Mistake Master

The Fundamental Theorem of Calculus and Accumulation Functions AB & BC

With the upper limit free to move, $g(x) = \int_{a}^{x} f(t)\,dt$ is a function of $x$: the signed area collected from $a$ to $x$, so $g(a) = 0$ and $g$ falls wherever $f$ is below the axis. The letter $t$ inside is a dummy and never meets the $x$ outside. The Fundamental Theorem of Calculus, Part 1, says that for continuous $f$ this $g$ is differentiable with $g'(x) = f(x)$: pushing the finish line forward by $h$ adds a strip of width $h$ and height about $f(x)$. The starting point $a$ does not appear in the conclusion, because moving it shifts $g$ by a constant.

The marks are lost when the limit is not simply $x$. A composite upper limit brings the chain rule, $\frac{d}{dx}\int_{a}^{u(x)} f(t)\,dt = f(u(x))\,u'(x)$, so the height is read at the finish line and the width is how fast the finish line travels. A variable lower limit flips the sign, giving $-f(v(x))\,v'(x)$, since advancing it removes area. When both limits move, the two contributions subtract, and a single-term answer is the signature of a moving edge that went unnoticed.

g(x) IS THE AREA COLLECTED FROM t = 1 UP TO t = x. x x x = 3, g(3) = 4 x = 5, g(5) = 10 MOVING THE RIGHT EDGE CHANGES THE AREA, SO g IS A FUNCTION OF x. THE LETTER t INSIDE IS A DUMMY. IT NEVER MEETS THE x OUTSIDE.
Drawn to scale at 48 px per unit across and 30 px per unit up, under $f(t) = \frac{t}{2} + 1$. The right-hand region is two and a half times the left-hand one, because the strip being added keeps getting taller.
THE UPPER LIMIT IS u(x), NOT x. THE EDGE MOVES AT u′(x). AREA = g(u(x)) NEW SLIVER HEIGHT f(u(x)) WIDTH u′(x) dx u(x) SO THE DERIVATIVE IS f(u(x)) TIMES u′(x). WITH u(x) = x THE FACTOR IS 1 AND THE CHAIN RULE IS INVISIBLE. WITH u(x) = x² IT IS 2x, AND DROPPING IT COSTS THE MARK.
The sliver is drawn wide enough to see. Its height is read at the finish line rather than at $x$, and its width is how fast the finish line is travelling.

The work

3 ways in · any order
Lesson
The Fundamental Theorem of Calculus and Accumulation Functions

Defines the accumulation function with a moving upper limit, derives the Fundamental Theorem Part 1 from the strip of new area, and then works the three cases that cost marks: a composite upper limit carrying a chain-rule factor, a variable lower limit carrying a minus sign, and both limits moving at once giving two terms.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the two failure modes of this topic: differentiating an accumulation function without the chain-rule factor its limit demands or without the sign flip a moving lower limit demands, and misreading the notation, including the dummy variable and the value of the accumulation at its own starting point.

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