The Fundamental Theorem of Calculus
The FTC is the hinge of the whole course: Part 1 says the derivative of an accumulated area is the curve's height, and Part 2 turns hard Riemann-sum limits into one subtraction, \(F(b) - F(a)\).
By the end you'll be able to evaluate \(\displaystyle\int_a^b f(x)\,dx\) as \(F(b) - F(a)\) with any antiderivative \(F\), and differentiate an integral with a variable upper limit using \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\).
Predict: the curve on top crosses zero at t = 2. When the sweep passes that point, what does the accumulation curve below do — keep rising, jump, or flatten out? Sweep x past 2 to check.
Top panel: \(f(t) = 4 - t^2\). Bottom panel: the accumulation function \(A(x) = \int_0^x f(t)\,dt\). As you sweep x right, area fills in on top and the point \((x, A(x))\) traces the curve below. The key link is in green: the height of f on top always equals the slope of A below — that's Part 1. Drag either panel or use the slider; hover or focus the dots for exact values.
x = 1.50 · area so far A(x) = 4.88 · slope of A at x = height of f = 1.75
The Fundamental Theorem of Calculus links differentiation and integration: Part 1 says the derivative of the accumulation function \(\int_a^x f(t)\,dt\) is \(f(x)\); Part 2 evaluates \(\int_a^b f(x)\,dx\) as \(F(b) - F(a)\) for any antiderivative \(F\).
Let \(f\) be continuous on \([a, b]\). Part 1: the accumulation function \(G(x) = \int_a^x f(t)\,dt\) is differentiable, and \(G'(x) = f(x)\) — differentiating an integral with a variable upper limit gives back the integrand. Part 2: if \(F\) is any antiderivative of \(f\), then \(\int_a^b f(x)\,dx = F(b) - F(a)\), often written \([F(x)]_a^b\). The two antiderivatives \(F\) and \(F + C\) give the same answer because the \(+C\) appears in both terms and cancels in the subtraction.
Part 1 says the rate at which accumulated area grows as you push the right edge outward is exactly the height of the curve there — accumulation and rate are two sides of one coin, which is what the green height-equals-slope link in the demo shows. Think odometer and speedometer: distance is accumulated speed, and the rate your odometer turns is your current speed. Part 2 is the workhorse consequence: instead of computing a hard limit of Riemann sums, find one antiderivative and subtract its values at the two endpoints.
Evaluate \(\displaystyle\int_1^3 x^2\,dx\). Step 1: an antiderivative of \(x^2\) is \(F(x) = x^3/3\). Step 2: apply Part 2: \(F(3) - F(1) = 27/3 - 1/3 = 9 - \tfrac{1}{3} = \) \(26/3\). Note the order: top limit minus bottom limit, always \(F(b) - F(a)\).
Find \(\dfrac{d}{dx}\displaystyle\int_0^x \cos t\,dt\). Part 1 answers instantly: the derivative of the accumulation function is the integrand evaluated at \(x\), so the answer is \(\cos x\) — no integration needed at all.
Evaluate \(\displaystyle\int_1^4 \sqrt{x}\,dx\). Step 1: an antiderivative of \(x^{1/2}\) is \(F(x) = \tfrac{2}{3}x^{3/2}\). Step 2: \(F(4) = \tfrac{2}{3}\cdot 8 = \tfrac{16}{3}\) and \(F(1) = \tfrac{2}{3}\). Step 3: \(F(4) - F(1) = \) ____
Reveal the answer
\(F(4) - F(1) = \tfrac{16}{3} - \tfrac{2}{3} = \) \(14/3\). If you'd used \(F(x) = \tfrac{2}{3}x^{3/2} + 5\) instead, both terms would grow by 5 and the difference would still be \(14/3\) — any antiderivative works.
More info — a variable limit with a chain in it, and when the theorem breaks
Part 1 as stated needs the upper limit to be plain \(x\). With a function up there, say \(\frac{d}{dx}\int_a^{u(x)} f(t)\,dt\), the chain rule joins in: the derivative is \(f(u(x))\cdot u'(x)\). For example, \(\frac{d}{dx}\int_0^{x^2} \cos t\,dt = \cos(x^2)\cdot 2x\) — compare that with the plain-\(x\) worked example above. Also remember the fine print from the formal statement: both parts require \(f\) continuous on the interval. An asymptote inside \([a, b]\) breaks Part 2 — blindly subtracting \(F(b) - F(a)\) across one gives nonsense. The OpenStax section in Dive deeper below works several examples of both situations.
Check your understanding
Use Part 2 of the Fundamental Theorem to evaluate \(\displaystyle\int_0^2 3x^2\,dx\).
What is \(\dfrac{d}{dx}\displaystyle\int_2^x \sqrt{1+t^3}\,dt\)?
\(f\) is continuous and crosses from positive to negative at \(x = c\). What does the accumulation function \(A(x) = \displaystyle\int_a^x f(t)\,dt\) do at \(x = c\)?
Two antiderivatives of \(f(x) = 2x\) are \(F(x) = x^2\) and \(G(x) = x^2 + 7\). Using each in Part 2 to evaluate \(\displaystyle\int_1^4 2x\,dx\) gives…
Recap
- Part 1: \(\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x)\) — the slope of the accumulation function is the height of the curve.
- Part 2: \(\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)\) for any antiderivative \(F\); the \(+C\) cancels in the subtraction.
- Order matters: it is \(F(b) - F(a)\), top limit minus bottom limit.
- Both parts require \(f\) continuous on \([a, b]\); an asymptote inside the interval breaks the theorem.
- A function in the upper limit brings the chain rule: \(\dfrac{d}{dx}\displaystyle\int_a^{u(x)} f(t)\,dt = f(u(x))\cdot u'(x)\).
Dive deeper
- OpenStax Calculus Volume 1 — 5.3 The Fundamental Theorem of Calculus State and apply both parts of the Fundamental Theorem of Calculus
Sources
- The Fundamental Theorem of Calculus