Find where the accumulation increases or decreases

A pond that rises and falls

A pond is 30 centimeters deep at \(t = 0\) hours, and rain and a drain change its depth. The upper graph shows the rate \(f(t)\) of this change in centimeters per hour, and the lower graph shows the depth \(D(t)\).

Drag \(t\) to the right. The region is teal where the rate is positive and red where it is negative. The lower graph is green where the depth rises and red where it falls.

At \(t = 1\), the rate is \(f(1) = 3\), above the axis, so the depth rises. At \(t = 3\), it is \(f(3) = -1\), below the axis, so the depth falls.

An orange dot marks each zero of the rate, at \(t = 2\) and \(t = 4\), where the depth turns around.

The sign of the rate decides

The rate is any function \(f\), and the amount \(F\) is the starting value plus the signed area under the curve.

Drag \(x\). The derivative of the amount is the rate, so the orange tangent on the lower graph has slope \(f(x)\), the height of the upper curve.

Where the upper curve is above the axis, the tangent slopes up and \(F\) increases. Where it is below, the tangent slopes down and \(F\) decreases.

Lift the lower graph. The colors do not change, because the sign of \(f\) decides whether \(F\) rises or falls, not its height.

Notation

1. The slope of the amount graph at \(x\) equals the height of the rate graph at \(x\).

   \[F'(x) = \cy{f(x)}\]

   This needs \(f\) to be continuous, so that \(F\) has a derivative.

2. If the rate is positive, the slope is positive, so \(F\) is increasing. The strip beside the edge is green because it adds area.

   \[\cy{f(x)} > 0 \;\Rightarrow\; \cg{F \text{ increasing}}\]

3. If the rate is negative, the slope is negative, so \(F\) is decreasing. The strip is red because it removes area.

   \[\cy{f(x)} < 0 \;\Rightarrow\; \cr{F \text{ decreasing}}\]

4. To find where \(F\) increases or decreases, solve \(f(x) = 0\). The edge is now on a zero, where the tangent is level. A continuous function changes sign only at a zero, so the sign at one point between two zeros holds on the whole interval.

5. For \(G(x) = cx + \int_a^x f(t)\,dt\), the derivative of \(cx\) is \(c\), so the slope of \(G\) is \(c + f(x)\). The region drops to the dashed line \(y = -c\), where its height is this sum, so compare the curve with that line, not the axis.

   \[\begin{aligned} &\frac{d}{dx}\left(cx + \int_a^x f(t)\,dt\right) \\ &\quad = c + \cy{f(x)} \end{aligned}\]

Example 1: a rate with two zeros

\(F(x) = \displaystyle\int_0^x (4 - t^2)\,dt\) for \(-3 \le x \le 3\). Find where \(F\) is increasing and where it is decreasing.

1. The slope of \(F\) is the integrand, so the orange tangent on the lower graph has slope \(4 - x^2\).

   \[F'(x) = 4 - x^2\]

2. The orange dots mark the zeros of the slope, \(x = \pm 2\), which split the interval into three parts.

   \[4 - x^2 = 0 \;\Rightarrow\; x = \pm 2\]

3. Test one point in each part. The tangent slopes down at \(x = -3\), up at \(x = 0\) and down at \(x = 3\).

   \[\begin{aligned} F'(-3) &= -5 \\ F'(0) &= 4 \\ F'(3) &= -5 \end{aligned}\]

4. \(F\) is decreasing on \([-3, -2]\), increasing on \([-2, 2]\) and decreasing on \([2, 3]\).

Example 2: a term outside the integral

\(G(x) = 2x + \displaystyle\int_1^x (t^2 - 9)\,dt\) for \(0 \le x \le 5\). Find where \(G\) is increasing and where it is decreasing.

1. The term \(2x\) has slope 2, and the integral has slope \(f(x) = x^2 - 9\), its integrand. Their sum is the slope of \(G\).

   \[G'(x) = 2 + (x^2 - 9) = x^2 - 7\]

2. A tempting wrong move: Use only the integrand. It is negative for \(x < 3\), so \(G\) decreases up to \(x = 3\).

   \[x^2 - 9 < 0 \text{ for } x < 3\]

   At \(x = 2.7\), the integrand is \(-1.71\), so the strip is red. But the term \(2x\) adds 2, so the slope of \(G\) is \(2 - 1.71 = 0.29\), and \(G\) already rises before \(x = 3\).

3. The slope is zero at \(x = \sqrt{7}\), about 2.65. There the curve crosses the dashed line \(y = -2\), because \(2 + f(x) = 0\) means \(f(x) = -2\).

   \[x^2 - 7 = 0 \;\Rightarrow\; x = \sqrt{7}\]

4. Test one point on each side. The slope is negative at \(x = 0\) and positive at \(x = 5\), so \(G\) is decreasing on \([0, \sqrt{7}]\) and increasing on \([\sqrt{7}, 5]\).

   \[G'(0) = -7, \qquad G'(5) = 18\]

Practice

\(Q(x) = \displaystyle\int_0^x (1 - 2\sin t)\,dt\) for \(0 \le x \le \pi\). Find where \(Q\) is increasing and where it is decreasing.

Find where the upper curve crosses the axis. Then test one point in each part.

Questions

The upper graph shows a rate \(f\) made of straight pieces through \((0, 2)\), \((2, 0)\), \((4, -3)\), \((6, 0)\) and \((8, 3)\). The amount is \(F(x) = \int_0^x f(t)\,dt\).

1. On which interval is \(F\) decreasing?

   • From 2 to 6
   • From 0 to 2
   • From 6 to 8
   Answer

   From 2 to 6

2. What is true of \(F\) at \(t = 4\)?

   • \(F\) is decreasing, and its slope is \(-3\).
   • \(F\) has a minimum.
   • \(F(4) = -3\).
   Answer

   \(F\) is decreasing, and its slope is \(-3\).

3. \(R(x) = 10 - 4x + \displaystyle\int_1^x t^2\,dt\) for \(x > 0\). Which statement is true?

   • \(R\) decreases on \((0, 2)\) and increases on \((2, \infty)\).
   • \(R\) increases for all \(x > 0\).
   • \(R\) decreases for all \(x > 0\).
   Answer

   \(R\) decreases on \((0, 2)\) and increases on \((2, \infty)\).

Summary

The depth rises while the rate is above the axis and falls while it is below, because the derivative of the accumulation is the rate.

You add the slope of any term outside the integral to the rate, find its zeros, and test the sign between them.

The sign of the rate answers any question about where a function defined by an integral increases or decreases.

This is one of five skills in topic 6.5 of Which & Why: AP Calculus.