Unit 6 · 15–20% of the exam Beta

Unit 6 practice test: Integration and Accumulation of Change

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value or an expression; sets share one graph or table.

Covers: Accumulating Change; Estimating Area with Riemann Sums; Riemann Sums, Sigma Notation and the Definite Integral; The Fundamental Theorem and Accumulation Functions; Behavior of Accumulation Functions; Properties of Definite Integrals; The Fundamental Theorem and Definite Integrals; Antiderivatives and Indefinite Integrals; Integrating by Substitution; Integrating by Long Division and Completing the Square; Choosing an Antidifferentiation Technique.

Graph

The graph of $f(t) = t^2 - 4t$ , with $g(x)$ $= \int_0^x f(t)\,dt$

Let

f(t) = t^2 - 4t

and

g(x)

= \int_0^x f(t)\,dt

for

0 \le x \le 6

1. On which interval is $g$ decreasing?

$(0, 4)$
$(0, 2)$
$(4, 6)$
$(2, 6)$

Show the answer

$g'(x) = f(x) = x(x - 4)$ , which is negative for $0 < x < 4$ . So $g$ decreases on $(0, 4)$ .

Correct: $(0, 4)$ : Correct: $g' = f < 0$ on $(0, 4)$ .
$(0, 2)$ : On $(0, 2)$ , $f$ is decreasing, which makes $g$ concave down. $g$ decreases wherever $f$ is negative: all of $(0, 4)$ .
$(4, 6)$ : There $f > 0$ , so $g$ increases.
$(2, 6)$ : On $(2, 6)$ , $f$ is increasing; that is where $g$ is concave up, not where it decreases.

2. At which $x$ does $g$ have a relative minimum, and why?

$x = 2$ , because $f$ reaches its minimum value at $x = 2$
$x = 0$ , because $g(0) = 0$ at the start of the interval
$x = 4$ , because $f(4) = 0$ and so $g'(4) = 0$
$x = 4$ : $g' = f$ changes from negative to positive

Show the answer

$g' = f$ is negative on $(0, 4)$ and positive on $(4, 6)$ , so $g$ has a relative minimum at $x = 4$ . A zero of $g'$ is not enough; the change of sign is the reason.

$x = 2$ , because $f$ reaches its minimum value at $x = 2$ : At $x = 2$ , $g' = f$ is most negative, so $g$ is falling fastest; $g$ keeps falling until $x = 4$ .
$x = 0$ , because $g(0) = 0$ at the start of the interval: $g(0) = 0$ is just the starting value; $g$ is decreasing right after 0.
$x = 4$ , because $f(4) = 0$ and so $g'(4) = 0$ : The location is right, but $f(4) = 0$ alone is not a reason: $g'$ must change sign there, from negative to positive.
Correct: $x = 4$ : $g' = f$ changes from negative to positive: Correct: $g$ stops falling and starts rising at $x = 4$ .

3. Where is $g'' = f'$ positive, so that the graph of $g$ bends upward?

$(0, 2)$
$(0, 4)$
$(2, 6)$
$(4, 6)$

Show the answer

$g''(x) = f'(x) = 2x - 4$ , positive for $x > 2$ . So $g$ is concave up on $(2, 6)$ .

$(0, 2)$ : On $(0, 2)$ , $f$ is decreasing, so $g$ is concave down.
$(0, 4)$ : On $(0, 4)$ , $f < 0$ , which makes $g$ decrease; concavity is a different question.
Correct: $(2, 6)$ : Correct: $f$ is increasing there, so $g'' = f' > 0$ .
$(4, 6)$ : $f > 0$ on $(4, 6)$ makes $g$ increase. Concavity of $g$ follows whether $f$ increases: from $x = 2$ on.

4. Is $g(4)$ positive, negative or zero?

zero, because $f$ has the same value 0 at both ends
negative: on $(0, 4)$ the graph of $f$ is below the axis
positive, because an area is a positive number
zero, because $f(4) = 0$ at the right end of $[0, 4]$

Show the answer

On $(0, 4)$ , $f(t) = t(t - 4) < 0$ , so the region lies below the axis and $g(4)$ $= \int_0^4 f(t)\,dt < 0$ .

zero, because $f$ has the same value 0 at both ends: Equal heights at the ends do not make the integral 0. The region in between lies below the axis.
Correct: negative: on $(0, 4)$ the graph of $f$ is below the axis: Correct: $f < 0$ on $(0, 4)$ , so the integral from 0 to 4 is negative.
positive, because an area is a positive number: The size of the region is positive, but it lies below the axis, so it counts as negative in $\int_0^4 f(t)\,dt$ .
zero, because $f(4) = 0$ at the right end of $[0, 4]$ : $f(4) = 0$ is the height at one point. $g(4)$ adds up $f$ over all of $[0, 4]$ .

Table

Pumping water into a pool

A pump changes the water in a pool at a constant rate during each period below. A negative rate means the pump is draining the pool. At

t = 0

the pool holds 8000 gallons.

Rate of change of the water in the pool, in gallons per hour, during each period of time t, in hours
Period (hours)	$0 \le t < 2$	$2 \le t < 5$	$5 \le t \le 6$
Rate (gallons per hour)	300	450	$-120$

5. How many gallons are in the pool at $t = 6$ hours?

1830 gallons
10,070 gallons
9950 gallons
9830 gallons

Show the answer

Amount at 6 = starting amount + net change $= 8000 + (600 + 1350 - 120)$ $= 9830$ gallons.

1830 gallons: That is only the net change. The pool started with 8000 gallons; add the change to the starting amount.
10,070 gallons: This adds the 120 gallons of the last hour, but the rate is negative then: the pump drains the pool.
9950 gallons: This stops at t = 5. The last hour drains 120 gallons.
Correct: 9830 gallons: Correct: 8000 + 600 + 1350 - 120.

6. Which expression gives the change in the amount of water, in gallons, from $t = 2$ to $t = 6$ ?

$(450 - 120) \cdot 4$
$450 \cdot (5 - 2) + 120 \cdot (6 - 5)$
$450 \cdot 3 + (-120) \cdot 1$
$450 \cdot 4 - 120$

Show the answer

Gallons = (gallons per hour) $\times$ (hours) for each period: $450(3) + (-120)(1)$ $= 1230$ gallons.

$(450 - 120) \cdot 4$ : Subtracting the rates first treats both as lasting 4 hours. Multiply each rate by its own time.
$450 \cdot (5 - 2) + 120 \cdot (6 - 5)$ : This adds the 120 gallons of the last hour. The rate is negative then: the pump drains the pool, so that term is subtracted.
Correct: $450 \cdot 3 + (-120) \cdot 1$ : Correct: each rate times the length of its period, with the draining hour negative.
$450 \cdot 4 - 120$ : The rate 450 lasts only 3 hours (from 2 to 5), not 4.

7. On a graph of the rate against time, the region for $5 \le t \le 6$ lies below the t-axis and has area 120. What does it mean?

The pool gains 120 gallons, since an area is positive
The pool loses 120 gallons from t = 5 to t = 6 hours
The pool holds 120 gallons at t = 6 hours
The rate of draining grows by 120 gallons per hour

Show the answer

Area under a rate graph is an amount of change; a region below the axis is a decrease. From $t = 5$ to $t = 6$ the pool loses 120 gallons.

The pool gains 120 gallons, since an area is positive: A region below the axis comes from a negative rate, so the amount decreases.
Correct: The pool loses 120 gallons from t = 5 to t = 6 hours: Correct: area means gallons, and below the axis means the amount goes down.
The pool holds 120 gallons at t = 6 hours: The area is a change over the hour, not the amount in the pool.
The rate of draining grows by 120 gallons per hour: That reads the area as a change in the rate. The area is gallons, a rate times a time: here, water that leaves the pool.

8. Find the right Riemann sum for $g(x) = x^2$ on $[0, 2]$ with four subintervals of equal width.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Width $0.5$ . Right ends: $0.5, 1, 1.5, 2$ . Sum $= 0.5(0.25 + 1 + 2.25 + 4)$ $= 0.5(7.5)$ $= 3.75$ .

Answer: 3.75

9. A home uses electric power at a rate $P(t)$ kilowatts, where $t$ is in hours. What does $\int_0^{24} P(t)\,dt$ mean?

the power, in kilowatts, that the home is using at t = 24 hours
the average power of the home over the day, in kilowatts
how fast the power changes during the 24 hours, in kilowatts per hour
the kilowatt-hours of energy the home uses during the 24 hours

Show the answer

$\int_0^{24} P(t)\,dt$ is the limit of sums of (kilowatts) $\times$ (hours): the energy, in kilowatt-hours, used from $t = 0$ to $t = 24$ .

the power, in kilowatts, that the home is using at t = 24 hours: That is $P(24)$ , one value of the rate. The integral accumulates the rate over the day.
the average power of the home over the day, in kilowatts: An average would divide the accumulated energy by 24 hours. The integral is the accumulated energy itself.
how fast the power changes during the 24 hours, in kilowatts per hour: That would be a rate of change of $P$ , its derivative. The integral of $P$ adds up $P$ times time.
Correct: the kilowatt-hours of energy the home uses during the 24 hours: Correct: kilowatts times hours is kilowatt-hours, the energy used over the day.

10. Let $F(x)$ $= \int_3^x (t^2 + 1)\,dt$ . Which statement is true?

$F(3) = 0$ and $F'(x) = 2x$
$F(3) = 10$ and $F'(x) = x^2 + 1$
$F(3) = 0$ and $F'(x) = x^2 + 1$
$F(0) = 0$ and $F'(x) = x^2 + 1$

Show the answer

$F(3)$ $= \int_3^3 (t^2 + 1)\,dt$ $= 0$ , and since $t^2 + 1$ is continuous, $F'(x) = x^2 + 1$ : $F$ is the antiderivative of $x^2 + 1$ that is 0 at $x = 3$ .

$F(3) = 0$ and $F'(x) = 2x$ : $2x$ is the derivative of the integrand. $F'$ is the integrand itself.
$F(3) = 10$ and $F'(x) = x^2 + 1$ : 10 is the integrand at 3. $F(3)$ $= \int_3^3 (t^2 + 1)\,dt$ $= 0$ .
Correct: $F(3) = 0$ and $F'(x) = x^2 + 1$ : Correct: equal limits give 0, and the Fundamental Theorem gives the integrand at $x$ .
$F(0) = 0$ and $F'(x) = x^2 + 1$ : $F$ is 0 where the limits match, at $x = 3$ , not at 0.

11. If $\int_1^4 f(x)\,dx$ $= 6$ , what is $\int_4^1 3f(x)\,dx$ ?

$18$
$-6$
$-9$
$-18$

Show the answer

$\int_4^1 3f(x)\,dx$ $= -3\int_1^4 f(x)\,dx$ $= -3(6)$ $= -18$ .

$18$ : The limits run from 4 to 1, which flips the sign.
$-6$ : The constant 3 multiplies the integral too.
$-9$ : The 3 multiplies the integral; it is not added to it. Then reverse the sign: -3(6).
Correct: $-18$ : Correct: the 3 multiplies, and reversing the limits flips the sign.

12. Find $\int_0^{\pi/2} \cos x\,dx$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{d}{dx}\sin x = \cos x$ , so $\int_0^{\pi/2} \cos x\,dx$ $= \sin x\Big|_0^{\pi/2}$ $= 1 - 0$ $= 1$ .

Answer: 1

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Practice the whole unit Take a timed unit test Print the unit sheet

Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.

The graph of f(t)=t2−4tf(t) = t^2 - 4t, with g(x)g(x)=∫0xf(t)dt= \int_0^x f(t)\,dt

Pumping water into a pool

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The graph of $f(t) = t^2 - 4t$ , with $g(x)$ $= \int_0^x f(t)\,dt$