A function can be defined by an integral. Given a continuous \(f\) and a number \(a\) in its interval, set

\[g(x) = \int_a^x f(t)\,dt,\]

and \(g\) is a perfectly ordinary function with a graph, a derivative, and extrema. The framework calls it an accumulation function and states plainly what makes it usable: graphical, numerical, analytical, and verbal representations of \(f\) provide information about \(g\).

Why \(g' = f\) is the Fundamental Theorem and is argued elsewhere. This article is about the consequence: the translation from a picture of \(f\) to the behaviour of \(g\).

The translation

Two lines do all the work. The Fundamental Theorem gives \(g'(x) = f(x)\), and differentiating again gives \(g''(x) = f'(x)\). Everything follows by applying the usual derivative tests to those.

about \(g\) read from \(f\)
\(g\) is increasing \(f\) is above the axis
\(g\) is decreasing \(f\) is below the axis
\(g\) has a local maximum \(f\) crosses from above to below
\(g\) has a local minimum \(f\) crosses from below to above
\(g\) is concave up \(f\) is increasing
\(g\) is concave down \(f\) is decreasing
\(g\) has an inflection point \(f\) has a local extremum or a corner
\(g(x) = 0\) the signed area from \(a\) to \(x\) cancels out

This is reading the graph of a derivative with the labels shifted one place: \(f\) plays for \(g\) the role \(f'\) plays for \(f\). The skill is the same one and the table is the same table.

The picture, and the lower limit

Top: a piecewise linear f, the format these questions arrive in, with the region between a and x shaded — dark where it counts positively, light where it counts negatively. Bottom: the graph of g for the current lower limit, with hollow dots at its local extrema and bars at its inflection points. Move x and watch g trace out. Then move a and watch the whole curve slide vertically while every marker stays exactly where it was, because changing the lower limit changes g by a constant and a constant has no effect on any derivative.

The integrand crosses the axis at \(x = 0.5\), \(3.5\), and \(7.25\), and those are exactly where \(g\) turns: down to up at the first, up to down at the second, down to up again at the third. It has corners at \(x = 2\) and \(x = 5\), where its slope jumps from \(+2\) to \(-2\) and from \(-2\) to \(+\tfrac43\), and those are exactly where \(g\) changes concavity.

Values of \(g\) come from geometry, not from an antiderivative — the framework says as much, that a definite integral can sometimes be evaluated using areas. To get \(g(3.5)\) with \(a = 0\), cut the region at every axis crossing and add the pieces with their signs. From \(0\) to \(0.5\) the graph is a triangle below the axis with base \(0.5\) and height \(1\), contributing \(-\tfrac14\). From \(0.5\) to \(2\) it is a triangle above the axis with base \(1.5\) and height \(3\), contributing \(\tfrac94\). From \(2\) to \(3.5\) it is a triangle above the axis with base \(1.5\) and height \(3\) again, contributing another \(\tfrac94\). Altogether

\[g(3.5) = -\tfrac14 + \tfrac94 + \tfrac94 = \tfrac{17}{4},\]

and no antiderivative of \(f\) was ever written down. Cutting at the crossings is the part that gets skipped, and skipping it turns the first triangle from \(-\tfrac14\) into \(+\tfrac14\).

The signed areas of the three linear pieces are \(2\), \(0\), and \(-3\). The middle one being zero is worth pausing on: \(f\) is not zero on \([2,5]\) and \(g\) is certainly not constant there, but the positive and negative parts cancel exactly, so \(g(5) = g(2)\). A definite integral of zero says the accumulation returned to where it started, not that nothing happened.

What the lower limit does, and does not do

Move \(a\) and the entire graph of \(g\) slides vertically. Not one marker moves.

The reason is one line of the framework’s own properties of definite integrals — the integral over adjacent intervals adds:

\[\int_{a_2}^{x} f = \int_{a_2}^{a_1} f + \int_{a_1}^{x} f.\]

The first term on the right does not involve \(x\). So changing the lower limit from \(a_1\) to \(a_2\) adds a constant to \(g\), and a constant has no derivative, which is why \(g'\), \(g''\), and every feature they control are untouched.

One thing does change: the value. With \(a = 0\) the maximum of \(g\) is \(\tfrac{17}{4}\) at \(x = 3.5\); with \(a = 2\) the same maximum, at the same place, is \(\tfrac94\). And \(g(a) = 0\) always, because the integral from a point to itself is zero — which also means the zeros of \(g\) move when \(a\) does, while its extrema do not.

Below the lower limit

If \(x < a\) the integral runs backwards, and reversing the limits reverses the sign:

\[\int_a^x f = -\int_x^a f.\]

So \(g\) is defined to the left of \(a\) too, and the translation table still holds there without modification — \(g\) still rises where \(f\) is positive. It is worth checking that on the tool by putting \(a\) to the right of \(x\), because the sign flip feels as though it ought to reverse something, and it does not.

On a free-response question the graph of \(f\) is given and \(g\) is defined in the stem, and the parts almost always run in this order: a value of \(g\), then \(g'\) or a statement about increase, then \(g''\) or a statement about concavity, then a justification. Each part is one line of the table above. Write \(g' = f\) and \(g'' = f'\) at the top of the page before reading part (a), and the rest of the question is arithmetic on a picture.