The definite integral of a continuous function on is zero even though the graph of…
MATHEMATICS · Single correct · AP · Calculus AB · Integration and Accumulation of Change · Integration and Accumulation of Change
Problem
The definite integral of a continuous function p on [a,b] is zero even though the graph of p intersects the x-axis only at two interior points c and d with a<c<d<b. Which of the following best explains how this can occur?
(a) Because the definite integral computes the average value of p over [a,b], it is zero whenever p takes both positive and negative values.
(b) Because a continuous function that crosses the x-axis twice must enclose equal positive and negative areas by symmetry, making the integral zero.
(c) Because the areas above and below the x-axis are equal in magnitude, their signed contributions cancel, so the integral is zero.
(d) Because the Fundamental Theorem of Calculus guarantees that the integral of a continuous function over any interval in which the function changes sign is zero.