Unit 5

The Integral

Antiderivatives, Riemann sums and the trapezoidal rule, the definite integral, both parts of the Fundamental Theorem, u-substitution, and motion.

Topics in this unit

Riemann Sums · Trapezoidal Rule · The Definite Integral · Properties of Definite Integrals · Accumulation & Net Change · The Fundamental Theorem (Part 1) · The Fundamental Theorem (Part 2) · Antiderivatives · u-Substitution (Indefinite) · u-Substitution (Definite) · Integration: Long Division & Completing the Square · Selecting an Antidifferentiation Technique · Initial Value Problems · Straight-Line Motion

Riemann Sums

Question 1

no calculatormedium

The figure shows

f (x) = x^{2}

[0, 4]

with

4

rectangles whose heights are set by the

left

endpoint of each subinterval (width

1

). What is this left Riemann sum?

Show the answer

The answer is B.

Left endpoints

0, 1, 2, 3

give heights

f = 0, 1, 4, 9

. With width

1

1 (0 + 1 + 4 + 9) = 14

Why not A:That uses the $right$ endpoints ( $f (1) + f (2) + f (3) + f (4) = 1 + 4 + 9 + 16$ ). Left endpoints give $f (0) + f (1) + f (2) + f (3)$ .
Why not C:That's the midpoint sum. Left endpoints are $0, 1, 2, 3$ , giving heights $0, 1, 4, 9$ .
Why not D: $16 = f (4)$ is just the tallest single rectangle's height, not the sum of all four areas.

Trapezoidal Rule

Question 2

no calculatorhard

For a

concave-down

function, how do the

midpoint

and

trapezoidal

Riemann approximations compare to the true integral?

Show the answer

The answer is D.

For a concave-down curve: midpoint rectangles (tangent-line equivalent) sit above the curve, so midpoint OVERestimates; the trapezoid chords sit below, so trapezoidal UNDERestimates.

Why not A:It's the reverse for concave down: the midpoint tangent-line rectangles lie above the curve (over), while the chords lie below (trapezoid under).
Why not B:Only the midpoint overestimates for concave down; the trapezoidal underestimates.
Why not C:The midpoint rule OVERestimates a concave-down function; only the trapezoidal underestimates.

The Definite Integral

Question 3

no calculatorhard

n \to \infty lim i = 1 \sum n (\frac{3 i}{n})^{2} \cdot \frac{3}{n}

equals

Show the answer

The answer is D.

This is the limit of a right Riemann sum for

f (x) = x^{2}

[0, 3]

with

Δ x = \frac{3}{n}

and

x_{i} = \frac{3 i}{n}

. So it equals

\int_{0}^{3} x^{2} d x = \frac{x ^{3}}{3}_{0}^{3} = 9

Why not A: $27 = x^{3}_{0}^{3}$ forgets to divide by $3$ in the antiderivative $\frac{x ^{3}}{3}$ .
Why not B: $3$ is the interval length $Δ x$ accumulates over, not the value of the integral.
Why not C: $\frac{27}{2}$ would be $\int_{0}^{3} x d x$ ; here the integrand is $x^{2}$ .

Question 4

no calculatormedium

The graph of

f

, consisting of two line segments and a semicircle, is shown below. What is the value of

\int_{0}^{8} f (x) d x

Show the answer

The answer is A.

Split at the corners. On

[0, 4]

the two segments form a triangle above the axis with base

4

and height

2

: area

\frac{1}{2} (4) (2) = 4

. On

[4, 8]

the graph is the lower half of a circle of radius

2

, so that region has area

\frac{1}{2} π (2)^{2} = 2 π

and lies below the axis, contributing

- 2 π

. Therefore

\int_{0}^{8} f (x) d x = 4 - 2 π

Why not B:The semicircle lies BELOW the $x$ -axis on $[4, 8]$ , so its area counts as $- 2 π$ , not $+ 2 π$ . The integral is $4 - 2 π$ .
Why not C:A semicircle of radius $2$ is HALF a disk: $\frac{1}{2} π (2)^{2} = 2 π$ , not $4 π$ . So the integral is $4 - 2 π$ .
Why not D:The two segments form one triangle with base $4$ and height $2$ , area $\frac{1}{2} (4) (2) = 4$ , not a $4 \times 2$ rectangle of area $8$ .

The Fundamental Theorem (Part 1)

Question 5

no calculatorhard

The graph of

f

is shown below. Which of the following could be the graph of

g (x) = \int_{0}^{x} f (t) d t

Show the answer

The answer is D.

By the Fundamental Theorem,

g^{'} (x) = f (x) = x^{2} - 1

and

g (0) = 0

, giving

g (x) = \frac{1}{3} x^{3} - x

. Since

f < 0

(- 1, 1)

g

decreases there, with a local max at

x = - 1

and a local min at

x = 1

, and it passes through the origin.

Why not A:This is the graph of $f$ itself, $y = x^{2} - 1$ , not its accumulation.
Why not B:This is $\frac{1}{3} x^{3}$ , the antiderivative of $x^{2}$ alone; it ignores the $- 1$ term and would be increasing everywhere, but $g$ must decrease where $f < 0$ .
Why not C:This is the sign-flip $y = x - \frac{1}{3} x^{3}$ , whose derivative is $1 - x^{2} = - f$ . It increases on $(- 1, 1)$ where $g$ should decrease.

The Fundamental Theorem (Part 2)

Question 6

no calculatormedium

f

is a linear function and

0 < a < b

, then

\int_{a}^{b} f^{''} (x) d x =

Show the answer

The answer is B.

A linear function has

f^{''} (x) = 0

for all

x

, so

\int_{a}^{b} f^{''} (x) d x = 0

. (Equivalently, by the FTC it equals

f^{'} (b) - f^{'} (a)

, and

f^{'}

is constant for a linear

f

Why not A:This would be $\int_{a}^{b} x d x$ ; here we integrate $f^{''}$ , and a linear $f$ has $f^{''} \equiv 0$ .
Why not C:This is $\int_{a}^{b} 1 d x$ ; but $f^{''} = 0$ for a linear function, so the integral is $0$ .
Why not D:Not an antiderivative outcome here; a linear function has $f^{''} (x) = 0$ , so $\int_{a}^{b} f^{''} d x = 0$ .

Antiderivatives

Question 7

no calculatormedium

f^{'} (x) = 4 x^{3}

and

f (1) = 5

, then

f (x) =

Show the answer

The answer is D.

f (x) = \int 4 x^{3} d x = x^{4} + C

. Using

f (1) = 5

1 + C = 5 \Rightarrow C = 4

, so

f (x) = x^{4} + 4

Why not A:You took $C = 5$ , but $f (1) = 1 + C = 5$ gives $C = 4$ .
Why not B:You used $C = 1$ ; substitute $x = 1$ into $x^{4} + C = 5$ to solve correctly.
Why not C: $12 x^{2}$ is the derivative of $4 x^{3}$ ; you differentiated instead of antidifferentiating.

u-Substitution (Indefinite)

Question 8

no calculatormedium

\int \frac{f ^{'} ( x )}{[ f ( x ) ] ^{2}} d x =

Show the answer

The answer is D.

Let

u = f (x)

, so

d u = f^{'} (x) d x

and the integral becomes

\int u^{- 2} d u = - u^{- 1} + C = - \frac{1}{f ( x )} + C

. Check by differentiating:

\frac{d}{d x} [- \frac{1}{f ( x )}] = \frac{f ^{'} ( x )}{[ f ( x ) ] ^{2}} .

Why not A:This treats the integrand as $[f (x)]^{2}$ and applies the power rule to $f$ directly, ignoring that $f^{'} (x)$ is the derivative factor needed for the substitution.
Why not B:This is the antiderivative of $\frac{f ^{'} ( x )}{f ( x )}$ (a first power in the denominator), not $\frac{f ^{'} ( x )}{[ f ( x ) ] ^{2}}$ .
Why not C:Correct substitution but a sign error: $\int u^{- 2} d u = - u^{- 1} + C$ , not $+ u^{- 1}$ . The antiderivative of $u^{- 2}$ is negative.

Straight-Line Motion

Question 9

calculatorhard

A particle has velocity

v (t) = t^{2} - 3 t + 1

for

0 \leq t \leq 3

. What total distance does it travel? Round to the nearest hundredth.

Show the answer

The answer is A.

v = 0

t = \frac{3 \pm 5}{2} \approx 0.38, 2.62

; summing

∣Δ s ∣

over the legs gives

\approx 2.23

Why not B:That's the displacement magnitude $∣ s (3) - s (0) ∣$ ; total distance counts the reversals at $t \approx 0.38, 2.62$ .
Why not C:The trips don't fully cancel; total distance is $\approx 2.23$ , not $0$ .
Why not D: $\int_{0}^{3} ∣ v ∣ d t \approx 2.23$ , not $4.5$ .

Question 10

no calculatorhard

A particle moves along a line with velocity

v (t) = 2 t - 4

for

0 \leq t \leq 3

. What is the total distance traveled by the particle over

[0, 3]

Show the answer

The answer is D.

v = 0

t = 2

, with

v < 0

[0, 2)

and

v > 0

(2, 3]

. Distance

= \int_{0}^{2} ∣ v ∣ + \int_{2}^{3} ∣ v ∣ = ∣ - 4 ∣ + ∣ 1 ∣ = 4 + 1 = 5

, where

\int_{0}^{2} (2 t - 4) d t = - 4

and

\int_{2}^{3} (2 t - 4) d t = 1

Why not A:This is $∣ displacement ∣ = ∣ - 3∣ = 3$ , not total distance; the path reverses at $t = 2$ , adding to $5$ .
Why not B:This is the (signed) displacement $\int_{0}^{3} v = - 3$ ; total distance uses $\int ∣ v ∣ = 5$ .
Why not C:This is the net of the two pieces $- 4 + 1$ in absolute value mishandled; total distance adds magnitudes: $4 + 1 = 5$ .

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