Fundamental theorem of calculus Study Guide
Study Guide
📖 Core Concepts
Fundamental Theorem of Calculus (FTC) – Connects differentiation (instantaneous rate) with integration (accumulated area). The two operations are inverses.
Area (or “integral”) function \(F(x)=\displaystyle\int{a}^{x}f(t)\,dt\).
Continuous \(f\) on \([a,b]\) ⇒ \(F\) is continuous on \([a,b]\) and differentiable on \((a,b)\).
First Part (FTC‑1): \(F'(x)=f(x)\).
Antiderivative – Any function \(G\) with \(G'(x)=f(x)\).
Second Part (FTC‑2): If \(G\) is an antiderivative of \(f\) on \([a,b]\), then \(\displaystyle\int{a}^{b} f(t)\,dt = G(b)-G(a)\).
Riemann integrable – A function whose upper and lower sums can be made arbitrarily close; the FTC‑2 still works even if \(f\) has a few discontinuities, as long as it’s Riemann integrable.
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📌 Must Remember
FTC‑1: For continuous \(f\), \(\displaystyle\frac{d}{dx}\int{a}^{x}f(t)\,dt = f(x)\).
FTC‑2: \(\displaystyle\int{a}^{b} f(t)\,dt = F(b)-F(a)\) for any antiderivative \(F\) of \(f\).
Uniform continuity of the area function \(F\) on a closed interval \([a,b]\).
Mean Value Theorem for integrals: \(\displaystyle\int{x}^{x+\Delta x} f(t)\,dt = f(c)\,\Delta x\) for some \(c\) in \((x,x+\Delta x)\).
When \(f\) is not continuous but Riemann integrable, FTC‑2 still holds.
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🔄 Key Processes
Using FTC‑1 to find a derivative of an integral
Define \(F(x)=\int{a}^{x}f(t)\,dt\).
Differentiate: \(F'(x)=f(x)\).
Evaluating a definite integral with FTC‑2
Find any antiderivative \(G\) of \(f\).
Compute \(G(b)-G(a)\).
Riemann‑sum proof sketch (FTC‑2)
Partition \([a,b]\) into \([x{i-1},xi]\) with widths \(\Delta xi\).
Apply mean‑value theorem: \(F(xi)-F(x{i-1}) = f(ci)\Delta xi\).
Sum: \(\sum f(ci)\Delta xi = F(b)-F(a)\).
Take limit as \(\max\Delta xi \to 0\) → \(\int{a}^{b} f(t)\,dt = F(b)-F(a)\).
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🔍 Key Comparisons
FTC‑1 vs. FTC‑2
FTC‑1: Differentiates an integral → gives the original integrand.
FTC‑2: Integrates a function by evaluating an antiderivative at the endpoints.
Continuous \(f\) vs. Riemann‑integrable \(f\)
Continuous → guarantees both parts of FTC automatically.
Riemann‑integrable (may have jump discontinuities) → still valid for FTC‑2, but FTC‑1 requires continuity for the derivative to equal \(f\).
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⚠️ Common Misunderstandings
“The derivative of any integral equals the integrand” – Only true when the upper limit is the variable and the integrand is continuous on the interval (FTC‑1).
Confusing antiderivative with indefinite integral – An indefinite integral \(\int f(x)\,dx\) denotes a family of antiderivatives; FTC‑2 uses any single antiderivative \(F\).
Assuming FTC works for improper integrals automatically – The theorem requires the function to be (properly) Riemann integrable on the closed interval.
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🧠 Mental Models / Intuition
“Area‑as‑a‑function”: Imagine the graph of \(f\) and a sliding vertical line at \(x\). The area swept from \(a\) to the line is \(F(x)\). Moving the line a tiny bit adds a thin rectangle of height \(f(x)\); the rectangle’s height is exactly the instantaneous rate of change of the accumulated area → \(F'(x)=f(x)\).
Inverse machines: Think of integration as a “store” that accumulates tiny pieces, and differentiation as a “scanner” that reads the current piece. The scanner (derivative) tells you exactly what is being stored at that point.
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🚩 Exceptions & Edge Cases
Discontinuous but integrable functions – FTC‑2 still works; however, FTC‑1 may fail at points of discontinuity (the derivative of the area function may not equal \(f\) there).
Endpoints – FTC‑1 guarantees differentiability only on the open interval \((a,b)\); at the exact endpoints the derivative may not be defined.
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📍 When to Use Which
Need a quick value of a definite integral? → Find any antiderivative (FTC‑2) and apply \(F(b)-F(a)\).
Given an integral with a variable upper limit and asked for its derivative? → Apply FTC‑1 directly: derivative equals the integrand (provided continuity).
Function is piecewise or has a few jump points? → Check Riemann integrability; use FTC‑2 for the integral, avoid FTC‑1 at the jumps.
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👀 Patterns to Recognize
Integral of a derivative → \(\displaystyle\int{a}^{b} F'(x)\,dx = F(b)-F(a)\).
Derivative of an integral with variable limit → \(\displaystyle\frac{d}{dx}\int{a}^{x} f(t)\,dt = f(x)\).
Repeated “+C” in indefinite integrals – any antiderivative works for FTC‑2; the constant cancels in \(F(b)-F(a)\).
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🗂️ Exam Traps
Choosing the wrong limit – Forgetting that FTC‑2 uses the same antiderivative for both limits; mixing different antiderivatives adds extra constants.
Assuming continuity automatically – A question may give a function with a removable discontinuity; you must verify continuity on the interval before applying FTC‑1.
Confusing \(\inta^b f(t)\,dt\) with \(\inta^x f(t)\,dt\) – The former is a number; the latter is a function of \(x\). Their derivatives behave differently.
Mis‑applying the Mean Value Theorem for integrals – The theorem guarantees a point \(c\) inside the subinterval, not at the endpoints; using endpoint values leads to wrong estimates.
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