Distinguish the graph's height from accumulated area
An account that starts in debt
An account starts with a balance of \(-8\) hundred dollars, so the owner owes money. The upper graph shows the rate \(b(t) = 6 - 2t\) at which the balance changes, in hundreds of dollars per month. The lower graph shows the balance \(B(t)\). The shaded region is teal above the axis and red below it, where it counts as negative.
Drag \(t\) to the right. The orange segment on the upper graph is the height of the rate, and the teal segment on the lower graph is the balance.
The rate is zero at \(t = 3\), but the balance there is 1, not 0.
The balance is zero at \(t = 2\) and \(t = 4\), where the rate is 2 and \(-2\), not 0.
Where the rate is zero, the balance turns from rising to falling. Where the balance is zero, the account owes nothing.
Two graphs, two questions
The orange segment reaches the upper curve at height \(f(x)\), and the teal segment reaches the lower curve at height \(F(x)\).
\(f(x)\) is a rate and \(F(x)\) is an amount, so they are different quantities with different zeros.
Drag \(x\) to where the upper curve meets the axis. The tangent on the lower graph is flat there, so \(F\) turns, at whatever height it has.
Move the starting value. The zeros of \(f\) stay, but the zeros of \(F\) move, because \(F\) depends on the starting value.
Notation
The height \(f(x)\) of the upper graph is the rate, and also the slope of the lower graph.
\[\cy{f(x)} = F'(x)\]The height \(F(x)\) of the lower graph is the amount, the starting value plus the signed area.
\[F(x) = F(a) + \cb{\int_a^x f(t)\,dt}\]Where \(f\) is zero, the slope of \(F\) is zero, so \(F\) has a flat tangent. At the first orange dot, that tangent is above the axis, not on it.
\[\cy{f(x)} = 0 \;\Rightarrow\; F'(x) = 0\]\(F\) is zero where the signed area cancels the starting value. At the teal dots, \(F\) is on its axis but \(f\) is not.
\[\begin{aligned} F(x) = 0 \;\Leftrightarrow\; &\cb{\int_a^x f(t)\,dt} \\ &= -F(a) \end{aligned}\]To find the zeros of \(f\), solve \(f(x) = 0\). To find the zeros of \(F\), solve \(F(x) = 0\), not \(f(x) = 0\).
Example 1: a zero of the rate and a zero of the amount
\(F(x) = \displaystyle\int_0^x (e^t - 2)\,dt\) for \(x \ge 0\). Find the zero of the integrand, and find the zeros of \(F\).
The integrand is zero when \(e^t = 2\), so \(f\) has one zero.
\[t = \ln 2\]As a check, \(\ln 2 \approx 0.693\).
\(F(0) = 0\), because the region from 0 to 0 has no width. This zero of \(F\) is at 0, not at \(\ln 2\).
Evaluate the antiderivative \(e^t - 2t\) from 0 to \(\ln 2\). The value is negative, so the flat tangent of \(F\) is below the axis, not on it.
\[\begin{aligned} F(\ln 2) &= (2 - 2\ln 2) - 1 \\ &= 1 - 2\ln 2 \approx \cr{-0.386} \end{aligned}\]After \(\ln 2\), the rate is positive, so \(F\) rises and returns to zero where \(e^x = 1 + 2x\). A calculator solves this equation.
\[\begin{aligned} F(x) &= e^x - 1 - 2x = 0 \\ &\Rightarrow\; x \approx 1.256 \end{aligned}\]
Example 2: the amount where the rate is zero
\(F(0) = -5\). The graph of \(f\) is a straight-line path through \((0, 0)\), \((2, 4)\) and \((4, -4)\). Find \(F(3)\). Is \(F\) zero where \(f\) is zero?
The line from \((2, 4)\) to \((4, -4)\) falls by 4 for each unit, so it crosses the axis at \(t = 3\).
\[f(3) = 0\]A tempting wrong move: The rate is zero at \(t = 3\), so \(F(3) = 0\).
\[f(3) = 0 \;\Rightarrow\; F(3) = 0\]The lower graph starts at \(-5\) and rises, so by \(t = 3\) it is above the axis. Its slope is zero there, so it turns, but it turns above the axis, not at the red dot.
Add the areas to the starting value. The triangles have height 4 and bases 2 and 1, so their areas are 4 and 2.
\[F(3) = -5 + \cb{4} + \cb{2} = 1\]\(F\) is 1 where \(f\) is zero, not 0. \(F\) crosses the axis earlier, between \(t = 2\) and \(t = 3\), because \(F(2) = -1\).
Practice
An account has \(B(0) = -12\) and rate \(b(t) = 8 - 2t\), for \(0 \le t \le 8\). Find the time when the rate is zero. Find the times when the balance is zero.
Drag t to see where each graph crosses its axis. The balance is −12 plus the integral of the rate, so solve B(t) = 0.
Solution
The rate is zero when \(8 - 2t = 0\).
\[t = 4\]The balance is the starting value plus the signed area from 0 to \(t\).
\[\begin{aligned} B(t) &= -12 + \int_0^t (8 - 2s)\,ds \\ &= -12 + 8t - t^2 \end{aligned}\]\(B(t)\) factors as \(-(t - 2)(t - 6)\), so the balance is zero at \(t = 2\) and \(t = 6\). At \(t = 4\), where the rate is zero, the balance is 4, its highest value.
\[\begin{aligned} -(t - 2)(t - 6) &= 0 \\ B(4) &= 4 \end{aligned}\]
Questions
The upper graph shows a rate \(f\), a straight-line path through \((0, 0)\), \((2, 4)\), \((4, 0)\), \((6, -4)\) and \((8, 0)\). The amount is \(F(x) = -3 + \int_0^x f(t)\,dt\).
What is \(F(4)\)?
- 5
- 0
- 8
Answer
5
How many times does \(F\) equal 0 for \(0 \le x \le 8\)?
- 2
- 3
- 1
Answer
2
\(H(x) = -4 + \displaystyle\int_0^x 3t^2\,dt\). For \(x > 0\), where is \(H(x) = 0\)?
- \(x = \sqrt[3]{4}\)
- \(x = \dfrac{2}{\sqrt{3}}\)
- \(x = 0\)
Answer
\(x = \sqrt[3]{4}\)
Summary
The rate is zero at \(t = 3\), where the balance turns at height 1, not at 0.
The balance is zero at \(t = 2\) and \(t = 4\), because there the signed area cancels the starting value \(-8\).
To find where an accumulation is zero, you solve \(F(x) = 0\), not \(f(x) = 0\), because \(f\) is the slope of \(F\), not its height.
This is one of five skills in topic 6.5 of Which & Why: AP Calculus.