Practice & Assess

Fundamental Theorem of Calculus

Link net area under f to an antiderivative — and see why A′(x) = f(x).

Fundamental Theorem

Link the area under f to an antiderivative — the M2 way.

f(x) = sin(x) + 2

Part 1. If

A (x) = \int_{a}^{x} f (t) d t

, then

A^{'} (x) = f (x)

Part 2.

\int_{a}^{b} f (x) d x = F (b) - F (a)

, where

F^{'} = f

Lower limit a1.0

Upper limit b8.0

A (b) = \int_{a}^{b} f (t) d t = F (b) - F (a)

A(b) = F(b) - F(a)14.6858

f(b)2.9894

A'(b) [= f(b) by FTC]2.9894

The amber tangent on the bottom graph has slope A'(b) = f(b).

The shaded region is the net area from a to b under y = f(x).
The bottom curve is the accumulation A(x) = ∫_a^x f. At x = b its height equals that area.
Steeper A means larger f: the instantaneous rate of area growth is exactly f(b).

f(x) = sin(x) + 2

Green shading = ∫_a^b f. Amber segment at b has length f(b) = A'(b).

A(x) = F(x) - F(a)

A(x)

Tangent: slope = f(b)

Height of A at b equals the green area above. The amber tangent's slope matches f(b) — that is FTC Part 1.