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1. Distance and displacement from velocity

Warm-up

Three quick questions from earlier lessons.

Discover

A cyclist rides at a steady 12 mph for 2 hours. On a graph of velocity against time, that is a horizontal line at 12 from $t = 0$ to $t = 2$ . What is the area of the rectangle under it, and what does it mean?

24: the distance in miles
14: the sum of the numbers

Key idea

Area under a rate graph gives the total change.

v(t) \ge 0

is a velocity, the area between its graph and the

t

-axis on

[a, b]

is the distance traveled. If

v

is sometimes negative, areas below the axis count negatively and the signed total is the displacement.

Worked example

$v(t) = 3t$ m/s. How far does the object go from $t = 0$ to $t = 4$ ?

Step 1The velocity grows steadily: v(t) = 3t.
Step 2The distance is the area under the graph from 0 to 4.
Step 3A triangle: ½ · 4 · 12 = 24 meters.

Base

4

, height

v(4) = 12

The region under v is a triangle.

Area

= \tfrac{1}{2} \cdot 4 \cdot 12 = 24

½ × base × height.

Distance:

24

Units: m/s × s.

Your turn, guided

Your turn: 30 mph for 2 hours, then 50 mph for 1 hour.

Distance:

110

miles

60 + 50.

Practice ★

A car drives at 30 mph for 2 hours, then at 57 mph until hour 5. How far does it travel, in miles?
A car drives at 28 mph for 3 hours, then at 34 mph until hour 4. How far does it travel, in miles?

From rest, a cart's velocity is $v(t) = 4t$ m/s. How far does it go in the first 6 seconds, in meters?
From rest, a cart's velocity is $v(t) = 9t$ m/s. How far does it go in the first 5 seconds, in meters?

Watch out!

Practice ★★

A particle's velocity is $v(t) = t - 1$ m/s. Find its displacement from $t = 0$ to $t = 8$ , in meters (areas below the axis count as negative).
A particle's velocity is $v(t) = t - 3$ m/s. Find its displacement from $t = 0$ to $t = 5$ , in meters (areas below the axis count as negative).

Exit ticket

Three questions to check you have got it.

From rest, a cart's velocity is $v(t) = 4t$ m/s. How far does it go in the first 6 seconds, in meters?
From rest, a cart's velocity is $v(t) = 9t$ m/s. How far does it go in the first 5 seconds, in meters?
From rest, a cart's velocity is $v(t) = 7t$ m/s. How far does it go in the first 3 seconds, in meters?

Summary card

Distance from velocity

Area under $v$ = distance (when $v \ge 0$ ).
Signed area (below the axis negative) = displacement.

Watch out: use the right area formula, ½ for triangles.

2. Riemann sums

Warm-up

Three quick questions from earlier lessons.

Discover

To estimate the area under $y = x^2$ on $[0, 2]$ , you use two rectangles of width 1 whose heights are taken at the left ends, $f(0)$ and $f(1)$ . Is the estimate too small or too big?

Too small
Too big

Key idea

Rectangles approximate area; thinner rectangles do better.

Step 12 rectangles: 2 · 1 + 2 · 2 = 6.
Step 24 rectangles: 1 + 1.25 + 2 + 3.25 = 7.5 — closer.
Step 3More and thinner rectangles approach the exact area, 28/3 ≈ 9.33.

Split

[a, b]

into

n

pieces of width

\Delta x = \dfrac{b - a}{n}

. A Riemann sum adds

f(x_i^*)\,\Delta x

, with

x_i^*

the left end, right end or midpoint of each piece. For increasing

f

, left sums underestimate and right sums overestimate.

Worked example

Right sum for $\displaystyle\int_0^3 x^2\,dx$ with 3 pieces.

\Delta x = 1

Width 1; right ends 1, 2, 3.

1 + 4 + 9 = 14

Heights f(1), f(2), f(3).

R_3 = 14

Compare with the exact value 9: an overestimate.

Your turn, guided

Your turn: the left sum for the same integral, 3 pieces.

L_3 = 5

f(0) + f(1) + f(2).

Practice ★

Estimate $\displaystyle\int_0^{ 5 } x^2\,dx$ with a left Riemann sum of 5 subintervals of width 1.
Estimate $\displaystyle\int_0^{ 6 } x^2\,dx$ with a left Riemann sum of 6 subintervals of width 1.
Estimate $\displaystyle\int_0^{ 4 } x^2\,dx$ with a left Riemann sum of 4 subintervals of width 1.

Watch out!

Practice ★★

Estimate $\displaystyle\int_0^4 (x^2 + 4)\,dx$ with a midpoint sum of 2 subintervals.
Estimate $\displaystyle\int_0^4 (x^2 + 2)\,dx$ with a midpoint sum of 2 subintervals.

$f$ is increasing on $[0, 7]$ . How does a left Riemann sum compare with $\displaystyle\int_0^{ 7 } f(x)\,dx$ ?
1. It underestimates: each rectangle uses the lowest height on its piece
2. It overestimates
3. It is exact

Exit ticket

Three questions to check you have got it.

Estimate $\displaystyle\int_0^{ 5 } x^2\,dx$ with a left Riemann sum of 5 subintervals of width 1.
Estimate $\displaystyle\int_0^{ 6 } x^2\,dx$ with a left Riemann sum of 6 subintervals of width 1.
Estimate $\displaystyle\int_0^{ 4 } x^2\,dx$ with a left Riemann sum of 4 subintervals of width 1.

Summary card

Riemann sums

$\Delta x = (b - a)/n$ ; add height × $\Delta x$ .
Left, right or midpoint heights.
Increasing $f$ : left under, right over.

Watch out: multiply by the width.

3. The definite integral and its properties

Warm-up

Three quick questions from earlier lessons.

Discover

$\displaystyle\int_0^2 (-3)\,dx$ : the graph $y = -3$ lies below the axis. What is the integral?

$-6$
$6$

Key idea

The definite integral is the limit of Riemann sums: a signed area.

\displaystyle\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum f(x_i^*)\Delta x

. Properties:

\displaystyle\int_b^a f = -\int_a^b f

;

\displaystyle\int_a^c f = \int_a^b f + \int_b^c f

;

\displaystyle\int_a^b (kf + g) = k\int_a^b f + \int_a^b g

;

\displaystyle\int_a^b c\,dx = c(b - a)

Worked example

$\displaystyle\int_0^4 f = 5$ and $\displaystyle\int_4^6 f = -2$ . Find $\displaystyle\int_0^6 \big(2f(x) - 1\big)\,dx$ .

\displaystyle\int_0^6 f = 5 + (-2) = 3

Join the intervals.

2 \cdot 3 - 1 \cdot 6

Linearity; the constant −1 over a length of 6.

= 0

Compute.

Your turn, guided

Your turn: with the same $f$ , find $\displaystyle\int_6^0 f(x)\,dx$ .

-3

Reverse the limits: change the sign.

Practice ★

$\displaystyle\int_1^3 f(x)\,dx = 7$ and $\displaystyle\int_3^7 f(x)\,dx = -4$ . Find $\displaystyle\int_7^1 f(x)\,dx$ .
$\displaystyle\int_1^3 f(x)\,dx = -7$ and $\displaystyle\int_3^7 f(x)\,dx = 8$ . Find $\displaystyle\int_7^1 f(x)\,dx$ .

Use geometry: find $\displaystyle\int_{-6}^{ 6 } \sqrt{ 36 - x^2}\,dx$ .
Use geometry: find $\displaystyle\int_{-1}^{ 1 } \sqrt{ 1 - x^2}\,dx$ .

Watch out!

Practice ★★

$\displaystyle\int_0^{ 4 } f(x)\,dx = 3$ . Find $\displaystyle\int_0^{ 4 } \big(3f(x) + 2\big)\,dx$ .
$\displaystyle\int_0^{ 9 } f(x)\,dx = 4$ . Find $\displaystyle\int_0^{ 9 } \big(3f(x) + 2\big)\,dx$ .

Exit ticket

Three questions to check you have got it.

$\displaystyle\int_1^3 f(x)\,dx = 7$ and $\displaystyle\int_3^7 f(x)\,dx = -4$ . Find $\displaystyle\int_7^1 f(x)\,dx$ .
$\displaystyle\int_1^3 f(x)\,dx = -7$ and $\displaystyle\int_3^7 f(x)\,dx = 8$ . Find $\displaystyle\int_7^1 f(x)\,dx$ .
$\displaystyle\int_1^3 f(x)\,dx = 5$ and $\displaystyle\int_3^7 f(x)\,dx = 7$ . Find $\displaystyle\int_7^1 f(x)\,dx$ .

Summary card

The definite integral

Signed area: below the axis is negative.
Reverse the limits: change the sign. Adjacent intervals add.
$\displaystyle\int_a^b c\,dx = c(b - a)$ .

Watch out: a constant is integrated over the whole length.

4. Antiderivatives

Warm-up

Three quick questions from earlier lessons.

Discover

Which function has derivative $3x^2$ ?

$x^3$
$6x$

Key idea

Run the derivative rules backwards. Antiderivatives differ by a constant $C$ .

\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n + 1} + C

(

n \ne -1

);

\displaystyle\int \dfrac{1}{x}\,dx = \ln|x| + C

;

\displaystyle\int e^x\,dx = e^x + C

;

\displaystyle\int \cos x\,dx = \sin x + C

;

\displaystyle\int \sin x\,dx = -\cos x + C

. A given value such as

F(0)

fixes

C

Worked example

Find $F$ with $F'(x) = 4x^3 + \sin x$ and $F(0) = 2$ .

F(x) = x^4 - \cos x + C

Term by term.

C = 3

F(0) = 0 − 1 + C = 2.

F(x) = x^4 - \cos x + 3

Check by differentiating.

Your turn, guided

Your turn: $F'(x) = 6x^2$ , $F(0) = 0$ .

F(x) = 2x^3

6x³/3.

Practice ★

Find the antiderivative $F$ of $f(x) = 4x^{ 3 } + 9$ with $F(0) = 0$ .
Find the antiderivative $F$ of $f(x) = 2x^{ 2 } + 8$ with $F(0) = 0$ .
Find the antiderivative $F$ of $f(x) = 5x^{ 4 } + 6$ with $F(0) = 0$ .

Watch out!

Practice ★★

Find the antiderivative $F$ of $f(x) = \cos x + e^x$ with $F(0) = 6$ .
Find the antiderivative $F$ of $f(x) = \cos x + e^x$ with $F(0) = 4$ .

Find the antiderivative $F$ of $f(x) = \dfrac{1}{x}$ , for $x > 0$ , with $F(1) = -9$ .
Find the antiderivative $F$ of $f(x) = \dfrac{1}{x}$ , for $x > 0$ , with $F(1) = 8$ .

Exit ticket

Three questions to check you have got it.

Find the antiderivative $F$ of $f(x) = 4x^{ 3 } + 9$ with $F(0) = 0$ .
Find the antiderivative $F$ of $f(x) = 2x^{ 2 } + 8$ with $F(0) = 0$ .
Find the antiderivative $F$ of $f(x) = 5x^{ 4 } + 6$ with $F(0) = 0$ .

Summary card

Antiderivatives

$x^n \to \dfrac{x^{n+1}}{n+1}$ ; $\tfrac{1}{x} \to \ln|x|$ ; $e^x \to e^x$ ; $\cos \to \sin$ ; $\sin \to -\cos$ .
Use a given value to find $C$ .

Watch out: check by differentiating.

5. The Fundamental Theorem of Calculus: evaluating integrals

Warm-up

Three quick questions from earlier lessons.

Discover

$v(t)$ is a velocity and $s(t)$ the position, so $s' = v$ . The area under $v$ on $[a, b]$ is the change in position. Which expression is it?

$s(b) - s(a)$
$s(a) - s(b)$

Key idea

No more limits of sums: use an antiderivative.

Fundamental Theorem of Calculus: if

f

is continuous on

[a, b]

and

F' = f

, then

\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)

Worked example

Evaluate $\displaystyle\int_1^3 (x^2 + 1)\,dx$ .

F(x) = \dfrac{x^3}{3} + x

An antiderivative.

F(3) - F(1) = 12 - \tfrac{4}{3}

F(3) = 12, F(1) = 4/3.

= \tfrac{32}{3}

Compute.

Your turn, guided

Your turn: $\displaystyle\int_0^2 3x^2\,dx$ .

= 8

F(x) = x³: F(2) − F(0).

Practice ★

Evaluate $\displaystyle\int_0^{ 5 } x^2\,dx$ .
Evaluate $\displaystyle\int_0^{ 9 } x^2\,dx$ .
Evaluate $\displaystyle\int_0^{ 8 } x^2\,dx$ .

Watch out!

Practice ★★

Evaluate $\displaystyle\int_0^{\pi/5} \cos x\,dx$ . Give an exact value.
Evaluate $\displaystyle\int_0^{\pi/3} \cos x\,dx$ . Give an exact value.

Evaluate $\displaystyle\int_{ -1 }^{ 4 } (2x + 9)\,dx$ .
Evaluate $\displaystyle\int_{ 1 }^{ 4 } (2x + 7)\,dx$ .

Boss problem. Find the area of the region between the parabola $y = 4 - x^2$ and the $x$ -axis.

Exit ticket

Three questions to check you have got it.

Evaluate $\displaystyle\int_{ -1 }^{ 4 } (2x + 9)\,dx$ .
Evaluate $\displaystyle\int_{ 1 }^{ 4 } (2x + 7)\,dx$ .
Evaluate $\displaystyle\int_{ -2 }^{ 5 } (2x + 8)\,dx$ .

Summary card

Fundamental Theorem (evaluation)

$\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$ , with $F' = f$ .

Watch out: upper minus lower; check the antiderivative's signs.

6. The average value of a function

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^2$ on $[0, 3]$ ; the blue line is $y = a$ . Move $a$ until the rectangle under the line on $[0, 3]$ has the same area as the region under the curve, $\displaystyle\int_0^3 x^2\,dx = 9$ .

The parabola y = x² on [0, 3] and a horizontal line y = a. At a = 3 the rectangle of width 3 under the line has area 9, the same as under the curve.

Which height $a$ works?

$a = 3$
$a = 4.5$

Key idea

The average value is the height of the rectangle with the same area.

The average value of

f

[a, b]

\dfrac{1}{b - a}\displaystyle\int_a^b f(x)\,dx

Worked example

Average value of $f(x) = 2x + 1$ on $[0, 4]$ .

\displaystyle\int_0^4 (2x + 1)\,dx = 20

F(x) = x² + x.

Average

= \tfrac{20}{4} = 5

Divide by the length 4.

Your turn, guided

Your turn: average of $f(x) = 3x^2$ on $[0, 2]$ .

Average

= 4

∫ = 8; divide by 2.

Practice ★

Find the average value of $f(x) = x^2$ on $[0, 3]$ .
Find the average value of $f(x) = x^2$ on $[0, 6]$ .
Find the average value of $f(x) = x^2$ on $[0, 2]$ .

Watch out!

Practice ★★

$v(t) = 3t + 6$ m/s. Find the average velocity on $[0, 8]$ , in m/s.
$v(t) = 3t + 1$ m/s. Find the average velocity on $[0, 7]$ , in m/s.

Find the average value of $f(x) = 6\sin x$ on $[0, \pi]$ . Give an exact value.

Exit ticket

Three questions to check you have got it.

Find the average value of $f(x) = x^2$ on $[0, 3]$ .
Find the average value of $f(x) = x^2$ on $[0, 6]$ .
Find the average value of $f(x) = x^2$ on $[0, 2]$ .

Summary card

Average value

$f_{\text{avg}} = \dfrac{1}{b - a}\displaystyle\int_a^b f(x)\,dx$ .

Watch out: divide by the length; do not just average the end values.

7. Accumulation functions: the Fundamental Theorem, part 2

Warm-up

Three quick questions from earlier lessons.

Discover

$g(x) = \displaystyle\int_0^x 2t\,dt$ is the area under $y = 2t$ from 0 to $x$ , which is $x^2$ . What is $g'(x)$ ?

$2x$
$x^2$

Key idea

The area function grows at the rate of the height.

Fundamental Theorem, part 2: if

f

is continuous,

g(x) = \displaystyle\int_a^x f(t)\,dt

has

g'(x) = f(x)

. With a variable upper limit

u(x)

\dfrac{d}{dx}\displaystyle\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x)

Worked example

$g(x) = \displaystyle\int_1^{x^3} \sqrt{t}\,dt$ . Find $g'(x)$ .

g(x) = F(x^3)

with

F' = \sqrt{t}

Upper limit u = x³.

g'(x) = \sqrt{x^3} \cdot 3x^2

Chain rule.

Your turn, guided

Your turn: $g(x) = \displaystyle\int_2^x (t^3 - t)\,dt$ .

g'(x) = x^3 - x

FTC part 2.

Practice ★

$g(x) = \displaystyle\int_{ 4 }^{x} (t^{ 2 } + 2)\,dt$ . Find $g'(x)$ .
$g(x) = \displaystyle\int_{ 1 }^{x} (t^{ 3 } + 2)\,dt$ . Find $g'(x)$ .
$g(x) = \displaystyle\int_{ 4 }^{x} (t^{ 3 } + 1)\,dt$ . Find $g'(x)$ .

Watch out!

Practice ★★

$g(x) = \displaystyle\int_0^{x^{ 4 } } \cos t\,dt$ . Find $g'(x)$ .
$g(x) = \displaystyle\int_0^{x^{ 3 } } \cos t\,dt$ . Find $g'(x)$ .

Here is the graph of $y = f(t)$ . Let $g(x) = \displaystyle\int_{-5}^{x} f(t)\,dt$ . Tap the point of this graph at the $x$ -value where $g$ has a local maximum.
Here is the graph of $y = f(t)$ . Let $g(x) = \displaystyle\int_{-5}^{x} f(t)\,dt$ . Tap the point of this graph at the $x$ -value where $g$ has a local maximum.

Exit ticket

Three questions to check you have got it.

$g(x) = \displaystyle\int_{ 4 }^{x} (t^{ 2 } + 2)\,dt$ . Find $g'(x)$ .
$g(x) = \displaystyle\int_{ 1 }^{x} (t^{ 3 } + 2)\,dt$ . Find $g'(x)$ .
$g(x) = \displaystyle\int_{ 4 }^{x} (t^{ 3 } + 1)\,dt$ . Find $g'(x)$ .

Summary card

Accumulation functions

$\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt = f(x)$ .
Upper limit $u(x)$ : multiply by $u'(x)$ .
$g$ rises where $f > 0$ , falls where $f < 0$ .

Watch out: the answer is the integrand, not its antiderivative.

8. Net change: an amount from its rate

Warm-up

Three quick questions from earlier lessons.

Discover

A bath holds 40 L. Over 10 minutes, $\displaystyle\int_0^{10} r(t)\,dt = 60$ L flow in. How much water is in the bath now?

100 L
60 L

Key idea

The integral of a rate is the net change of the amount.

Net change:

F(b) = F(a) + \displaystyle\int_a^b F'(t)\,dt

. Starting value plus accumulated change.

Worked example

A town has 5000 people; it grows at $r(t) = 200 + 30t$ people per year. Population after 4 years?

\displaystyle\int_0^4 (200 + 30t)\,dt = 800 + 240 = 1040

Accumulated change.

5000 + 1040 = 6040

Add the start.

Your turn, guided

Your turn: $F(2) = 7$ and $\displaystyle\int_2^5 F'(x)\,dx = -3$ . Find $F(5)$ .

F(5) = 4

7 + (−3).

Practice ★

$F' = f$ , $F(0) = -1$ and $\displaystyle\int_0^{ 3 } f(x)\,dx = 0$ . Find $F(3)$ .
$F' = f$ , $F(0) = -2$ and $\displaystyle\int_0^{ 6 } f(x)\,dx = -9$ . Find $F(6)$ .
$F' = f$ , $F(0) = 8$ and $\displaystyle\int_0^{ 4 } f(x)\,dx = -2$ . Find $F(4)$ .

Watch out!

Practice ★★

A tank holds 37 L. Water flows in at $r(t) = 5 + 1t$ L/min. How much water is in the tank after 4 minutes, in liters?
A tank holds 57 L. Water flows in at $r(t) = 6 + 3t$ L/min. How much water is in the tank after 5 minutes, in liters?

$v(t)$ is a particle's velocity in m/s. What does $\displaystyle\int_0^{ 7 } v(t)\,dt$ give?
1. Its displacement (change in position) from $t = 0$ to $t = 7$ , in meters
2. The total distance it travels, in meters
3. Its velocity at $t = 7$

Exit ticket

Three questions to check you have got it.

A tank holds 37 L. Water flows in at $r(t) = 5 + 1t$ L/min. How much water is in the tank after 4 minutes, in liters?
A tank holds 57 L. Water flows in at $r(t) = 6 + 3t$ L/min. How much water is in the tank after 5 minutes, in liters?
A tank holds 38 L. Water flows in at $r(t) = 4 + 1t$ L/min. How much water is in the tank after 2 minutes, in liters?

Summary card

Net change

$F(b) = F(a) + \displaystyle\int_a^b F'(t)\,dt$ : start + accumulated change.

Watch out: a rate at one moment is not an amount.

Answers: Integrals and accumulation

1. Distance and displacement from velocity

$231$
$118$

$72$
$\frac{225}{2}$

$24$
$- \frac{5}{2}$

$72$
$\frac{225}{2}$
$\frac{63}{2}$

2. Riemann sums

$30$
$55$
$14$

$36$
$28$

It underestimates: each rectangle uses the lowest height on its piece

$30$
$55$
$14$

3. The definite integral and its properties

$-3$
$-1$

$18 \pi$
$\frac{\pi}{2}$

$17$
$30$

$-3$
$-1$
$-12$

4. Antiderivatives

$x^{4} + 9 x$
$\frac{2 x^{3}}{3} + 8 x$
$x^{5} + 6 x$

$e^{x} + \sin{\left(x \right)} + 5$
$e^{x} + \sin{\left(x \right)} + 3$

$\ln{\left(x \right)} - 9$
$\ln{\left(x \right)} + 8$

$x^{4} + 9 x$
$\frac{2 x^{3}}{3} + 8 x$
$x^{5} + 6 x$

5. The Fundamental Theorem of Calculus: evaluating integrals

$\frac{125}{3}$
$243$
$\frac{512}{3}$

$\sqrt{\frac{5}{8} - \frac{\sqrt{5}}{8}}$
$\frac{\sqrt{3}}{2}$

$60$
$36$

$\frac{32}{3}$

$60$
$36$
$77$

6. The average value of a function

$3$
$12$
$\frac{4}{3}$

$18$
$\frac{23}{2}$

$\frac{12}{\pi}$

$3$
$12$
$\frac{4}{3}$

7. Accumulation functions: the Fundamental Theorem, part 2

$x^{2} + 2$
$x^{3} + 2$
$x^{3} + 1$

$4 x^{3} \cos{\left(x^{4} \right)}$
$3 x^{2} \cos{\left(x^{3} \right)}$

$(3, 0)$
$(1, 0)$

$x^{2} + 2$
$x^{3} + 2$
$x^{3} + 1$

8. Net change: an amount from its rate

$-1$
$-11$
$6$

$65$
$\frac{249}{2}$

Its displacement (change in position) from $t = 0$ to $t = 7$ , in meters

$65$
$\frac{249}{2}$
$48$