Wemazi

Topic 6.5

Calculate an accumulation value

Builds on defining a function by an integral with a moving upper limit. 8 minutes.

A tank that fills and drains

A tank holds 20 liters of water at time \(t = 0\) minutes. The upper graph shows the rate \(f(t)\) at which water enters the tank, in liters per minute. Where the curve is below the axis, the rate is negative, so water flows out.

Drag \(t\) to the right. The shaded region fills the space between the curve and the axis, from 0 to \(t\). It is teal above the axis and red below it, where the area counts as negative.

The lower graph shows the amount of water \(G(t)\) in liters, which is the starting amount 20 plus the signed area of the shaded region.

At \(t = 2\), the shaded region is a triangle with base 2 and height 6, so its area is 6 and the tank holds 26 liters. At \(t = 6\), the tank holds 44 liters. From 6 to 7, a red triangle with area 1.5 removes water, which leaves 42.5 liters.

Any rate, any starting value

The rate is any function \(f\), and the amount is a function \(F\). The starting point is \(a\), and the known value there is \(F(a)\).

Drag \(x\). While the edge moves through a teal part, the lower graph rises, because area is added. Through a red part it falls, because that area counts as negative.

Drag the starting value. The whole lower graph moves up or down and keeps its shape, because the shaded area does not depend on it.

At every \(x\), the lower graph is the starting value plus the signed area from \(a\) to \(x\).

Notation

  1. The shaded region runs from the starting point \(a\) to \(x\), and its signed area is a definite integral with \(x\) as the upper limit.

    \[\cb{\int_a^x f(t)\,dt}\]

  2. The amount at \(x\) is the starting value plus the signed area.

    \[F(x) = F(a) + \cb{\int_a^x f(t)\,dt}\]

  3. At \(x = a\), the shaded region has no width, so its area is 0 and \(F(a)\) is the starting value.

    \[\int_a^a f(t)\,dt = 0\]

  4. Where \(f\) is positive the amount rises, and where \(f\) is negative it falls.

    The rate \(f\) must be continuous between \(a\) and \(x\), so that the area is defined.

When it fails

The known value belongs to one input. Suppose the tank holds 20 liters at \(t = 2\), not at \(t = 0\), so the area counts from \(t = 2\). From 2 to 4 the region is a rectangle with area \(2 \cdot 6 = 12\), so the amount at \(t = 4\) is \(20 + 12 = 32\) liters.

If you count from 0 instead, you add the first triangle's 6 liters, which the 20 already includes, and reach 38.

Example 1: a starting point away from zero

\(F(1) = 4\), and the rate of change of \(F\) is \(f(t) = 2t - 2\). Find \(F(4)\).

  1. The known value is at \(t = 1\), so the area starts there. The line crosses the axis at \(t = 1\), because \(f(1) = 0\).

    \[f(1) = 2(1) - 2 = 0\]

  2. From 1 to 4, the region is a triangle with base 3 and height \(f(4) = 6\).

    \[f(4) = 2(4) - 2 = \cy{6}\]

  3. The area of a triangle is half the base times the height.

    \[\cb{\int_1^4 f(t)\,dt} = \tfrac12 (3)(6) = \cb{9}\]

  4. The line is above the axis, so the area is positive. Add it to the starting value.

    \[F(4) = 4 + 9 = 13\]

Example 2: a rate that changes sign

\(F(0) = 9\), and the rate of change of \(F\) is \(f(t) = t - 4\). Find \(F(6)\).

  1. The line crosses the axis at \(t = 4\). It is below the axis before 4 and above it after.

    \[\begin{aligned} f(0) &= -4 \\ f(4) &= 0 \\ f(6) &= 2 \end{aligned}\]

  2. The red triangle from 0 to 4 has base 4 and height 4.

    \[\tfrac12 (4)(4) = \cr{8}\]

  3. The teal triangle from 4 to 6 has base 2 and height 2.

    \[\tfrac12 (2)(2) = \cb{2}\]

  4. A tempting wrong move: Add both areas to the starting value.

    \[F(6) = 9 + 8 + 2 = 19\]

    The red triangle lies below the axis, so it counts as negative. The amount falls from 9 to 1 by \(t = 4\), and only 2 more units of area follow, so it cannot reach 19.

  5. Subtract the red area and add the teal area.

    \[F(6) = 9 - \cr{8} + \cb{2} = 3\]

Practice

\(G(0) = 5\), and the rate of change of \(G\) is \(g\). The graph of \(g\) is the horizontal line \(g(t) = 3\) from 0 to 2, then a straight line through \((3, 0)\) and \((4, -3)\). Find \(G(4)\).

Find where the graph crosses the axis, and decide which pieces of area lie below it.

Solution

  1. From 0 to 2, the region is a rectangle with width 2 and height 3.

    \[\int_0^2 g(t)\,dt = 6\]

  2. From 2 to 3, the region is a teal triangle with base 1 and height 3. The red triangle from 3 to 4 has the same size and counts as negative.

    \[\begin{aligned} \int_2^3 g(t)\,dt &= 1.5 \\ \int_3^4 g(t)\,dt &= -1.5 \end{aligned}\]

  3. Add the three signed areas to the starting value.

    \[G(4) = 5 + 6 + 1.5 - 1.5 = 11\]

Questions

The figure shows a rate \(f\) from 0 to 6. The graph rises from \((0, 0)\) to \((2, 4)\), stays at 4 until \(t = 4\), then falls in a straight line through \((5, 0)\) to \((6, -4)\). The amount \(F\) has \(F(0) = 3\).

  1. What is \(F(2)\)?

    • 7
    • 4
    • 3
    Answer

    7

  2. What is \(F(6)\)?

    • 15
    • 19
    • 12
    Answer

    15

  3. \(F(1) = 6\) and \(F'(x) = 2x + 3\). What is \(F(3)\)?

    • 20
    • 24
    • 14
    Answer

    20

Summary

The amount is the known starting value plus the signed area from the starting point, where area below the axis counts as negative.

This works because the area so far is the change in the amount.

You use it when a graph or a formula gives a rate and the amount is known at one input.

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