Answer key · Unit 15

Calculus.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Average rate of change

A line has one slope, and a curve does not. Measure the average rate between two points with a secant, and see what happens as the two points close in.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
From a line to a curve
Practice · Calculate
Question
What is the average rate of change of $y = x^2$ ⁠ between $x = 1$ ⁠ and $x = 3$ ⁠?

Correct answer
4 ± 0.01
Setup, choices and feedback
The Scene at this step
Point x
1fixed
Run h
2fixed
The secant / The tangent
The secant
Other settings
Mark secant/nodes/graph/props/markSecant = false
Other settings
Mark slope/nodes/graph/props/markSlope = false
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Rise ÷ run, the slope of the secant; NaN when run is 0/nodes/graph/outputs/secantSlope
Wrong answers it recognises
- 8 ± 0.01
  Misconception: Gives the change in y as the rate, without dividing by the change in x.
  Feedback from the catalogue
  A rate is a change per unit of $x$ ⁠: divide the change in $y$ ⁠ by the change in $x$ ⁠. Between $x = 1$ ⁠ and $x = 3$ ⁠ the parabola rises $8$ ⁠, but over $2$ ⁠ units, so the average rate is $8 \div 2 = 4$ ⁠.
- 0.25 ± 0.01
  Misconception: Divides the change in x by the change in y.
  Feedback from the catalogue
  Slope and rate are the change in $y$ ⁠ divided by the change in $x$ ⁠, just as for a line: rise on top, run underneath. The other way round gives the upside-down fraction.
Feedback for a right answer
$\dfrac{9 - 1}{3 - 1} = \dfrac{8}{2} = 4$ ⁠. That is the slope of the secant between the two points. Choose The tangent under Look inside and compare it with the tangent at $x = 1$ ⁠.

Feedback for a wrong answer
Not yet. Divide the change in $y$ ⁠ by the change in $x$ ⁠.

Hint
$f(1) = 1$ ⁠ and $f(3) = 9$ ⁠. How much did $y$ ⁠ change, and how much did $x$ ⁠?

Worked solution
1. At the first point: $f(1) = 1^2 = 1$ ⁠.
2. At the second: $f(3) = 3^2 = 9$ ⁠.
3. The change in $y$ ⁠: $9 - 1 = 8$ ⁠. The change in $x$ ⁠: $3 - 1 = 2$ ⁠.
4. The average rate: $\dfrac{8}{2} = 4$ ⁠.
$8$ ⁠ is only the change in $y$ ⁠, not divided. $0.25$ ⁠ is $\dfrac{2}{8}$ ⁠, the fraction upside down.
Step 2 of 4
Open this step
Bring the points together
Practice · Predict, then look
Question
Now $h = 2$ ⁠ and the secant’s slope is $4$ ⁠. Shrink Run h to $0.1$ ⁠, so that the second point almost touches the first. What happens to the secant’s slope?

Correct answer
- Correct It closes in on $2$ ⁠
Setup, choices and feedback
The Scene at this step
Point x
1fixed
Run h
2
The secant / The tangent
The secant
Choices
- Correct It closes in on $2$ ⁠
- Wrong It closes in on $0$ ⁠, because the rise almost vanishes
  Misconception: Thinks the secant's slope goes to zero as the two points close in, because rise and run both vanish.
  Feedback from the catalogue
  The rise and the run both shrink to zero, but the ratio between them does not: on $y = x^2$ ⁠ at $x = 1$ ⁠ it is $2 + h$ ⁠, which closes in on $2$ ⁠. That is the idea of a derivative: the ratio of two very small things can be a perfectly ordinary number.
- Wrong It stays $4$ ⁠
  Feedback
  On a line the slope does not depend on $h$ ⁠, but on a curve it does: the shorter the stretch, the closer the average is to the steepness at the point itself.
- Wrong It grows without limit
  Feedback
  The rise shrinks too, along with $h$ ⁠: the ratio between them stays an ordinary number.
Once the prediction is locked in, the Scene changes
Run h
0.1
The secant / The tangent
The tangent
The Scene then shows: 2 · The slope of the tangent at pointX, the derivative 2a·x + b/nodes/graph/outputs/tangentSlope
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At $h = 0.1$ ⁠ the slope is $2.1$ ⁠, and at $h = 0.01$ ⁠ it is $2.01$ ⁠: each time $2 + h$ ⁠. The secant turns into the tangent, whose slope at $x = 1$ ⁠ is exactly $2$ ⁠.

Worked solution
1. The secant’s slope from $x = 1$ ⁠ to $1 + h$ ⁠: $\dfrac{(1 + h)^2 - 1}{h}$ ⁠.
2. Expand: $(1 + h)^2 - 1 = 2h + h^2$ ⁠.
3. Divide by $h$ ⁠: $2 + h$ ⁠.
4. At $h = 2$ ⁠ that is $4$ ⁠, at $h = 0.1$ ⁠ it is $2.1$ ⁠, and as $h$ ⁠ shrinks to zero it closes in on $2$ ⁠.
The rise does vanish, but so does the run, and the ratio between them settles on $2$ ⁠: the slope of the tangent.
Step 3 of 4
Open this step
Either side of the vertex
Practice · Predict, then look
Question
What is the average rate of change of $y = x^2$ ⁠ between $x = -1$ ⁠ and $x = 1$ ⁠?

Correct answer
- Correct $0$ ⁠
Setup, choices and feedback
The Scene at this step
Point x
-1fixed
Run h
2fixed
The secant / The tangent
The secant
Other settings
Mark secant/nodes/graph/props/markSecant = false
Other settings
Mark slope/nodes/graph/props/markSlope = false
Choices
- Correct $0$ ⁠
- Wrong $1$ ⁠
  Misconception: Averages the y-values instead of finding the average rate of change.
  Feedback from the catalogue
  The average rate of change is how much $y$ ⁠ changed divided by how much $x$ ⁠ changed, not the average of the $y$ ⁠ values. From $x = -1$ ⁠ to $x = 1$ ⁠, $y$ ⁠ starts at $1$ ⁠ and ends at $1$ ⁠: the change is $0$ ⁠, so the average rate is $0$ ⁠.
- Wrong $2$ ⁠
  Feedback
  $2$ ⁠ is the slope of the tangent at $x = 1$ ⁠ alone. Here you average over the whole stretch, and for part of it the parabola goes down.
- Wrong It cannot be done, because the curve goes down and then up
  Feedback
  It can: the average does not care what happened along the way, only about the start and the end.
Once the prediction is locked in, the Scene changes
Other settings
Mark secant/nodes/graph/props/markSecant = true
Other settings
Mark slope/nodes/graph/props/markSlope = true
The Scene then shows: 0 · Rise ÷ run, the slope of the secant; NaN when run is 0/nodes/graph/outputs/secantSlope
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$f(-1) = 1$ ⁠ and $f(1) = 1$ ⁠: the same height at both ends, so $\dfrac{1 - 1}{2} = 0$ ⁠. The secant lies flat, even though the parabola went down and up along the way.

Worked solution
1. $f(-1) = (-1)^2 = 1$ ⁠ and $f(1) = 1^2 = 1$ ⁠.
2. The change in $y$ ⁠: $1 - 1 = 0$ ⁠. The change in $x$ ⁠: $1 - (-1) = 2$ ⁠.
3. The average rate: $\dfrac{0}{2} = 0$ ⁠.
The average hides the path: the fall to the vertex and the climb out of it cancel. $1$ ⁠ is the average of the $y$ ⁠ values, not of the change.
Step 4 of 4
Open this step
Look for a rate
Practice · Reach the target
Question
Move Point x so that the secant with $h = 1$ ⁠ has slope $5$ ⁠, then press Check.

Goal
- 5 ± 0.01
  At the start of the step: -3
Setup, choices and feedback
The Scene at this step
Point x
-2
Run h
1fixed
The secant / The tangent
The secant
Read from the Scene at
Rise ÷ run, the slope of the secant; NaN when run is 0/nodes/graph/outputs/secantSlope
One setting that reaches it
Point x
2
Feedback for a right answer
$(x + 1)^2 - x^2 = 2x + 1 = 5$ ⁠, so $x = 2$ ⁠: the secant from $(2, 4)$ ⁠ to $(3, 9)$ ⁠ rises $5$ ⁠ over one step.

Feedback for a wrong answer
Not yet. The further right the point, the steeper the secant.

Hint
The slope from $x$ ⁠ to $x + 1$ ⁠ is $(x + 1)^2 - x^2$ ⁠. Simplify and solve.

Worked solution
1. The secant’s slope from $x$ ⁠ to $x + 1$ ⁠: $\dfrac{(x + 1)^2 - x^2}{1}$ ⁠.
2. Expand: $x^2 + 2x + 1 - x^2 = 2x + 1$ ⁠.
3. Solve $2x + 1 = 5$ ⁠: $x = 2$ ⁠.
4. Set Point x to $2$ ⁠. Check: $f(3) - f(2) = 9 - 4 = 5$ ⁠. ✓

Lesson 2 of 4 · About 25 minutes

The derivative

The tangent at a point and its slope, the derivative. Why the derivative of x² is 2x, the rule for any quadratic function, and why the slope is zero at the vertex.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
The tangent
Practice · Calculate
Question
What is the slope of the tangent to $y = x^2$ ⁠ at $x = 3$ ⁠?

Correct answer
6 ± 0.01
Setup, choices and feedback
The Scene at this step
Point x
3fixed
a
1fixed
b
0fixed
The curve / The tangent
The curve
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The slope of the tangent at pointX, the derivative 2a·x + b/nodes/graph/outputs/tangentSlope
Wrong answers it recognises
- 9 ± 0.01
  Misconception: Gives the value of the function (or the function itself) instead of the slope.
  Feedback from the catalogue
  $f(x)$ ⁠ says how high the curve is; $f'(x)$ ⁠ says how steep it is. At $x = 3$ ⁠ on $y = x^2$ ⁠ the height is $9$ ⁠, but the tangent rises $6$ ⁠ for every step. The derivative of $x^2$ ⁠ is $2x$ ⁠, not $x^2$ ⁠.
- 3 ± 0.01
  Misconception: Lowers the power but drops the factor in front: the derivative of x² taken as x.
  Feedback from the catalogue
  When you differentiate, the power comes down in front as a factor: $x^2$ ⁠ gives $2x$ ⁠, and $3x^2$ ⁠ gives $3 \cdot 2x = 6x$ ⁠. You can check with a secant: on $y = x^2$ ⁠ the slope from $x$ ⁠ to $x + h$ ⁠ is $2x + h$ ⁠, which closes in on $2x$ ⁠.
Feedback for a right answer
$2 \cdot 3 = 6$ ⁠: the secant from $3$ ⁠ to $3 + h$ ⁠ has slope $6 + h$ ⁠, which closes in on $6$ ⁠. Choose The tangent under Look inside to see it.

Feedback for a wrong answer
Not yet. The question is about the steepness at the point, not the height.

Hint
The slope of the secant from $x$ ⁠ to $x + h$ ⁠ on $y = x^2$ ⁠ is $2x + h$ ⁠. What happens as $h$ ⁠ shrinks to zero?

Worked solution
1. The secant from $x = 3$ ⁠ to $3 + h$ ⁠: $\dfrac{(3 + h)^2 - 9}{h} = \dfrac{6h + h^2}{h} = 6 + h$ ⁠.
2. As $h$ ⁠ shrinks to zero, the slope closes in on $6$ ⁠.
3. In short: the derivative of $x^2$ ⁠ is $2x$ ⁠, and at $x = 3$ ⁠ it is $6$ ⁠.
$9$ ⁠ is the height of the curve at $x = 3$ ⁠, not the slope. $3$ ⁠ forgets the $2$ ⁠ that comes down from the power.
Step 2 of 4
Open this step
The derivative as a function
Practice · Choose
Question
The slopes of the tangents to $y = x^2$ ⁠ at $x = 1$ ⁠, $2$ ⁠ and $3$ ⁠ are $2$ ⁠, $4$ ⁠ and $6$ ⁠. What is the derivative of $x^2$ ⁠?

Correct answer
- Correct $2x$ ⁠
Setup, choices and feedback
The Scene at this step
Point x
1
a
1fixed
b
0fixed
The curve / The tangent
The tangent
Choices
Shown in a random order.
- Correct $2x$ ⁠
- Wrong $x^2$ ⁠
  Misconception: Gives the value of the function (or the function itself) instead of the slope.
  Feedback from the catalogue
  $f(x)$ ⁠ says how high the curve is; $f'(x)$ ⁠ says how steep it is. At $x = 3$ ⁠ on $y = x^2$ ⁠ the height is $9$ ⁠, but the tangent rises $6$ ⁠ for every step. The derivative of $x^2$ ⁠ is $2x$ ⁠, not $x^2$ ⁠.
- Wrong $x$ ⁠
  Misconception: Lowers the power but drops the factor in front: the derivative of x² taken as x.
  Feedback from the catalogue
  When you differentiate, the power comes down in front as a factor: $x^2$ ⁠ gives $2x$ ⁠, and $3x^2$ ⁠ gives $3 \cdot 2x = 6x$ ⁠. You can check with a secant: on $y = x^2$ ⁠ the slope from $x$ ⁠ to $x + h$ ⁠ is $2x + h$ ⁠, which closes in on $2x$ ⁠.
- Wrong $2$ ⁠
  Feedback
  $2$ ⁠ is the slope only at $x = 1$ ⁠. At other points the tangent is steeper, so the derivative changes with $x$ ⁠.
Feedback for a right answer
At every point the slope is twice the $x$ ⁠: $f'(x) = 2x$ ⁠. At $x = 0$ ⁠ it is $0$ ⁠, at the vertex, and to the left of it negative. Move Point x and check.

Hint
Each slope is how many times the point’s $x$ ⁠?

Worked solution
1. The table: $x = 1 \to 2$ ⁠, $x = 2 \to 4$ ⁠, $x = 3 \to 6$ ⁠. Each slope is $2$ ⁠ times $x$ ⁠.
2. The working confirms it: the secant from $x$ ⁠ to $x + h$ ⁠ has slope $\dfrac{(x + h)^2 - x^2}{h} = 2x + h$ ⁠, which closes in on $2x$ ⁠.
3. So $f'(x) = 2x$ ⁠.
$x^2$ ⁠ is the height, not the slope; at $x = 3$ ⁠ it gives $9$ ⁠, not $6$ ⁠.
Step 3 of 4
Open this step
Where is the tangent level?
Practice · Reach the target
Question
The board shows $y = x^2 - 4x + 3$ ⁠. Move Point x to where the tangent lies level, then press Check.

Goal
- 0 ± 0.01
  At the start of the step: -4
Setup, choices and feedback
The Scene at this step
Point x
0
a
1fixed
b
-4fixed
The curve / The tangent
The tangent
Other settings
C/nodes/graph/props/c = 3
Read from the Scene at
The slope of the tangent at pointX, the derivative 2a·x + b/nodes/graph/outputs/tangentSlope
One setting that reaches it
Point x
2
Feedback for a right answer
At $x = 2$ ⁠ the tangent is level: $f'(x) = 2x - 4$ ⁠, and $f'(2) = 0$ ⁠. That is the vertex, $(2, -1)$ ⁠, the parabola’s minimum point.

Feedback for a wrong answer
Not yet. If the tangent slopes down, move right; if it slopes up, move left.

Hint
The tangent is level at the vertex. In standard form the vertex is at $x = -\dfrac{b}{2a}$ ⁠.

Worked solution
1. The derivative: $f'(x) = 2x - 4$ ⁠.
2. The tangent is level where $f'(x) = 0$ ⁠: $2x - 4 = 0$ ⁠, that is, $x = 2$ ⁠.
3. Set Point x to $2$ ⁠.
4. That is exactly the vertex from the quadratic functions Unit: $x = -\dfrac{b}{2a} = -\dfrac{-4}{2} = 2$ ⁠, and $f(2) = -1$ ⁠.
Step 4 of 4
Open this step
The rule
Practice · Calculate
Question
What is the slope of $y = 3x^2 - 2x + 5$ ⁠ at $x = 1$ ⁠?

Correct answer
4 ± 0.01
Setup, choices and feedback
The Scene at this step
Point x
1fixed
a
3fixed
b
-2fixed
The curve / The tangent
The curve
Other settings
C/nodes/graph/props/c = 5
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The slope of the tangent at pointX, the derivative 2a·x + b/nodes/graph/outputs/tangentSlope
Wrong answers it recognises
- 9 ± 0.01
  Misconception: Keeps the constant term when differentiating.
  Feedback from the catalogue
  A constant only lifts or lowers the whole curve; it does not change how steep it is. So its derivative is $0$ ⁠: the derivative of $3x^2 - 2x + 5$ ⁠ is $6x - 2$ ⁠, without the $5$ ⁠.
- 1 ± 0.01
  Misconception: Lowers the power but drops the factor in front: the derivative of x² taken as x.
  Feedback from the catalogue
  When you differentiate, the power comes down in front as a factor: $x^2$ ⁠ gives $2x$ ⁠, and $3x^2$ ⁠ gives $3 \cdot 2x = 6x$ ⁠. You can check with a secant: on $y = x^2$ ⁠ the slope from $x$ ⁠ to $x + h$ ⁠ is $2x + h$ ⁠, which closes in on $2x$ ⁠.
- 6 ± 0.01
  Misconception: Gives the value of the function (or the function itself) instead of the slope.
  Feedback from the catalogue
  $f(x)$ ⁠ says how high the curve is; $f'(x)$ ⁠ says how steep it is. At $x = 3$ ⁠ on $y = x^2$ ⁠ the height is $9$ ⁠, but the tangent rises $6$ ⁠ for every step. The derivative of $x^2$ ⁠ is $2x$ ⁠, not $x^2$ ⁠.
Feedback for a right answer
$f'(x) = 6x - 2$ ⁠, and $f'(1) = 4$ ⁠. Choose The tangent under Look inside and check.

Feedback for a wrong answer
Not yet. Differentiate first, and substitute only after that.

Hint
Differentiate with the rule $f'(x) = 2ax + b$ ⁠, then substitute $x = 1$ ⁠.

Worked solution
1. Differentiate term by term: $3x^2 \to 6x$ ⁠, $-2x \to -2$ ⁠, $5 \to 0$ ⁠.
2. $f'(x) = 6x - 2$ ⁠.
3. Substitute $x = 1$ ⁠: $f'(1) = 6 - 2 = 4$ ⁠.
$9$ ⁠ keeps the $5$ ⁠ in the derivative. $1$ ⁠ differentiates $3x^2$ ⁠ to $3x$ ⁠, without the $2$ ⁠. And $6$ ⁠ is $f(1)$ ⁠, the height, not the slope.

Lesson 3 of 4 · About 25 minutes

The area under a curve

The area under a line is a triangle, and under a curve you estimate it with narrow rectangles. The antiderivative gives it exactly, and area below the axis counts as negative.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
The area under a line
Practice · Calculate
Question
What is the area under the line $y = x$ ⁠, above the $x$ ⁠-axis, between $x = 0$ ⁠ and $x = 4$ ⁠?

Correct answer
8 ± 0.01
Setup, choices and feedback
The Scene at this step
Up to x
4fixed
Slices
0fixed
The curve / The area
The curve
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The signed area between the curve and the x-axis from areaFrom to areaTo, the integral/nodes/graph/outputs/area
Wrong answers it recognises
- 16 ± 0.01
  Misconception: Takes base × height for a triangle, forgetting the half.
  Feedback from the catalogue
  The area of a triangle is half the base times the height: it is half the rectangle around it. Under $y = x$ ⁠ from $0$ ⁠ to $4$ ⁠: $\tfrac{1}{2} \cdot 4 \cdot 4 = 8$ ⁠, not $16$ ⁠.
Feedback for a right answer
$\tfrac{1}{2} \cdot 4 \cdot 4 = 8$ ⁠. Choose The area under Look inside and see the triangle.

Hint
It is a triangle: base $4$ ⁠ along the $x$ ⁠-axis, and height $4$ ⁠ at $x = 4$ ⁠.

Worked solution
1. Between the line $y = x$ ⁠ and the $x$ ⁠-axis, from $0$ ⁠ to $4$ ⁠, there is a right-angled triangle.
2. The base: $4$ ⁠. The height: $y = 4$ ⁠ at $x = 4$ ⁠.
3. The area: $\tfrac{1}{2} \cdot 4 \cdot 4 = 8$ ⁠.
$16$ ⁠ is the area of the whole $4$ ⁠ by $4$ ⁠ square; the triangle is half of it.
Step 2 of 5
Open this step
Narrow rectangles
Practice · Calculate
Question
Cut the area under $y = x^2$ ⁠ from $0$ ⁠ to $3$ ⁠ into three rectangles $1$ ⁠ wide, each as tall as the curve at its left edge. What do their areas add up to?

Correct answer
5 ± 0.01
Setup, choices and feedback
The Scene at this step
Up to x
3fixed
Slices
3fixed
The curve / The area
The curve
Other settings
A/nodes/graph/props/a = 1
Other settings
B/nodes/graph/props/b = 0
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The left Riemann sum with `slices` slices; NaN when slices is 0/nodes/graph/outputs/sliceSum
Wrong answers it recognises
- 14 ± 0.01
  Misconception: Uses the height at each rectangle's right edge when the rectangles are drawn from the left edge.
  Feedback from the catalogue
  Here each rectangle is as tall as the curve at its left edge: at $x = 0$ ⁠, $1$ ⁠ and $2$ ⁠, that is, $0$ ⁠, $1$ ⁠ and $4$ ⁠. Heights from the right edge, $1 + 4 + 9 = 14$ ⁠, are a different choice, which gives an estimate that is too high.
- 9 ± 0.01
  Misconception: Thinks a sum of rectangles already is the exact area.
  Feedback from the catalogue
  A flat-topped rectangle does not cover a curve exactly: pieces are left out (or poke out). The sum of the rectangles is an estimate, and it closes in on the exact area only as the rectangles get narrower and narrower.
Feedback for a right answer
$1 \cdot 0 + 1 \cdot 1 + 1 \cdot 4 = 5$ ⁠. Choose The area under Look inside and see how much area the rectangles leave outside the curve.

Feedback for a wrong answer
Not yet. Each rectangle: width $1$ ⁠ times the height of the curve at its left edge.

Hint
The left edges are at $x = 0$ ⁠, $1$ ⁠ and $2$ ⁠. How high is the curve at each?

Worked solution
1. The left edges: $x = 0$ ⁠, $1$ ⁠, $2$ ⁠.
2. The heights: $0^2 = 0$ ⁠, $1^2 = 1$ ⁠, $2^2 = 4$ ⁠.
3. Each is $1$ ⁠ wide, so the sum is $0 + 1 + 4 = 5$ ⁠.
$14$ ⁠ takes heights from the right edge ( $1 + 4 + 9$ ⁠). $9$ ⁠ is the exact area, which the rectangles only estimate.
Step 3 of 5
Open this step
More rectangles
Practice · Predict, then look
Question
Three rectangles give $5$ ⁠. What happens to the sum when the same stretch is cut into $30$ ⁠ narrow rectangles?

Correct answer
- Correct It grows and closes in on the exact area, but stays below it
Setup, choices and feedback
The Scene at this step
Up to x
3fixed
Slices
3
The curve / The area
The area
Other settings
A/nodes/graph/props/a = 1
Other settings
B/nodes/graph/props/b = 0
Choices
- Correct It grows and closes in on the exact area, but stays below it
- Wrong It goes past the exact area
  Misconception: Thinks narrower rectangles push the sum past the exact area.
  Feedback from the catalogue
  On a rising curve, every rectangle whose height comes from its left edge is lower than the curve, narrow or wide. Narrower rectangles leave out less area, so the sum closes in on the exact area from below, without passing it.
- Wrong It stays $5$ ⁠, because it is the same area
  Misconception: Thinks a sum of rectangles already is the exact area.
  Feedback from the catalogue
  A flat-topped rectangle does not cover a curve exactly: pieces are left out (or poke out). The sum of the rectangles is an estimate, and it closes in on the exact area only as the rectangles get narrower and narrower.
Once the prediction is locked in, the Scene changes
Slices
30
The Scene then shows: 8.555 · The left Riemann sum with `slices` slices; NaN when slices is 0/nodes/graph/outputs/sliceSum
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
With $30$ ⁠ rectangles the sum is about $8.56$ ⁠, much closer to the exact area of $9$ ⁠. Each rectangle is still lower than the curve, but the pieces left out are smaller.

Worked solution
1. The curve $y = x^2$ ⁠ rises from $0$ ⁠ to $3$ ⁠, so every rectangle, whose height comes from its left edge, is lower than the curve across its whole width. The sum is always less than the exact area.
2. Narrower rectangles leave out smaller pieces, so the sum grows.
3. With $30$ ⁠ rectangles $0.1$ ⁠ wide: the sum is about $8.56$ ⁠. The exact area is $9$ ⁠.
As the rectangles get narrower and narrower, the sum comes as close to $9$ ⁠ as you like: that is the integral.
Step 4 of 5
Open this step
The antiderivative
Practice · Calculate
Question
Use the antiderivative $\dfrac{x^3}{3}$ ⁠: what is the exact area under $y = x^2$ ⁠ from $0$ ⁠ to $3$ ⁠?

Correct answer
9 ± 0.01
Setup, choices and feedback
The Scene at this step
Up to x
3fixed
Slices
0fixed
The curve / The area
The curve
Other settings
A/nodes/graph/props/a = 1
Other settings
B/nodes/graph/props/b = 0
Accepted within
± 0.01
Computed from the Scene at this step’s setting
The signed area between the curve and the x-axis from areaFrom to areaTo, the integral/nodes/graph/outputs/area
Wrong answers it recognises
- 27 ± 0.01
  Misconception: Raises the power when integrating but forgets to divide by the new power.
  Feedback from the catalogue
  The antiderivative of $x^2$ ⁠ is $\dfrac{x^3}{3}$ ⁠, not $x^3$ ⁠: the derivative of $x^3$ ⁠ is $3x^2$ ⁠, three times too big. So divide by the new power: $\dfrac{3^3}{3} = 9$ ⁠.
- 6 ± 0.01
  Misconception: Differentiates instead of finding the antiderivative.
  Feedback from the catalogue
  For an area you need the reverse of differentiating: a function whose derivative is $x^2$ ⁠. Differentiating $x^2$ ⁠ gives $2x$ ⁠, a slope, not an area. The power has to go up, not down.
Feedback for a right answer
$\dfrac{3^3}{3} - 0 = 9$ ⁠. Choose The area under Look inside.

Feedback for a wrong answer
Not yet. The antiderivative at the right end, minus the antiderivative at the left end.

Hint
Substitute $3$ ⁠ and $0$ ⁠ into $\dfrac{x^3}{3}$ ⁠ and subtract.

Worked solution
1. The antiderivative of $x^2$ ⁠ is $\dfrac{x^3}{3}$ ⁠: its derivative is $\dfrac{3x^2}{3} = x^2$ ⁠. ✓
2. At the right end: $\dfrac{3^3}{3} = \dfrac{27}{3} = 9$ ⁠.
3. At the left end: $\dfrac{0^3}{3} = 0$ ⁠.
4. The area: $9 - 0 = 9$ ⁠.
$27$ ⁠ forgets to divide by $3$ ⁠. $6$ ⁠ differentiates $x^2$ ⁠ to $2x$ ⁠ instead of finding an antiderivative.
Step 5 of 5
Open this step
Below the axis
Practice · Choose
Question
What is the integral of $y = x - 2$ ⁠ from $x = 0$ ⁠ to $x = 4$ ⁠?

Correct answer
- Correct $0$ ⁠
Setup, choices and feedback
The Scene at this step
Up to x
4fixed
Slices
0fixed
The curve / The area
The curve
Other settings
C/nodes/graph/props/c = -2
Choices
Shown in a random order.
- Correct $0$ ⁠
- Wrong $4$ ⁠
  Misconception: Counts area below the x-axis as positive in an integral.
  Feedback from the catalogue
  The integral counts area below the $x$ ⁠-axis as negative: there $y$ ⁠ is negative, and width times a negative height is a negative number. Under $y = x - 2$ ⁠ from $0$ ⁠ to $4$ ⁠ there is a triangle of $-2$ ⁠ and a triangle of $2$ ⁠, which add up to $0$ ⁠.
- Wrong $-2$ ⁠
  Feedback
  $-2$ ⁠ is only the triangle below the axis, from $0$ ⁠ to $2$ ⁠. From $2$ ⁠ to $4$ ⁠ there is another triangle, above the axis.
- Wrong $8$ ⁠
  Feedback
  $8$ ⁠ is the area under $y = x$ ⁠. Here the line is $2$ ⁠ lower.
Feedback for a right answer
From $0$ ⁠ to $2$ ⁠ a triangle below the axis, $-2$ ⁠, and from $2$ ⁠ to $4$ ⁠ a triangle above it, $+2$ ⁠. Together $0$ ⁠. Choose The area under Look inside and see the pink and the blue.

Hint
The line crosses the $x$ ⁠-axis at $x = 2$ ⁠. Work out each triangle on its own, with its sign.

Worked solution
1. $x - 2 = 0$ ⁠ at $x = 2$ ⁠: there the line passes from below the axis to above it.
2. From $0$ ⁠ to $2$ ⁠: a triangle with base $2$ ⁠ and height $2$ ⁠, below the axis: $-\tfrac{1}{2} \cdot 2 \cdot 2 = -2$ ⁠.
3. From $2$ ⁠ to $4$ ⁠: the same triangle above the axis: $+2$ ⁠.
4. The integral: $-2 + 2 = 0$ ⁠.
Check with the antiderivative $\dfrac{x^2}{2} - 2x$ ⁠: at $4$ ⁠ it is $8 - 8 = 0$ ⁠, and at $0$ ⁠ it is $0$ ⁠. If you answered $4$ ⁠, you counted the area below the axis as positive too.

Lesson 4 of 4 · About 20 minutes

Slope and area undo each other

One curve with a hill and a dip, a tangent moving along it and area piling up beneath it. The fundamental theorem joins the two questions, because the rate at which the area grows is the height of the curve.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
A curve with a hill
Practice · Reach the target
Question
Move Point x until the tangent lies level on top of the hill, then press Check.

Goal
- 1 ± 0.01
  At the start of the step: 2.4
Setup, choices and feedback
The Scene at this step
Slice width
0.5
Point x
2.4
Thinner slices
Slices
Read from the Scene at
Tangent x/nodes/board/props/tangentX
One setting that reaches it
Slice width
0.05
Point x
1
Feedback for a right answer
At $x = 1$ ⁠ the slope is $0.6 \cdot 0 \cdot (-2) = 0$ ⁠: the tangent is level, and on both sides of it the curve is lower. That is a local maximum point.

Feedback for a wrong answer
Not yet. If the tangent tilts up, the top is still ahead; if it tilts down, you have passed it.

Hint
The derivative $0.6(x - 1)(x - 3)$ ⁠ is zero in two places. The hill is the first.

Worked solution
1. At the top the tangent lies level: $f'(x) = 0$ ⁠.
2. $0.6(x - 1)(x - 3) = 0$ ⁠ at $x = 1$ ⁠ or $x = 3$ ⁠.
3. To the left of $1$ ⁠ the slope is positive (the curve rises), and between $1$ ⁠ and $3$ ⁠ negative (it falls): at $x = 1$ ⁠ it stops rising. That is the hill.
4. Set Point x to $1$ ⁠. At $x = 3$ ⁠ is the dip.
Step 2 of 4
Open this step
Downhill
Practice · Choose
Question
What is the sign of the slope between the top of the hill ( $x = 1$ ⁠) and the bottom of the dip ( $x = 3$ ⁠)?

Correct answer
- Correct Negative
Setup, choices and feedback
The Scene at this step
Slice width
0.5
Point x
2
Thinner slices
Slices
Choices
Shown in a random order.
- Correct Negative
- Wrong Positive, because the curve is above the $x$ ⁠-axis
  Misconception: Reads the sign of the slope from the height of the curve: above the axis means a positive slope.
  Feedback from the catalogue
  Height and slope are two different things: a curve can be high above the axis and still be going down. The slope depends only on the direction: rising from left to right, positive; falling, negative.
- Wrong Zero
  Feedback
  The slope is zero only at the two points themselves, $x = 1$ ⁠ and $x = 3$ ⁠. Between them the tangent tilts.
Feedback for a right answer
Between them the curve goes down from the hill into the dip, so the slope is negative. At $x = 2$ ⁠, for example: $0.6 \cdot 1 \cdot (-1) = -0.6$ ⁠.

Hint
Move Point x between $1$ ⁠ and $3$ ⁠ and see which way the tangent tilts.

Worked solution
1. The derivative: $f'(x) = 0.6(x - 1)(x - 3)$ ⁠.
2. Between $1$ ⁠ and $3$ ⁠: $x - 1$ ⁠ is positive and $x - 3$ ⁠ negative, so the product is negative.
3. For example $f'(2) = 0.6 \cdot 1 \cdot (-1) = -0.6$ ⁠.
The whole curve is above the axis, but that says the height is positive, not the slope: it is high and going down.
Step 3 of 4
Open this step
Narrow slices
Practice · Predict, then look
Question
With slices $1$ ⁠ wide, the slice sum up to $x = 4$ ⁠ is $3.2$ ⁠, and the exact area is $3.6$ ⁠. What happens to the sum at a Slice width of $0.1$ ⁠?

Correct answer
- Correct It closes in on $3.6$ ⁠
Setup, choices and feedback
The Scene at this step
Slice width
1
Point x
4fixed
Thinner slices
Slices
Choices
- Correct It closes in on $3.6$ ⁠
- Wrong It is exactly $3.6$ ⁠
  Misconception: Thinks a sum of rectangles already is the exact area.
  Feedback from the catalogue
  A flat-topped rectangle does not cover a curve exactly: pieces are left out (or poke out). The sum of the rectangles is an estimate, and it closes in on the exact area only as the rectangles get narrower and narrower.
- Wrong It moves away from $3.6$ ⁠
  Feedback
  A narrow slice hugs the curve more closely than a wide one, so each slice’s error is smaller.
Once the prediction is locked in, the Scene changes
Slice width
0.1
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At $0.1$ ⁠ the sum is $3.56$ ⁠: the error dropped from $0.4$ ⁠ to $0.04$ ⁠, ten times smaller, like the width. It reaches zero only in the limit, as the width shrinks to zero.

Worked solution
1. At width $1$ ⁠ the error is $3.6 - 3.2 = 0.4$ ⁠.
2. Each slice’s mistake is a thin sliver between its flat top and the curve, and it shrinks along with the width.
3. At width $0.1$ ⁠ the sum is $3.56$ ⁠, and the error $0.04$ ⁠: $10$ ⁠ times smaller.
The sum is never exactly $3.6$ ⁠ at any width; $3.6$ ⁠ is the limit, the integral.
Step 4 of 4
Open this step
The fundamental theorem
Practice · Choose
Question
The area $A(x)$ ⁠ under the curve, from $0$ ⁠ up to the point, grows as the point moves right. At what rate?

Correct answer
- Correct At the rate $f(x)$ ⁠, the height of the curve at the point
Setup, choices and feedback
The Scene at this step
Slice width
0.5
Point x
2.4
Thinner slices
Exact area
Choices
Shown in a random order.
- Correct At the rate $f(x)$ ⁠, the height of the curve at the point
- Wrong At the rate $f'(x)$ ⁠, the slope of the curve at the point
  Misconception: Thinks the area under a curve grows at the rate of the curve's slope.
  Feedback from the catalogue
  Each small step $dx$ ⁠ to the right adds a strip $dx$ ⁠ wide and $f(x)$ ⁠ tall, an area of $f(x) \cdot dx$ ⁠. So the area grows at the rate $f(x)$ ⁠, the height of the curve, not its slope. That is the fundamental theorem.
- Wrong At the rate $A(x)$ ⁠, the area already collected
  Feedback
  The area already collected does not affect how much is being added now: what is added is a strip as tall as the curve at the point.
Feedback for a right answer
A step $dx$ ⁠ adds a strip $f(x)$ ⁠ tall, that is, $f(x)\,dx$ ⁠ of area. So $A'(x) = f(x)$ ⁠: the slope of $A$ ⁠ is the height of $f$ ⁠. On the board, the rise of the pink triangle equals the mint rod.

Hint
A small step to the right adds a thin strip. How tall is it?

Worked solution
1. When the point moves a small step $dx$ ⁠ to the right, a thin strip $dx$ ⁠ wide is added.
2. The strip is as tall as the curve there, $f(x)$ ⁠, so the area grows by $f(x) \cdot dx$ ⁠.
3. The rate: $\dfrac{f(x)\,dx}{dx} = f(x)$ ⁠. That is, $A'(x) = f(x)$ ⁠.
That is the fundamental theorem: the derivative of the area is the curve. And so, the other way round, you find areas with an antiderivative. On the hill, at $x = 1$ ⁠, the slope of $f$ ⁠ is $0$ ⁠, yet the area is still growing, and fast, because the curve is high there.

Across the unit · 16 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Average rate of change

From a line to a curve

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Bring the points together

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Either side of the vertex

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Look for a rate

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The derivative

The tangent

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The derivative as a function

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Where is the tangent level?

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

The rule

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The area under a curve

The area under a line

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Narrow rectangles

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

More rectangles

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution