A function can be defined by an integral. Let

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

where \(f\) is continuous. The Fundamental Theorem gives \(g'(x)=f(x)\), and differentiating again gives \(g''(x)=f'(x)\). Those two equations turn a graph of \(f\) into information about the behavior of \(g\).

The translation

About \(g\) Read from \(f\)
\(g\) is increasing \(f>0\)
\(g\) is decreasing \(f<0\)
\(g\) has a local maximum \(f\) crosses from positive to negative
\(g\) has a local minimum \(f\) crosses from negative to positive
\(g\) is concave up \(f\) is increasing
\(g\) is concave down \(f\) is decreasing
\(g\) changes concavity \(f'\) changes sign
\(g(x)=0\) the signed area from \(a\) to \(x\) is zero

This is the same reasoning used when reading the graph of a derivative, and here \(f\) plays the role of \(g'\).

Reading values from geometry

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 top graph shows \(f\), with the region between \(a\) and \(x\) shaded according to sign, and the lower graph shows the corresponding accumulation function.

Suppose \(a=0\). To compute \(g(3.5) = \textstyle\int_0^{3.5}f(t)\,dt\), add the signed areas. In the displayed example, the first small triangle lies below the axis and contributes \(-\tfrac14\), and the next two triangles lie above the axis and each contribute \(\tfrac94\), so

\[g(3.5) = -\frac14+\frac94+\frac94 = \frac{17}{4}\]

No antiderivative is needed, and the graph is enough. A definite integral of zero also does not mean nothing happened. If positive and negative areas cancel, the accumulation can return to an earlier value even while the function changes throughout the interval.

What changing the lower limit does

Changing \(a\) shifts the entire accumulation function vertically, and it does not change its derivative. To see why, compare \(g_1(x) = \textstyle\int_{a_1}^{x}f(t)\,dt\) with \(g_2(x) = \textstyle\int_{a_2}^{x}f(t)\,dt\). Using additivity,

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

The first term is constant with respect to \(x\), so \(g_2\) differs from \(g_1\) by a constant. That means the location of extrema and inflection points does not change, though the function values do change. Also, \(g(a)=0\) for any lower limit \(a\), because an integral from a point to itself is zero.

When \(x\) is less than \(a\)

The accumulation function is still defined when the upper limit lies to the left of the lower limit. Reversing the bounds changes the sign, so \(\textstyle\int_a^x f(t)\,dt = -\int_x^a f(t)\,dt\). The derivative relationship remains \(g'(x)=f(x)\), so \(g\) still increases wherever \(f\) is positive and decreases wherever \(f\) is negative, even to the left of \(a\).

A reliable setup on a free-response problem is to write \(g'=f\) and \(g''=f'\) before answering anything else. Then separate two kinds of questions. Questions about values of \(g\) use signed area, and questions about increase, decrease, extrema, and concavity use the derivative relationships.