AP Calculus AB Practice Exam 1: answers and scoring guide

The question

Question: Water flows into a tank at r(t) = 3t² + 2t litres per minute.

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 t³ + t².

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

That is 36 minus 0: 36.

So 36 litres 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 36.

Right!

Let's look back.

First: the net change.

Last: check.

At t = 1 water comes in at r(1) = 3 + 2 = 5 litres per minute.

It drains out at 10 litres per minute, which is more than 5.

So the amount of water is decreasing at t = 1.

A(t) = 50 + ∫₀^t(r(s) − 10) ds.

At t = 3: ∫₀³r(s) ds = 36 (part (a)), and 10 × 3 = 30 litres drain out.

A(3) = 50 + 36 − 30 = 56 litres.

A′(t) = r(t) − 10 = 3t² + 2t − 10. A′(t) = 0 when t = (−2 + √124)/6 ≈ 1.523 (the other answer is negative).

A′(t) < 0 before 1.523 and A′(t) > 0 after it: the water goes down, then up.

So the amount is least at t ≈ 1.523 minutes.

At rest means v(t) = 0: t² − 4t + 3 = 0.

(t − 1)(t − 3) = 0, so t = 1 and t = 3 seconds.

v(4) = 16 − 16 + 3 = 3, which is positive.

a(t) = v′(t) = 2t − 4, so a(4) = 4, which is positive.

The velocity and the acceleration have the same sign, so the particle is speeding up.

The question

Question: find the displacement and the total distance for v(t) = t² − 4t + 3 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 − 2 t² + 3t.

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

That is 20/3 minus 0: 20/3.

So the displacement is 20/3 ft.

2 Where it turns around

It turns around where v changes sign.

First, where is v = 0?

To factor t² − 4t + 3, the rule: we need two numbers that multiply to 3 and add to −4.

They are −1 and −3: they multiply to 3 and add to −4.

So we write it like this.

v is 0 at t = 1 and t = 3.

Draw v.

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

From 1 to 3, v is negative: it moves back.

From 3 to 5, v is positive: it moves forward.

3 The total distance

Call the antiderivative F.

Work it out at 1 and 3 too.

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

4/3 minus 0 is 4/3.

From 1 to 3: the same antiderivative at 3, minus at 1.

0 minus 4/3 is −4/3.

From 3 to 5: the same antiderivative at 5, minus at 3.

20/3 minus 0 is 20/3.

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: 28/3 ft.

4 Check

Let's check: the pieces with their signs add up to 20/3, the displacement.

Right!

Let's look back.

First: the net change.

Then: where it turns around.

Then: the total distance.

Last: check.

Position: x(5) = 2 + 203 = 263 feet.

Distance: the particle turns at t = 1 and t = 3. ∫₀¹v dt = 43, ∫₁³v dt = −43, ∫₃⁵v dt = 20/3.

Total distance = 43 + 43 + 203 = 283 feet.

The question

Question: find the derivative of f(x) = x³ − 3 x² − 9x + 5.

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: 5.

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 −12.

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

A calculator gives about −12.

Very close to −12.

Right!

Let's look back.

First: the rule.

Then: each term.

Last: check.

f′(x) = 3x² − 6x − 9 = 3(x − 3)(x + 1) = 0, so x = −1 or x = 3.

f′ changes from positive to negative at x = −1: a relative maximum.

f′ changes from negative to positive at x = 3: a relative minimum.

f″(x) = 6x − 6.

f″(x) > 0 when x > 1, so the graph is concave up for x > 1.

Check the ends and the critical point inside: f(0) = 5, f(3) = 27 − 27 − 27 + 5 = −22, f(4) = 64 − 48 − 36 + 5 = −15.

The smallest is −22, so the absolute minimum is −22, at x = 3.

At (1, 2): dy/dx = 6 × 1² × 2 = 12.

Tangent line: y − 2 = 12(x − 1).

At x = 1.1: y ≈ 2 + 12 × 0.1 = 3.2.

d²y/dx² = 12xy + 6x² (dy/dx) = 12xy + 6x² (6x² y).

At (1, 2): 12 × 2 + 6 × 12 = 24 + 72 = 96, which is positive.

The curve is concave up there, so the tangent line lies below it: the estimate is too small.

The question

Question: solve dy/dx = 6x²y, with y(0) = 2.

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

y(0) = 2 is the starting value: when x is 0, y is 2.

The right side splits into an x part, 6x², 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: 2x³.

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^(2x³ + C) = e^C × e^(2x³).

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 = 2.

The power is 0 there, and e⁰ = 1.

So A = 2.

So y = 2e^(2x³).

5 Check

Let's check with a calculator.

At x = 1, our answer gives y ≈ 14.778.

The slope between two very close points: about 88.67.

The right side at that point: about 88.67.

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 = 4x − x² 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 = 4.

The area we want is shaded: under the curve, above the x-axis, from 0 to 4.

This area is the integral from 0 to 4.

2 The antiderivative

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

Now we have F(x) = 2 x² − x³ over 3.

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 4 on the bar.

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

First the powers and the times.

Then each bracket: the first is 32/3, the second is 0.

Careful!

Top first, then take away the bottom.

The other way gives the wrong sign.

So the area is 32/3 square units.

4 Check

Let's check F: its derivative is 4x − x², 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 + 4x from x = 0 to x = 4 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 = 4, 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 + 4x.

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 4: 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 4 of (−x² + 4x)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 4, then take away the same with 0.

So the volume is 512π over 15 cubic units.

That is about 107.23.

7 Check

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

That gives about 107.23, 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.

The whole area is 323, so each part has area 16/3.

∫₀^k(4x − x²) dx = 16/3.

The question

Question: find the derivative of ∫ from 1 to x of (t² − 4) 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 − 4.

Careful!

The derivative of the area is just f at x.

The bottom number 1 does not matter.

3 Check

Let's check at x = 2.

Our answer gives 0.

Now the slope of the area function F, the integral itself, between x = 2 and 2.001, with a calculator: about 0.002.

The same!

Right!

Let's look back.

First: the rule.

Then: put x in place of t.

Last: check.

g′(x) = x² − 4 = 0 when x = 2 (x = −2 is not in the interval).

g′ changes from negative to positive at x = 2, so g has a relative minimum at x = 2.

The question

Question: find the integral from 1 to 3 of x² − 4.

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 − 4x.

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 1 and 3 on the bar.

Put in 3 first, then take away the same with 1.

First the powers and the times.

Then each bracket: the first is −3, the second is −11/3.

Careful!

Top first, then take away the bottom.

The other way gives the wrong sign.

So the integral is 2/3.

3 Check

Let's check F: its derivative is x² − 4, the function we started with.

Right!

Let's look back.

First: the antiderivative.

Then: top minus bottom.

Last: check.

g(1) = ∫ from 1 to 1 = 0, so the limit has the form 0/0.

L'Hôpital's rule: the limit = g′(1)/1 = (1 − 4)/1.

The limit is −3.