Accumulation Functions and the Second FTC
- Differentiate an accumulation function, including one with a variable expression as its upper limit
- Analyze g(x) = ∫ₐˣ f(t) dt for increase, decrease, extrema and concavity using the graph of f
- Evaluate an accumulation function at a point by reading areas off the graph of f
An integral with a variable limit is a function
g(x) = ∫ₐˣ f(t) dt takes an input x and returns the accumulated signed area from a to x. The Second Fundamental Theorem says that differentiating it recovers the integrand: g′(x) = f(x). So the graph of f is the graph of g′, and every tool from Unit 5 applies immediately — g increases where f is positive, g has a relative maximum where f crosses from positive to negative, g is concave up where f is increasing.
A variable upper limit brings a chain rule
If the upper limit is an expression rather than x alone, the chain rule appears: d/dx ∫ₐ^(u(x)) f(t) dt = f(u(x)) · u′(x). For ∫₀^(x²) sin t dt the derivative is sin(x²) · 2x, and dropping the 2x is the standard error. If the lower limit is the variable one, swap the limits and pick up a minus sign: d/dx ∫ₓᵇ f(t) dt = −f(x).
Reading values off the graph of f
To evaluate g at a point, add up signed areas: regions where f is above the axis count positive, below count negative. g(a) = 0 always, since the interval has collapsed. Exam graphs of f are built from triangles, rectangles and quarter-circles precisely so the areas are computable by geometry rather than antidifferentiation — recognize the shapes and the arithmetic is short.
Let g(x) = ∫₀ˣ f(t) dt, where the graph of f consists of a triangle above the axis on [0, 4] with height 3 at t = 2, and a triangle below the axis on [4, 6] with depth 2 at t = 5. Find g(4), g(6), and locate any relative maximum of g.
- 1.g(4) is the area of the first triangle: base 4, height 3, so (1/2)(4)(3) = 6.
- 2.On [4, 6], f is below the axis, so that area counts negative: (1/2)(2)(2) = 2, contributing −2.
- 3.g(6) = 6 + (−2) = 4.
- 4.g′(x) = f(x), so g has critical points where f = 0 — at x = 4 (and at x = 0, an endpoint).
- 5.f changes from positive to negative at x = 4, so g′ changes from positive to negative there: g has a relative maximum at x = 4, with value g(4) = 6.
Every question about g on these problems is really a Unit 5 question about a function whose derivative graph you were handed. Write "g′ = f" at the top of the page and the translation table does the rest.
If h(x) = ∫₁^(x³) e^t dt, what is h′(x)?
g(x) = ∫₀ˣ f(t) dt, and f is positive and decreasing on (0, 5). What is true of g there?
The graph-of-f accumulation question appears on nearly every exam and carries several points at once: values of g, an extremum with justification, concavity, and often a tangent line to g. Getting g′ = f stated early sets up all of them.
Answer the 2 checkpoints as you read.
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