AP Calculus AB Practice Exam 2: answers and scoring guide

The question

Question: Sand flows into a pile at r(t) = 16 - t² cubic metres per hour.

How much flows in from t = 0 to t = 3?

1 The net change

r(t) is a rate: how fast it flows in.

The rule: the amount is the integral of the rate.

Here it goes from 0 to 3.

The antiderivative: add 1 to each power, divide by the new power.

Now we have 16t − t³ over 3.

Put in 3, then take away the same with 0.

That is 39 minus 0: 39.

So 39 cubic metres flow in.

Careful!

r is the rate, not the amount.

The amount is the integral of the rate.

2 Check

Let's check with a calculator's integral: about 39.

Right!

Let's look back.

First: the net change.

Last: check.

r′(t) = −2t, so r′(2) = −4 cubic metres per hour per hour.

At t = 2 hours, the rate at which sand is added is going down by 4 cubic metres per hour, each hour.

The question

Question: find the average value of f(x) = 16 − x² on [0, 4].

1 The formula

The average value: the area under the curve, divided by the width of the interval.

Here the width is 4 minus 0: 4.

Think of it as the height of a rectangle with the same area as under the curve.

2 The integral

The antiderivative: add 1 to each power, divide by the new power.

Now we have 16x − x³ over 3.

Put in 4, then take away the same with 0.

That is 128/3 minus 0: 128/3.

3 Divide by the width

Divide by the width, 4.

The average value is 32/3.

Careful!

128/3 is the area.

To get the average height, divide it by the width, 4.

The right way: average = 128/3 over 4 = 32/3.

4 Check

Let's check: a rectangle 32/3 high and 4 wide has area 128/3: the same as the integral.

Right!

Let's look back.

First: the formula.

Then: the integral.

Then: divide by the width.

Last: check.

Sand at t = 4 = 20 + ∫₀⁴r(t) dt.

The integral is 1283 (part (c): 4 hours × the average rate 323).

So the pile holds 20 + 1283 = 1883 ≈ 62.667 cubic metres.

The question

Question: solve t² − 5t + 4 = 0.

1 Factor

Factoring goes backwards.

We need two numbers.

They multiply to 4, and they add to −5.

Look at the sign in front of 4: it is plus.

So the two numbers have the same sign.

Now the sign in front of 5t: it is minus.

So both are minus.

We list the pairs that make 4, smallest first.

For each pair, we check two things: they multiply to 4, and they add to 5.

1 × 4 = 4.

1 + 4 = 5.

Both work!

So the two numbers are 1 and 4.

With the signs: −1 and −4.

So it is (t − 1)(t − 4) = 0.

2 One bracket is 0

Two numbers multiply to 0.

That only happens if one of them is 0.

So one bracket must be 0.

We make each bracket 0.

Two small equations.

In t − 1 = 0, we add 1 on both sides.

In t − 4 = 0, we add 4 on both sides.

So t = 1 or t = 4.

3 Check

Let's check: put t = 1 into the left side.

We get 0, the same as the right side.

It works!

Now put t = 4 into the left side.

We get 0 again, the same as the right side.

So both answers are right!

Let's look back.

First: factor.

Then: one bracket is 0.

Last: check.

v(3) = 9 − 15 + 4 = −2, which is negative.

a(t) = v′(t) = 2t − 5, so a(3) = 1, which is positive.

The velocity and the acceleration have different signs, so the particle is slowing down.

The question

Question: find the displacement and the total distance for v(t) = t² − 5t + 4 from t = 0 to t = 5.

Displacement is where it ends up, compared with the start.

Distance counts every step, forward and back.

1 The net change

The displacement is the integral of v: in a tiny time it moves about v times that time.

Going back counts as negative.

The antiderivative: add 1 to each power, divide by the new power.

Now we have t³ over 3 − 5t² over 2 + 4t.

Put in 5, then take away the same with 0.

That is −5/6 minus 0: −5/6.

So the displacement is −5/6 m.

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

To factor t² − 5t + 4, the rule: we need two numbers that multiply to 4 and add to −5.

They are −1 and −4: they multiply to 4 and add to −5.

So we write it like this.

v is 0 at t = 1 and t = 4.

Draw v.

From 0 to 1, v is positive: it moves forward.

From 1 to 4, v is negative: it moves back.

From 4 to 5, v is positive: it moves forward.

3 The total distance

Call the antiderivative F.

Work it out at 1 and 4 too.

From 0 to 1: the same antiderivative at 1, minus at 0.

11/6 minus 0 is 11/6.

From 1 to 4: the same antiderivative at 4, minus at 1.

−8/3 minus 11/6 is −9/2.

From 4 to 5: the same antiderivative at 5, minus at 4.

−5/6 minus −8/3 is 11/6.

Careful!

The integral of v is the displacement.

Going back cancels going forward.

For the distance, add each piece without its sign.

Distance counts every piece as positive.

Add them: 49/6 m.

4 Check

Let's check: the pieces with their signs add up to −5/6, the displacement.

Right!

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

x(5) = x(0) + displacement = 3 + (−56) (part (c)).

x(5) = 136 ≈ 2.167, so the particle is at x = 13/6.

The question

Question: find the derivative of f(x) = x³ + 3 x² − 9x − 4.

1 The rule

d/dx means: the derivative of.

Here, of x to the power n.

The rule: the power n comes down in front, and the new power is n − 1.

We write the derivative as f′(x), "f prime of x".

2 Each term

First: x³.

Its power is 3.

The power comes down in front.

And the new power is 3 − 1.

3 − 1 is 2.

So x³ gives 3 x².

Next: 3 x².

Its power is 2.

The power comes down in front: 3 times 2.

And the new power is 2 − 1.

2 − 1 is 1.

3 times 2 is 6.

So 3 x² gives 6x.

Next: −9x.

x alone gives 1: its graph goes up 1 for each 1 across, so its slope is 1.

9 times 1 is 9.

Last: −4.

A number alone gives 0.

Its graph is a flat line, and a flat line has slope 0.

Careful!

A number alone does not stay.

It gives 0.

We put the terms together: f′(x) = 3 x² + 6x − 9.

3 Check

Let's check: the derivative is the slope.

At x = 1, the rule gives 0.

Now the slope between two very close points: x = 1 and x = 1.001.

A calculator gives about 0.01.

Very close to 0.

Right!

Let's look back.

First: the rule.

Then: each term.

Last: check.

The question

Question: find the maximum and minimum points of f(x) = x³ + 3x² − 9x − 4.

The derivative is the slope.

Where it is positive, the graph goes up.

Where it is negative, the graph goes down.

1 Find f′(x)

Use the power rule on each term.

Now we have f′(x) = 3x² + 6x − 9.

2 Solve f′(x) = 0

At a turning point the graph is flat for a moment: the slope is 0.

So we solve f′(x) = 0.

That is 3x² + 6x − 9 = 0.

Take out the 3.

To factor x² + 2x − 3, the rule: we need two numbers that multiply to −3 and add to 2.

They are 3 and −1: they multiply to −3 and add to 2.

So we write it as 3(x + 3)(x − 1).

A product is 0 only when one of its parts is 0.

So x = −3 or x = 1.

Careful!

f′ = 0 alone does not tell max or min.

The signs on each side tell us.

3 Sign table

A sign table.

The numbers −3 and 1 cut the line into 3 parts.

Test one number in each part: −4, −2, 2.

Any number in a part gives the same sign, so we pick easy ones.

f′(−4) is 15: positive.

So f goes up there.

f′(−2) is −9: negative.

So f goes down there.

f′(2) is 15: positive.

So f goes up there.

4 Max and min

At x = −3, f goes up, then down.

That is a maximum.

At x = 1, f goes down, then up.

That is a minimum.

Its y: put −3 into f.

We get 23.

Its y: put 1 into f.

We get −9.

So: a maximum at (−3, 23), and a minimum at (1, −9).

5 Check

Let's check with the graph.

Up, then down, then up: just like the table.

Right!

Let's look back.

First: find f′(x).

Then: solve f′(x) = 0.

Then: sign table.

Then: max and min.

Last: check.

The question

Question: find the concavity and the inflection points of f(x) = x³ + 3x² − 9x − 4.

Concave up means the curve bends up, like a cup.

Concave down bends down, like a cap.

f″ is the slope of the slope.

When f″ is positive, the slope keeps growing, so the curve turns up: a cup.

When f″ is negative: a cap.

Careful!

Concave up is about the bending, not about going up.

A cup can go down on one side.

1 Find f″(x)

First f′: the power rule on each term.

Now we have f′(x) = 3x² + 6x − 9.

Then f″: the power rule again, on f′.

Now we have f″(x) = 6x + 6.

2 Solve f″(x) = 0

The bending can only change where f″ is 0.

So we solve f″(x) = 0.

That is 6x + 6 = 0.

Take away 6 on both sides, so they stay equal.

Divide by 6: x = −1.

3 Sign table for f″

A sign table for f″.

The number −1 cuts the line into 2 parts.

In each part, f″ keeps one sign.

So one test number per part is enough.

Take easy ones: −2, 0.

f″(−2) is −6: negative.

So f bends down there: a cap.

f″(0) is 6: positive.

So f bends up there: a cup.

4 Inflection points

At x = −1, the bending changes: an inflection point.

Its y: put −1 into f.

We get 7.

f is concave up where f″ is positive: x > −1.

f is concave down where f″ is negative: x < −1.

And the inflection point is (−1, 7).

5 Check

Let's check with the graph.

A cap, then a cup.

It changes at the inflection point.

Right!

Let's look back.

First: find f″(x).

Then: solve f″(x) = 0.

Then: sign table for f″.

Then: inflection points.

Last: check.

The question

Question: find the absolute maximum and minimum of f(x) = x³ + 3x² − 9x − 4 on [−4, 2].

On a closed interval, the highest and lowest values are at an end, or inside.

Inside, a top or a bottom is where the curve turns.

There it is flat: the slope is 0.

So we test the ends and those points.

1 Critical points

Use the power rule on each term.

Now we have f′(x) = 3x² + 6x − 9.

Critical points: where the slope is 0.

Solve f′(x) = 0.

That is 3x² + 6x − 9 = 0.

Take out the 3.

To factor x² + 2x − 3, the rule: we need two numbers that multiply to −3 and add to 2.

They are 3 and −1: they multiply to −3 and add to 2.

So we write it as 3(x + 3)(x − 1).

A product is 0 only when one of its parts is 0.

So x = −3 or x = 1.

2 Keep the ones inside

x = −3 is inside the interval.

Keep it.

x = 1 is inside the interval.

Keep it.

3 Test every candidate

The left end: f of −4 is 16.

The critical point: f of −3 is 23.

The critical point: f of 1 is −9.

The right end: f of 2 is −2.

4 Biggest and smallest

The biggest value is 23, at x = −3.

That is the absolute maximum.

The smallest value is −9, at x = 1.

That is the absolute minimum.

5 Check

Let's check one more point.

Every x inside must give a value between the min and the max.

Take the middle, x = −1: it gives 7.

Between them.

Right!

Let's look back.

First: critical points.

Then: keep the ones inside.

Then: test every candidate.

Then: biggest and smallest.

Last: check.

d²y/dx² = 12x² y + 4x³ × dy/dx (the product rule).

Put in dy/dx = 4x³ y: d²y/dx² = 12x² y + 4x³ × 4x³ y = 12x² y + 16x⁶ y.

The slope at (1, 2) is dy/dx = 4 × 1³ × 2 = 8.

The tangent line: y = 2 + 8(x − 1).

At x = 1.1: y ≈ 2 + 8 × 0.1 = 2.8.

The question

Question: solve dy/dx = 4x³y, with y(0) = 3.

The x parts on one side, the y parts on the other.

y(0) = 3 is the starting value: when x is 0, y is 3.

The right side splits into an x part, 4x³, times a y part, y.

That is a separable equation.

Our goal: each side with only one letter, so we can integrate each side.

1 Separate x and y

Divide both sides by y, so y is only on the left.

Now multiply both sides by dx.

All the y parts are with dy, all the x parts with dx.

2 Integrate both sides

Put an integral sign on both sides.

On the left: ln |y|.

On the right: x⁴.

Each side would get its own number; the two join into one, so we write one + C.

The integral of 1/y is the natural log of the size of y.

3 Solve for y

To undo ln, use e on both sides: e and ln undo each other, so e to the ln of |y| is just |y|.

A plus in the power splits into a times: e^(x⁴ + C) = e^C × e^(x⁴).

Careful!

The C is not added at the end.

e^C is a number that multiplies.

e^C is just some number.

With the ± from |y|, call it A: any number.

4 Use the starting value

Put in x = 0 and y = 3.

The power is 0 there, and e⁰ = 1.

So A = 3.

So y = 3e^(x⁴).

5 Check

Let's check with a calculator.

At x = 1, our answer gives y ≈ 8.155.

The slope between two very close points: about 32.62.

The right side at that point: about 32.62.

The same slope.

Right!

Let's look back.

First: separate x and y.

Then: integrate both sides.

Then: solve for y.

Then: use the starting value.

Last: check.

The question

Question: find the area between y = −x² + 3x and the x-axis.

1 Draw the region

Draw the curve first.

Where does it cross the x-axis?

Where y is 0.

Factor it.

Every term has −x, so we take it out.

So it crosses at x = 0 and x = 3.

The area we want is shaded: under the curve, above the x-axis, from 0 to 3.

This area is the integral from 0 to 3.

2 The antiderivative

First the antiderivative: add 1 to each power, divide by the new power.

Now we have F(x) = −x³ over 3 + 3x² over 2.

No + C here: in F(b) minus F(a), the two Cs cancel.

3 Top minus bottom

The rule: put in the top number, then take away the bottom number.

Write F(x) with 0 and 3 on the bar.

Put in 3 first, then take away the same with 0.

First the powers and the times.

Then each bracket: the first is 9/2, the second is 0.

Careful!

Top first, then take away the bottom.

The other way gives the wrong sign.

So the area is 9/2 square units.

4 Check

Let's check F: its derivative is − x² + 3x, the function we started with.

Right!

Let's look back.

First: draw the region.

Then: the antiderivative.

Then: top minus bottom.

Last: check.

The question

Question: find the volume when y = −x² + 3x from x = 0 to x = 3 turns around the x-axis.

We turn a flat region around the x-axis, and slice the solid it makes.

1 Draw and turn it

Draw the curve first.

The region under the curve, from x = 0 to x = 3, is shaded.

2 One slice

Cut the solid into thin slices.

Each slice is a disk: a very flat cylinder.

Its radius is the height of the curve: r = −x² + 3x.

Its thickness is a tiny dx.

A cylinder is its circle's area times its thickness.

So one slice is π r² dx.

3 Write the integral

Add up all the slices from 0 to 3: that is an integral.

π is just a number, so it waits in front.

Careful!

Square the radius!

A circle's area is π r squared, not π r.

The right way: V = π∫ from 0 to 3 of (−x² + 3x)² dx.

4 Square it

Square the radius.

A bracket squared is the bracket times itself: multiply every part by every part.

Put it in the integral.

5 The antiderivative

The antiderivative: add 1 to each power, divide by the new power.

6 Top minus bottom

Put in 3, then take away the same with 0.

So the volume is 81π over 10 cubic units.

That is about 25.45.

7 Check

Let's check with a calculator: it adds up 1000 thin disks.

That gives about 25.45, the same as our answer.

Right!

Let's look back.

First: draw and turn it.

Then: one slice.

Then: write the integral.

Then: square it.

Then: the antiderivative.

Then: top minus bottom.

Last: check.

Each square has side 3x − x² (the height of R at x), so its area is (3x − x²)^2.

Volume = ∫₀³(3x − x²)² dx.

(It is the integral of part (b) without the π: 81/10.)

The question

Question: find the derivative of ∫ from 2 to x of (t² − 9) dt.

1 The rule

The rule: the derivative of the area from a to x is the function inside the integral, at x.

Just put x in place of t.

2 Put x in place of t

Put x in place of t.

So the derivative is x² − 9.

Careful!

The derivative of the area is just f at x.

The bottom number 2 does not matter.

3 Check

Let's check at x = 3.

Our answer gives 0.

Now the slope of the area function F, the integral itself, between x = 3 and 3.001, with a calculator: about 0.003.

The same!

Right!

Let's look back.

First: the rule.

Then: put x in place of t.

Last: check.

g′(x) = x² − 9 = 0 at x = 3 (x = −3 is not in 0 < x < 6).

g′(x) < 0 for 0 < x < 3 and g′(x) > 0 for 3 < x < 6, so g has a relative minimum at x = 3.

The question

Question: find the integral from 2 to 5 of x² − 9.

An integral from a to b gives one number.

1 The antiderivative

First the antiderivative: add 1 to each power, divide by the new power.

Now we have F(x) = x³ over 3 − 9x.

No + C here: in F(b) minus F(a), the two Cs cancel.

2 Top minus bottom

The rule: put in the top number, then take away the bottom number.

Write F(x) with 2 and 5 on the bar.

Put in 5 first, then take away the same with 2.

First the powers and the times.

Then each bracket: the first is −10/3, the second is −46/3.

Careful!

Top first, then take away the bottom.

The other way gives the wrong sign.

So the integral is 12.

3 Check

Let's check F: its derivative is x² − 9, the function we started with.

Right!

Let's look back.

First: the antiderivative.

Then: top minus bottom.

Last: check.

g(5) = 12 (part (c)).

g′(5) = 5² − 9 = 16 (part (a)).

The tangent line: y = 12 + 16(x − 5).