A tank’s rate of change of water volume is given by a differentiable function…
MATHEMATICS · Single correct · A Level · Mathematics · Integration · Integration
Problem
A tank’s rate of change of water volume is given by a differentiable function r(t) measured in liters per minute. Over certain intervals, r(t) attains negative values.
Which of the following best explains why portions of the graph that lie below the t-axis decrease the total change in water volume when the definite integral ∫abr(t)dt is computed?
(a) Because area is always taken as positive, portions of the graph below the t-axis contribute a positive amount, so they do not reduce the net change.
(b) Since the limits of integration dt are positive, multiplying a negative value of r(t) by a positive dt yields a negative product which is then multiplied by another negative when summed, turning it positive and leaving the total unchanged.
(c) Negative values of r(t) indicate water is leaving the tank, so the signed area below the axis subtracts from the total accumulation.
(d) The Fundamental Theorem of Calculus accumulates only where r(t)>0; any interval on which r(t)<0 is ignored, so those segments do not affect the net change.