Here is . Drag x: the top pane shades ∫₀ˣ f, the bottom pane plots that area as the curve g(x). Then flip on show the sliver and shrink h — watch the difference quotient close in on f(x). That convergence is Part 1 of the theorem, happening live.
The sliver of new area is a near-rectangle: height f(x), width h. Divide by h and only the height survives — shrink h to watch the gap die.
The single most-tested skill in this unit. You're shown the graph of f — segments and a semicircle, so every area is exact geometry — and asked everything about g(x) = ∫₀ˣ f. The translation rules: f's sign drives g up or down; f's direction bends g. Scrub across and read the story.
The way the rubric wants it — a candidates table. Step through it.
The multiple-choice version of the same skill. You're shown f — pick the graph of g(x) = ∫₀ˣ f. The distractors are the classic wrong answers; the behavior dictionary is your whole toolkit. Sign drives direction, crossings make extrema, and remember the anchor: g(0) = 0.
Part 1 says accumulation undoes differentiation. Read it backward and definite integrals become two evaluations and a subtraction: find any antiderivative F, then ∫ₐᵇ f = F(b) − F(a). Step through the recipe — and notice where the +C goes.
Why does F(b) − F(a) compute an area? Because they are the same number seen on two graphs. Below: f = cos x and its antiderivative F = sin x. Drag the bounds — the signed area under cos between a and b is always, exactly, the rise of sin from a to b. Red area? Then sin is falling by the same amount. No textbook draws this; it's the whole theorem.
Push b through a red region and watch both numbers fall together: negative area on the top graph IS descent on the bottom one. F(b) − F(a) isn't a trick — it's reading the accumulated area off the antiderivative's own axis.
A few rules let you cut, glue, flip, and scale integrals without computing anything. The figure gives two facts: the area above on [0, 3] is 5, the area below on [3, 7] is 3. Everything else follows — answer each check.
Everything about g is read off f — one derivative down from where your instincts look.
| f is… | so g is… |
|---|---|
| positive | increasing |
| negative | decreasing |
| crossing + → − | at a local max |
| crossing − → + | at a local min |
| increasing | concave up |
| decreasing | concave down |
The most common exam confusion: g(x) is an area, f(x) is g's slope there. A big f means g is climbing fast — it says nothing about whether g is big.
When the upper bound is a function of x, the answer is f at that bound times the bound's derivative. Free multiple-choice points.
Any antiderivative works — the +C appears in both F(b) and F(a) and cancels in the subtraction.
Five checks in the exam's own voice. Try each before you reveal the worked answer.
Eight fast multiple-choice checks across the whole lesson.
g′ IS f. To find where g increases or peaks, read the sign of f itself. Reaching for f′ is answering a question one derivative too deep — f′ only enters for g's concavity.
d/dx ∫₀^(x²) f(t)dt is f(x²)·2x, not f(x²). The moment the bound isn't a bare x, the bound's derivative comes along.
∫ₐᵇ f = F(b) − F(a) — both evaluations, every time. Computing F(b) alone is only right when F(a) happens to be 0, and the exam makes sure it isn't.
If g(x) = ∫₂ˣ f, then g(2) = 0 — an integral over an empty interval. It's the freebie candidates-table entry people leave blank.
“g is largest where f is largest” is false — g is largest where f finishes crossing from + to −. Height of f sets g's slope, not g's size.
A semicircle of radius 2 below the axis contributes −2π, not +2π. Geometry gives the magnitude; the position gives the sign.