\)-th coefficient is \(\frac{f^{(n)}(0)}{n!}\), so you'll compute repeated derivatives. And a differential equation is an equation about a derivative.
- First derivative: \(\frac{d}{dx}\sin x = \cos x\).
- Keep going: \(\frac{d}{dx}\cos x = -\sin x\), then \(\frac{d}{dx}(-\sin x) = -\cos x\), then back to \(\sin x\).
- Notice the cycle of 4: \(\sin \to \cos \to -\sin \to -\cos \to \sin \dots\) This four-step pattern is exactly what generates the alternating signs in the Taylor series.
- Evaluate at \(0\): \(\sin 0 = 0,\ \cos 0 = 1,\ -\sin 0 = 0,\ -\cos 0 = -1\) — giving \(0, 1, 0, -1, \dots\)
Try it 1 — Differentiate \(f(x) = x^4 - 3x^2 + 5\).
Power rule term by term: \(f'(x) = 4x^3 - 6x\). (The constant \(5\) has derivative \(0\).)
Try it 2 — Use the chain rule on \(g(x) = e^{2x}\).
Outside derivative \(e^{2x}\) times inside derivative \(2\): \(g'(x) = 2e^{2x}\). (This is why \(\frac{dy}{dx} = 2y\) has solution \(y = Ce^{2x}\) — a Module 5 idea.)
Exponential & logarithm functions
What it is: \(e^x\) is the function that equals its own derivative, and \(\ln x\) is its inverse, undoing it: \(\ln(e^x) = x\) and \(e^{\ln x} = x\). The headline calculus facts are \(\frac{d}{dx}e^x = e^x\), \(\frac{d}{dx}\ln x = \frac1x\), and \(\int \frac1x\,dx = \ln|x| + C\). Why you need it: exponential growth/decay and logistic models are differential equations, and \(e^x\) is the very first function the Taylor-series lab builds. Logarithms show up the moment you integrate \(\frac1x\).
- Undo the logarithm. A log equation becomes an exponential one: \(\ln x = 3\) means \(x = e^3\).
- Estimate the value. \(e \approx 2.718\), so \(e^3 \approx 20.09\).
- Check. \(\ln(e^3) = 3\) by the inverse rule. ✓
Key laws to keep handy: \(\ln(ab) = \ln a + \ln b\), \(\ln\!\frac{a}{b} = \ln a - \ln b\), and \(\ln(a^k) = k\ln a\). They turn products into sums, which is exactly what makes them useful when separating variables.
Try it 1 — Solve \(e^x = 7\) for \(x\) (exact form).
Take \(\ln\) of both sides: \(x = \ln 7\) (about \(1.946\)). The \(\ln\) undoes the \(e^x\).
Try it 2 — Find \(\frac{d}{dx}\ln x\) and \(\int \frac{1}{x}\,dx\).
\(\frac{d}{dx}\ln x = \frac1x\); going the other way, \(\int \frac1x\,dx = \ln|x| + C\). They're a derivative–antiderivative pair.
Quick Readiness Check
Six short questions spanning all four skills. Try each one first, then reveal the answer to check yourself.
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Find \(\int (4x^3 + 1)\,dx\).
Show answer
Show answer\(x^4 + x + C\). Don't forget the \(+C\)! Check: derivative is \(4x^3 + 1\).
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Does \(\sum_{n=0}^{\infty}\left(\tfrac{2}{3}\right)^n\) converge, and to what?
Show answer
Show answer\(a=1,\ r=\tfrac23,\ |r|<1\): converges to \(\dfrac{1}{1-\frac23} = 3\).
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True or false: if the terms \(a_n \to 0\), then \(\sum a_n\) must converge.
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Show answerFalse. The harmonic series \(\sum\tfrac1n\) has terms \(\to 0\) yet diverges.
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Differentiate \(f(x) = x^3 - 2x\).
Show answer
Show answer\(f'(x) = 3x^2 - 2\) by the power rule.
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What is \(\frac{d}{dx}e^x\)?
Show answer
Show answer\(e^x\) — it is its own derivative. (That property is what makes it special in this module.)
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Solve \(e^x = 1\).
Show answer
Show answer\(x = \ln 1 = 0\). (And notice \(e^0 = 1\) — the value the Taylor lab always nails at the center.)
Floor poured. When the skills above feel steady, step into the module — the Visual Lab lets you see a polynomial built term by term until it becomes \(e^x\), \(\sin x\), or \(\cos x\).