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Calculus Flashcards

A-Level Calculus — Flashcard Deck

15 flashcards with spaced repetition. Press Space to flip, then rate your recall (1—4).

What These Flashcards Cover

These flashcards test your fluency with the core techniques and applications of calculus at A-Level. You should be able to differentiate and integrate standard functions, apply calculus to real-world problems, and solve first-order differential equations.

Key areas:

  • Differentiation from first principles: dydx=lim⁡δx→0f(x+δx)−f(x)δx\frac{dy}{dx} = \lim_{\delta x \to 0} \frac{f(x+\delta x) - f(x)}{\delta x}. Know how to apply this to simple polynomials and understand the gradient interpretation.
  • Standard derivatives and rules: ddx(xn)=nxn−1\frac{d}{dx}(x^{n}) = nx^{n-1}. Chain rule: dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}. Product rule: ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}. Quotient rule: ddx(uv)=vdudx−udvdxv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}.
  • Standard integrals: ∫xn dx=xn+1n+1+C\int x^{n}\,dx = \frac{x^{n+1}}{n+1} + C (for n≠−1n \neq -1). ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln|x| + C. ∫ekx dx=1kekx+C\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C. ∫cos⁡(kx) dx=1ksin⁡(kx)+C\int \cos(kx)\,dx = \frac{1}{k}\sin(kx) + C.
  • Applications of differentiation: Finding stationary points (dydx=0\frac{dy}{dx} = 0) and classifying them using the second derivative test (d2ydx2>0\frac{d^{2}y}{dx^{2}} > 0 minimum, <0< 0 maximum). Tangents and normals. Connected rates of change.
  • Applications of integration: Area under a curve A=∫aby dxA = \int_{a}^{b} y\,dx. Area between curves A=∫ab(y2−y1) dxA = \int_{a}^{b}(y_{2} - y_{1})\,dx. Volumes of revolution: V=π∫aby2 dxV = \pi\int_{a}^{b} y^{2}\,dx (about x-axis).
  • Differential equations: Separable equations: separate variables and integrate both sides. dydx=ky\frac{dy}{dx} = ky has solution y=Aekxy = Ae^{kx}. Use boundary conditions to find the constant AA.

Intuition

Differentiation answers the question “how fast is this changing right now?” Think of it as zooming in on a curve until it looks straight — the gradient of that straight line is the derivative. Integration answers the opposite question: “given the rate of change, how much has accumulated?” It sums up infinitely many tiny contributions.

The fundamental theorem of calculus links the two: differentiation and integration are inverse operations. This is why finding the area under a curve uses the antiderivative.

Common Pitfalls

  1. Forgetting the constant of integration. Every indefinite integral needs +C+C. Forgetting it loses a mark even if the rest of your working is correct.
  2. Applying the chain rule incorrectly. When differentiating sin⁡(3x)\sin(3x), the answer is 3cos⁡(3x)3\cos(3x), not cos⁡(3x)\cos(3x). Always multiply by the derivative of the inner function.
  3. Mixing up the product and chain rules. The product rule is for products of two functions: (uv)′=u′v+uv′(uv)' = u'v + uv'. The chain rule is for composite functions: f(g(x))′=f′(g(x))g′(x)f(g(x))' = f'(g(x))g'(x). These are different operations.
  4. Incorrect limits on definite integrals. The area under a curve from x=ax = a to x=bx = b requires ∫ab\int_{a}^{b}. If you swap the limits, you get the negative of the area.

Cross-References