Mistake Master

Using Accumulation Functions and Definite Integrals in Applied Contexts AB & BC

An applied integral has the units of its integrand times those of its variable, so a rate in gallons per minute integrated over minutes gives gallons and the "per" cancels; an answer that still carries a "per" has confused a total with a rate. $R(6) = 20$ gallons per minute describes one instant while $\int_0^6 R = 84$ gallons describes the whole interval. Amounts are built as $A(b) = A(a) + \int_a^b A'$, and an accumulation function $G(x) = \int_2^x f$ satisfies $G(2) = 0$ because an interval of zero width accumulates nothing.

With an inflow $R$ and an outflow $W$, the change in the amount is $\int (R - W)$, and three different questions live in one setup: how much entered ($\int R$), by how much the amount changed ($\int (R - W)$), and how much is present at the end (the initial amount plus that change). Since $A' = R - W$, the amount is greatest where the two rates cross, which for $R = 8 + 2t$ and $W = 4t$ is $t = 4$. An interpretation earns its point by naming the quantity, the interval and the units: $84$ gallons of water entered during the first six minutes.

TWO RATES IN GALLONS PER MINUTE, OVER SIX MINUTES. +16 GAL W(t) = 4t, OUT R(t) = 8 + 2t, IN THEY CROSS AT t = 4 BEFORE THE CROSSING THE TANK GAINS 16 GAL AFTER IT, THE TANK LOSES 4 GAL THE AMOUNT IS GREATEST WHERE THE RATES CROSS, NOT WHERE EITHER RATE IS LARGEST. STARTING AT 50 GAL, THE TANK PEAKS AT 66.
Drawn to scale at $55$ px per minute and $8$ px per gallon-per-minute. Both rates are straight lines, so the shaded regions are exact triangles of area $16$ and $4$. The tank rises from $50$ to $66$ gallons and falls back to $62$.
FOUR CORRECT NUMBERS FROM ONE TANK PROBLEM. THE NUMBER WHAT IT IS WHAT IT IS NOT 20 THE INFLOW RATE AT t = 6 AN AMOUNT OF WATER 84 GALLONS THAT ENTERED WHAT THE TANK HOLDS 12 THE CHANGE IN THE AMOUNT THE FINAL AMOUNT 62 GALLONS AT t = 6 HOW MUCH ENTERED EVERY NUMBER IN THE LEFT COLUMN IS CORRECT ARITHMETIC. THE POINT IS EARNED OR LOST IN THE COLUMN NEXT TO IT.
All four rows come from the same problem: water in at $8 + 2t$, out at $4t$, starting from $50$ gallons. Nothing separates them except which question was asked, which is why an interpretation has to name the quantity, the interval and the units.

The work

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Lesson
Using Accumulation Functions and Definite Integrals in Applied Contexts

Reads the units of an applied integral off the integrand and the variable, adds the initial condition that turns accumulated change into an amount, separates the three questions hiding in an inflow-and-outflow problem, locates the maximum where the rates cross, and gives a three-part template for the interpretation sentence.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items on applied accumulation: distinguishing a rate at an instant from a total over an interval, adding the starting amount, handling competing inflow and outflow rates, locating where an amount peaks, and stating in words what an integral represents.

Not yet available · 10 items
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears the misconception and moves you to the next.

Take the diagnostic to identify your misconceptions