Curriculum

Mathematics

Unit 2

0/14

Unit 5

0/12

Text lesson

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals

Mathematics

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals

Overview

Integrate exponential and logarithmic functions and use integration to find a curve from its slope.
Use the differential, integration by substitution and integration by parts.
Evaluate definite integrals with the fundamental theorem of calculus, including odd and even functions.
Use the definite integral to find areas of plane regions and volumes of solids of revolution.

🌳 Concept tree

Lesson 1 Integration Techniques (Substitution and By Parts) and Definite Integrals
├── 3-1 (a) Indefinite integrals of exponential functions

│ ├── Reminder

│ └── Rules for ex

├── 3-1 (b) Indefinite integrals of logarithmic functions

│ ├── Rules

│ └── Geometric and physical applications

├── 3-2 (a) The differential and the standard integrals

│ ├── The differential

│ └── Standard integrals (reverse of the standard derivatives)

├── 3-2 (b) Integration by substitution

│ ├── Idea

│ ├── When the integrand has an extra factor of $x$

│ └── Two rules proved by substitution

├── 3-2 (c) Integration by parts

│ ├── Rule

│ └── How to choose y and dz

├── 3-2 (d) Applications of the indefinite integral

│ └── Finding a function from its rate of change

├── 3-3 (a) The definite integral and the fundamental theorem of calculus

│ ├── Meaning

│ └── Fundamental theorem of calculus

├── 3-3 (b) Properties of the definite integral

│ ├── Rules

│ └── The modulus function

├── 3-3 (c) Definite integral by substitution
odd and even functions

│ ├── Substitution in a definite integral

│ └── Odd and even functions

├── 3-4 (a) Area of a plane region

│ ├── Rule

│ ├── Method when the curve crosses the x-axis

│ └── Application→cost of covering an arch

└── 3-4 (b) Volumes of solids of revolution

├── Solid of revolution

├── Rule→revolution about the x-axis

└── Volume formulas proved by integration

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 1 / 14

3-1 (a) Indefinite integrals of exponential functions

Reminder

Comparison aspect	Main idea	Keywords
Antiderivative	of $f$ : a function $F$ with $F^{'} (x) = f (x)$ .	function
Indefinite integral	the whole family of antiderivatives, written $\int f (x) d x = F (x) + c$ where $c$ is an arbitrary constant. The curves $y = F (x) + c$ are parallel curves with equal slopes at the same $x$ .	wholefamilyantiderivativeswritten
Properties	∫[f(x)±g(x)] dx=∫f(x) dx±∫g(x) dx and ∫k f(x) dx=k∫f(x) dx
•	$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$ for $n \neq - 1$ .

Rules for ex

$\int e^{x} d x = e^{x} + c$ .
$\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$ for $a \neq 0$ .
∫f′(x) ef(x) dx=ef(x)+c.
$\int a^{x} d x = \frac{a^{x}}{\ln a} + c$ for $a > 0$ , $a \neq 1$ .

✍️ Worked examples

Example 1 Find (a) $\int e^{4 x} d x$ (b) $\int (e^{x} + 1)^{2} d x$ (c) $\int \sin x e^{\cos x} d x$ .
Solution:
(a) $\int e^{4 x} d x = \frac{1}{4} e^{4 x} + c$ .
(b) Expand first (ex+1)2=e2x+2ex+1.
$\int (e^{2 x} + 2 e^{x} + 1) d x = \frac{1}{2} e^{2 x} + 2 e^{x} + x + c$ .
(c) Put $f (x) = \cos x$ , so $f^{'} (x) = - \sin x$ .
$\int \sin x e^{\cos x} d x = - \int e^{\cos x} (- \sin x) d x = - e^{\cos x} + c$ .

Integration Techniques (Substitution→a lathe turning the curve y = √x into a vase shows integration techniques and definite integrals: the volume of revolution is 8π.: 3-1 (a) Indefinite integrals of exponential functions

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 2 / 14

3-1 (b) Indefinite integrals of logarithmic functions

Rules

∫1x dx=ln⁡∣x∣+c(x≠0)
∫f′(x)f(x) dx=ln⁡∣f(x)∣+c
$\int \tan x d x = - \ln ∣ \cos x ∣ + c = \ln ∣ \sec x ∣ + c$ .
$\int \cot x d x = \ln ∣ \sin x ∣ + c$ .

✍️ Worked examples

Example 2 Find (a) $\int (3 x^{2} + \frac{5}{x}) d x$ (b) $\int \frac{7}{x \ln 3} d x$ (c) $\int \frac{2 x + 3}{x^{2} + 3 x - 2} d x$ .
Solution:
(a) $\int 3 x^{2} d x + 5 \int \frac{1}{x} d x = x^{3} + 5 \ln ∣ x ∣ + c$ .
(b) $\frac{7}{\ln 3} \int \frac{1}{x} d x = \frac{7}{\ln 3} \ln ∣ x ∣ + c$ .
(c) The derivative of the denominator $x^{2} + 3 x - 2$ is $2 x + 3$ , which is the numerator.
So the integral is $\ln ∣ x^{2} + 3 x - 2 ∣ + c$ .

If the slope of the tangent at any point $(x, y)$ is $g (x)$ , then the curve is $y = \int g (x) d x$ , and the point given on the curve fixes $c$ .
If the rate of change $\frac{d A}{d t}$ is given, then $A = \int \frac{d A}{d t} d t$ , and the starting value fixes $c$ .

✍️ Worked examples

Example 3 The slope of the tangent to a curve at any point is $3 + \frac{2}{x}$ , and the curve passes through $(e, 3 e + 5)$ . Find its equation.
Solution:
$y = \int (3 + \frac{2}{x}) d x = 3 x + 2 \ln x + c$ (for $x > 0$ ).
Substitute (e,3e+5) 3e+5=3e+2ln⁡e+c=3e+2+c.
So $c = 3$ .
The curve is $y = 3 x + 2 \ln x + 3$ .
Example 4 The area of a lamina changes at the rate $\frac{d A}{d t} = e^{- t / 10}$ cm $^{2}$ /s, and its area at $t = 0$ is $80$ cm $^{2}$ . Find its area after 10 seconds.
Solution:
$A = \int e^{- t / 10} d t = - 10 e^{- t / 10} + c$ .
At t=0 80=−10+c, so c=90.
$A = 90 - 10 e^{- t / 10}$ .
At t=10 A=90−10e−1≈86.3 cm2.

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 3 / 14

3-2 (a) The differential and the standard integrals

The differential

If $y = f (x)$ is differentiable, the differential of $y$ is
$d y = f^{'} (x) d x$
The differential of $x$ is $d x = Δ x$ .
It comes from $\frac{Δ y}{Δ x} \to f^{'} (x)$ as $Δ x \to 0$ .
For a product of two functions z and ℓ of x d(zℓ)=z dℓ+ℓ dz.

✍️ Worked examples

Example 5 Find the differential of (a) $y = e^{2 x - 3}$ (b) $V = \frac{4}{3} π r^{3}$ with respect to $r$ .
Solution:
(a) $d y = y^{'} d x = 2 e^{2 x - 3} d x$ .
(b) $d V = V^{'} d r = 4 π r^{2} d r$ .

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c$ , $n \neq - 1$ .
$\int \cos x d x = \sin x + c$ and $\int \sin x d x = - \cos x + c$ .
$\int \sec^{2} x d x = \tan x + c$ .
$\int e^{x} d x = e^{x} + c$ and $\int a^{x} d x = \frac{a^{x}}{\ln a} + c$ .
$\int \frac{1}{x} d x = \ln ∣ x ∣ + c$ .

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 4 / 14

3-2 (b) Integration by substitution

Idea

Used for integrals of the form $\int f (g (x)) g^{'} (x) d x$ .
If $z = g (x)$ then $d z = g^{'} (x) d x$ , so
$\int f (g (x)) g^{'} (x) d x = \int f (z) d z$
Steps choose z; find dz; rewrite the whole integral in z (including x and dx); integrate; return to x.
Choose $z$ so that the new integral is simpler and a standard one; if the first choice fails, try another.

✍️ Worked examples

Find (a) $\int x^{3} (2 x^{4} - 7)^{5} d x$ (b) $\int \frac{x + 4}{(x^{2} + 8 x)^{3}} d x$ .
Solution:
(a) Put $z = 2 x^{4} - 7$ , so $d z = 8 x^{3} d x$ .
$\int z^{5} \cdot \frac{d z}{8} = \frac{1}{8} \cdot \frac{z^{6}}{6} + c$ .
$= \frac{(2 x^{4} - 7)^{6}}{48} + c$ .
(b) Put $z = x^{2} + 8 x$ , so $d z = (2 x + 8) d x = 2 (x + 4) d x$ .
$\int \frac{x + 4}{z^{3}} d x = \frac{1}{2} \int z^{- 3} d z = \frac{1}{2} \cdot \frac{z^{- 2}}{- 2} + c$ .
$= - \frac{1}{4 (x^{2} + 8 x)^{2}} + c$ .

Put $z$ equal to the inner expression and write $x$ in terms of $z$ too.

✍️ Worked examples

Example 7 Find (a) $\int x (x + 4)^{5} d x$ (b) $\int x \sqrt{x - 1} d x$ .
Solution:
(a) Put $z = x + 4$ , so $x = z - 4$ and $d x = d z$ .
$\int (z - 4) z^{5} d z = \int (z^{6} - 4 z^{5}) d z = \frac{z^{7}}{7} - \frac{2 z^{6}}{3} + c$ .
$= \frac{z^{6} (3 z - 14)}{21} + c = \frac{(x + 4)^{6} (3 x - 2)}{21} + c$ .
(b) Put $z^{2} = x - 1$ , so $x = z^{2} + 1$ and $d x = 2 z d z$ .
$\int (z^{2} + 1) z \cdot 2 z d z = 2 \int (z^{4} + z^{2}) d z = \frac{2 z^{5}}{5} + \frac{2 z^{3}}{3} + c$ .
$= \frac{2}{15} z^{3} (3 z^{2} + 5) + c = \frac{2}{15} (x - 1)^{3 / 2} (3 x + 2) + c$ .
Example 8 Find (a) $\int \frac{1}{1 + \sqrt{x}} d x$ (b) $\int 6 x e^{x^{2}} d x$ .
Solution:
(a) Put $z = 1 + \sqrt{x}$ , so $x = (z - 1)^{2}$ and $d x = 2 (z - 1) d z$ .
$\int \frac{2 (z - 1)}{z} d z = 2 \int (1 - \frac{1}{z}) d z = 2 z - 2 \ln ∣ z ∣ + c$ .
$= 2 (1 + \sqrt{x}) - 2 \ln (1 + \sqrt{x}) + c$ .
(b) Put $z = x^{2}$ , so $d z = 2 x d x$ .
$3 \int e^{z} d z = 3 e^{z} + c = 3 e^{x^{2}} + c$ .
Example 9 Find (a) $\int \frac{x}{3 x^{2} - 1} d x$ (b) $\int \frac{\sqrt{\ln x}}{x} d x$ .
Solution:
(a) Put $z = 3 x^{2} - 1$ , so $d z = 6 x d x$ .
$\frac{1}{6} \int \frac{d z}{z} = \frac{1}{6} \ln ∣ z ∣ + c = \frac{1}{6} \ln ∣ 3 x^{2} - 1 ∣ + c$ .
(b) Put $z = \ln x$ , so $d z = \frac{1}{x} d x$ .
$\int z^{1 / 2} d z = \frac{2}{3} z^{3 / 2} + c = \frac{2}{3} (\ln x)^{3 / 2} + c$ .

Two rules proved by substitution

Put z=f(x) in ∫[f(x)]nf′(x) dxit becomes ∫zn dz=zn+1n+1+c, so ∫[f(x)]nf′(x) dx=[f(x)]n+1n+1+c(n≠−1)

Put z=f(x) in ∫f′(x)f(x) dxit becomes ∫dzz=ln⁡∣z∣+c.

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 5 / 14

3-2 (c) Integration by parts

Rule

Comparison aspect	Main idea	Keywords
Start from the derivative of a product	ddx(yz)=ydzdx+zdydx	ydzdxzdydx
Integrate both sides and rearrange	∫y dz=yz−∫z dy
•	It is used for the integral of a product of two functions when neither is the derivative of the other	usedintegralproductfunctions

Choose $y$ so that $d y$ is simpler than $y$ (for example $y = x$ or $y = \ln x$ ).
Choose $d z$ so that it is easy to integrate to get $z$ .
The integral $\int z d y$ must be easier than the one you started with.
Adding a constant to $z$ does not change the final result.

✍️ Worked examples

Example 10 Find (a) $\int x e^{x} d x$ (b) $\int x^{2} e^{x} d x$ .
Solution:
(a) Let $y = x$ and $d z = e^{x} d x$ , so $d y = d x$ and $z = e^{x}$ .
$\int x e^{x} d x = x e^{x} - \int e^{x} d x = x e^{x} - e^{x} + c = e^{x} (x - 1) + c$ .
(b) Let $y = x^{2}$ and $d z = e^{x} d x$ , so $d y = 2 x d x$ and $z = e^{x}$ .
$\int x^{2} e^{x} d x = x^{2} e^{x} - 2 \int x e^{x} d x$ .
Use part (a) =x2ex−2ex(x−1)+c.
$= e^{x} (x^{2} - 2 x + 2) + c$ .
Example 11 Find (a) $\int \ln x d x$ (b) $\int x \ln x d x$ .
Solution:
(a) Let $y = \ln x$ and $d z = d x$ , so $d y = \frac{1}{x} d x$ and $z = x$ .
$\int \ln x d x = x \ln x - \int x \cdot \frac{1}{x} d x = x \ln x - x + c = x (\ln x - 1) + c$ .
(b) Let $y = \ln x$ and $d z = x d x$ , so $d y = \frac{1}{x} d x$ and $z = \frac{x^{2}}{2}$ .
$\int x \ln x d x = \frac{x^{2}}{2} \ln x - \int \frac{x^{2}}{2} \cdot \frac{1}{x} d x = \frac{x^{2}}{2} \ln x - \frac{x^{2}}{4} + c$ .
$= \frac{x^{2}}{4} (2 \ln x - 1) + c$ .
Example 12 Find $\int \frac{x e^{x}}{(x + 1)^{2}} d x$ .
Solution:
Let $y = x e^{x}$ and $d z = (x + 1)^{- 2} d x$ .
Then $d y = (x e^{x} + e^{x}) d x = e^{x} (x + 1) d x$ and $z = - \frac{1}{x + 1}$ .
$\int = - \frac{x e^{x}}{x + 1} + \int \frac{e^{x} (x + 1)}{x + 1} d x = - \frac{x e^{x}}{x + 1} + e^{x} + c$ .
$= \frac{e^{x} (- x + x + 1)}{x + 1} + c = \frac{e^{x}}{x + 1} + c$ .

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 6 / 14

3-2 (d) Applications of the indefinite integral

Finding a function from its rate of change

If $g (x)$ is the slope of the tangent (or a rate of change or a marginal function) of $f$ , then $f (x) = \int g (x) d x$ .
The integral has an arbitrary constant, found from a given point on the curve.

✍️ Worked examples

Example 13 The slope of the tangent to a curve at any point is $\frac{x e^{x}}{(x + 1)^{2}}$ and the curve passes through $(0, 1)$ . Find its equation.
Solution:
By Example 12, $f (x) = \frac{e^{x}}{x + 1} + c$ .
Substitute (0,1) 1=11+c, so c=0.
The curve is $f (x) = \frac{e^{x}}{x + 1}$ .

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 7 / 14

3-3 (a) The definite integral and the fundamental theorem of calculus

Meaning

The indefinite integral contains an arbitrary constant, so it does not give a definite number.
The difference between its values at $x = b$ and $x = a$ does not depend on the constant:
$(F (b) + c) - (F (a) + c) = F (b) - F (a)$
This number is the definite integral.

If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ on $[a, b]$ :
$\int_{a}^{b} f (x) d x = F (b) - F (a) = [F (x)]_{a}^{b}$
$a$ and $b$ are the lower and upper limits of the integral.
The value is a real number that depends only on $f$ , $a$ and $b$ ; the name of the variable does not matter: $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ .
You may drop the constant $c$ when evaluating, because it cancels in $F (b) - F (a)$ .
Every rule of indefinite integration and every standard integral still applies.
A continuous function on $[a, b]$ is integrable on $[a, b]$ .

Point of comparison	Indefinite integral	Definite integral
Result	A family of functions $F (x) + c$	One real number
Constant $c$	Present	Not needed
Limits	None	Lower $a$ and upper $b$

✍️ Worked examples

Find $\int_{- 2}^{4} (3 x^{2} - 2) d x$ .
Solution:
$f$ is a continuous polynomial.
$\int_{- 2}^{4} (3 x^{2} - 2) d x = [x^{3} - 2 x]_{- 2}^{4}$ .
$= (64 - 8) - (- 8 + 4)$ .
$= 56 + 4 = 60$ .

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 8 / 14

3-3 (b) Properties of the definite integral

Rules

For a continuous $f$ on $[a, b]$ and $c \in] a, b [$ :
$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$ .
$\int_{a}^{a} f (x) d x = 0$ .
$\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ .
$\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x$ .
$\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x$ .

✍️ Worked examples

Example 15 If $f$ is continuous on $R$ , $\int_{1}^{3} f (x) d x = 6$ and $\int_{5}^{3} f (x) d x = - 14$ , find $\int_{1}^{5} f (x) d x$ .
Solution:
$\int_{1}^{5} f (x) d x = \int_{1}^{3} f (x) d x + \int_{3}^{5} f (x) d x$ .
$\int_{3}^{5} f (x) d x = - \int_{5}^{3} f (x) d x = 14$ .
$\int_{1}^{5} f (x) d x = 6 + 14 = 20$ .

Write $∣ x - 3 ∣ = 3 - x$ for $x < 3$ and $x - 3$ for $x \geq 3$ , then split the integral at $x = 3$ .

✍️ Worked examples

Example 16 Find $\int_{1}^{5} ∣ x - 3 ∣ d x$ .
Solution:
$\int_{1}^{5} ∣ x - 3 ∣ d x = \int_{1}^{3} (3 - x) d x + \int_{3}^{5} (x - 3) d x$ .
$\int_{1}^{3} (3 - x) d x = [3 x - \frac{x^{2}}{2}]_{1}^{3} = (9 - 4.5) - (3 - 0.5) = 2$ .
$\int_{3}^{5} (x - 3) d x = [\frac{x^{2}}{2} - 3 x]_{3}^{5} = (12.5 - 15) - (4.5 - 9) = 2$ .
The total is $2 + 2 = 4$ (two triangles, each of base 2 and height 2).

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 9 / 14

3-3 (c) Definite integral by substitution, odd and even functions

Substitution in a definite integral

Method 1find the indefinite integral by substitution, return to x, then use the limits of x.

Method 2change the limits with the substitution, so you do not need to return to x.

✍️ Worked examples

Find $\int_{0}^{3} x \sqrt{x^{2} + 3} d x$ .
Solution:
Put $z = x^{2} + 3$ , so $d z = 2 x d x$ .
When $x = 0$ , $z = 3$ ; when $x = 3$ , $z = 12$ .
$\int_{3}^{12} \frac{1}{2} z^{1 / 2} d z = \frac{1}{2} \cdot \frac{2}{3} [z^{3 / 2}]_{3}^{12} = \frac{1}{3} (1 2^{3 / 2} - 3^{3 / 2})$ .
$1 2^{3 / 2} = 24 \sqrt{3}$ and $3^{3 / 2} = 3 \sqrt{3}$ .
The value is $\frac{1}{3} (24 \sqrt{3} - 3 \sqrt{3}) = 7 \sqrt{3}$ .

If f is continuous and odd on [−a,a] (that is f(−x)=−f(x))∫−aaf(x) dx=0

If f is continuous and even on [−a,a] (that is f(−x)=f(x))∫−aaf(x) dx=2∫0af(x) dx

✍️ Worked examples

Find (a) $\int_{- 2}^{2} \frac{x^{3} - 3 x}{x^{2} + 1} d x$ (b) $\int_{- 3}^{3} (x^{2} - 1) d x$ .
Solution:
(a) $f (- x) = \frac{- x^{3} + 3 x}{x^{2} + 1} = - f (x)$ , so $f$ is odd.
The integral over $[- 2, 2]$ equals $0$ .
(b) $f (x) = x^{2} - 1$ is an even polynomial.
$\int_{- 3}^{3} (x^{2} - 1) d x = 2 \int_{0}^{3} (x^{2} - 1) d x = 2 [\frac{x^{3}}{3} - x]_{0}^{3}$ .
$= 2 (9 - 3) = 12$ .

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 10 / 14

3-4 (a) Area of a plane region

Rule

If $f$ is continuous on $[a, b]$ and $f (x) \geq 0$ , the area bounded by the curve $y = f (x)$ , the $x$ -axis and the lines $x = a$ , $x = b$ is
$A = \int_{a}^{b} f (x) d x$
In general the area is $A = \int_{a}^{b} ∣ f (x) ∣ d x$ .
The area is measured in square units, and it is never negative.

Find the zeros of $f$ by putting $f (x) = 0$ .
Study the sign of $f$ on each sub-interval between the zeros and the limits.
Integrate on each sub-interval separately, taking the absolute value of each integral where $f < 0$ .
Add the areas.

✍️ Worked examples

Example 19 Find the area bounded by $f (x) = 1 + 4 x - x^{2}$ , the $x$ -axis and the lines $x = 1$ and $x = 4$ .
Solution:
$f (x) > 0$ on $[1, 4]$ (its zeros are $2 \pm \sqrt{5}$ , i.e. about $- 0.24$ and $4.24$ , both outside $[1, 4]$ ).
$A = \int_{1}^{4} (1 + 4 x - x^{2}) d x = [x + 2 x^{2} - \frac{x^{3}}{3}]_{1}^{4}$ .
$= (4 + 32 - \frac{64}{3}) - (1 + 2 - \frac{1}{3})$ .
$= \frac{44}{3} - \frac{8}{3} = 12$ square units.
Example 20 Find the area bounded by $f (x) = \sqrt{2 x + 2}$ , the $x$ -axis and the line $x = 1$ .
Solution:
Zero of f 2x+2=0 gives x=−1, and f>0 for x>−1.
$A = \int_{- 1}^{1} \sqrt{2 x + 2} d x = [\frac{(2 x + 2)^{3 / 2}}{3}]_{- 1}^{1}$ .
$= \frac{4^{3 / 2}}{3} - 0 = \frac{8}{3}$ square units.
Example 21 Find the area of the region above the $x$ -axis bounded by $f (x) = x^{3} - 4 x$ and the $x$ -axis, for $x \leq 3$ .
Solution:
$f (x) = x (x - 2) (x + 2)$ , so the zeros are $- 2, 0, 2$ .
$f > 0$ on $[- 2, 0]$ and on $[2, 3]$ .
$A_{1} = \int_{- 2}^{0} (x^{3} - 4 x) d x = [\frac{x^{4}}{4} - 2 x^{2}]_{- 2}^{0} = 0 - (4 - 8) = 4$ .
$A_{2} = \int_{2}^{3} (x^{3} - 4 x) d x = [\frac{x^{4}}{4} - 2 x^{2}]_{2}^{3} = (\frac{81}{4} - 18) - (4 - 8) = \frac{9}{4} + 4 = \frac{25}{4}$ .
$A = 4 + \frac{25}{4} = \frac{41}{4}$ square units.
Example 22 Find the area bounded by $f (x) = x^{3} - 4 x$ , the $x$ -axis and the lines $x = - 2$ and $x = 1$ .
Solution:
$f \geq 0$ on $[- 2, 0]$ and $f \leq 0$ on $[0, 1]$ .
$A_{1} = \int_{- 2}^{0} (x^{3} - 4 x) d x = 4$ .
$\int_{0}^{1} (x^{3} - 4 x) d x = [\frac{x^{4}}{4} - 2 x^{2}]_{0}^{1} = \frac{1}{4} - 2 = - \frac{7}{4}$ , so $A_{2} = ∣ - \frac{7}{4} ∣ = \frac{7}{4}$ .
$A = 4 + \frac{7}{4} = \frac{23}{4}$ square units.

Application: cost of covering an arch

✍️ Worked examples

Example 23 A hotel entrance arch has equation $y = - \frac{1}{2} (x - 1) (x - 7)$ with $x$ in metres. Find the cost of covering it with glass at 1500 LE per square metre.
Solution:
Cost = area $\times$ 1500.
The zeros are $x = 1$ and $x = 7$ , and $y > 0$ between them.
$A = \int_{1}^{7} - \frac{1}{2} (x^{2} - 8 x + 7) d x = - \frac{1}{2} [\frac{x^{3}}{3} - 4 x^{2} + 7 x]_{1}^{7}$ .
$[\dots]_{1}^{7} = (\frac{343}{3} - 196 + 49) - (\frac{1}{3} - 4 + 7) = - \frac{98}{3} - \frac{10}{3} = - 36$ .
$A = - \frac{1}{2} (- 36) = 18$ square metres.
Cost $= 18 \times 1500 = 27 000$ LE.

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 11 / 14

3-4 (b) Volumes of solids of revolution

Solid of revolution

A solid of revolution is generated when a plane region makes a complete revolution about a fixed line in its plane, called the axis of revolution.
Examples a right circular cone, a right circular cylinder and a sphere.

Rule: revolution about the x-axis

If $f$ is continuous on $[a, b]$ and $f (x) \geq 0$ , the region bounded by $y = f (x)$ , the $x$ -axis and the lines $x = a$ , $x = b$ generates a solid of volume
$V = π \int_{a}^{b} [f (x)]^{2} d x (cubic units)$

✍️ Worked examples

Example 24 Find the volume generated by revolving the region bounded by $f (x) = x^{2} + 1$ , the $x$ -axis and the lines $x = - 1$ and $x = 1$ about the $x$ -axis.
Solution:
$V = π \int_{- 1}^{1} (x^{2} + 1)^{2} d x = π \int_{- 1}^{1} (x^{4} + 2 x^{2} + 1) d x$ .
The integrand is even, so $V = 2 π [\frac{x^{5}}{5} + \frac{2 x^{3}}{3} + x]_{0}^{1}$ .
$= 2 π (\frac{1}{5} + \frac{2}{3} + 1) = 2 π \cdot \frac{28}{15} = \frac{56 π}{15}$ cubic units.
Example 25 Find the volume generated by revolving the region bounded by $f (x) = x$ , the $x$ -axis and the line $x = 3$ about the $x$ -axis, and name the solid.
Solution:
$V = π \int_{0}^{3} x^{2} d x = π [\frac{x^{3}}{3}]_{0}^{3} = 9 π$ cubic units.
The solid is a right circular cone with $r = 3$ and $h = 3$ .
Check 13πr2h=13π(9)(3)=9π.

Volume formulas proved by integration

Conethe line

y = \frac{r}{h} x

for

0 \leq x \leq h

revolves to a cone, so

V = π \int_{0}^{h} \frac{r^{2}}{h^{2}} x^{2} d x = π \frac{r^{2}}{h^{2}} \cdot \frac{h^{3}}{3} = \frac{1}{3} π r^{2} h

Spherethe circle

x^{2} + y^{2} = r^{2}

, so

y^{2} = r^{2} - x^{2}

and

V = π \int_{- r}^{r} (r^{2} - x^{2}) d x = 2 π [r^{2} x - \frac{x^{3}}{3}]_{0}^{r} = \frac{4}{3} π r^{3}

Cylinderthe line

y = r

for

0 \leq x \leq h

gives

V = π \int_{0}^{h} r^{2} d x = π r^{2} h

✍️ Worked examples

Example 26 Find the volume generated by revolving the region bounded by the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ about the $x$ -axis.
Solution:
$y^{2} = b^{2} (1 - \frac{x^{2}}{a^{2}})$ , and the limits are $y = 0$ , i.e. $x = \pm a$ .
$V = π \int_{- a}^{a} b^{2} (1 - \frac{x^{2}}{a^{2}}) d x = 2 π b^{2} \int_{0}^{a} (1 - \frac{x^{2}}{a^{2}}) d x$ (even function).
$= 2 π b^{2} [x - \frac{x^{3}}{3 a^{2}}]_{0}^{a} = 2 π b^{2} \cdot \frac{2 a}{3}$ .
$V = \frac{4}{3} π a b^{2}$ cubic units.
Example 27 Find the volume generated by revolving the region bounded by $y = 2 x - x^{2}$ and the $x$ -axis about the $x$ -axis.
Solution:
The curve meets the $x$ -axis where $2 x - x^{2} = 0$ , i.e. $x = 0$ and $x = 2$ .
$V = π \int_{0}^{2} (2 x - x^{2})^{2} d x = π \int_{0}^{2} (4 x^{2} - 4 x^{3} + x^{4}) d x$ .
$= π [\frac{4 x^{3}}{3} - x^{4} + \frac{x^{5}}{5}]_{0}^{2}$ .
$= π (\frac{32}{3} - 16 + \frac{32}{5}) = \frac{16 π}{15}$ cubic units.

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 12 / 14

Key Scientific Tables & Anatomical Diagrams Extracted from Official Curricula

💡 Core Diagrams & Reference Data→All data, reference comparison tables, and anatomical figures have been accurately digitized from official curriculum textbooks.

🔬 integration_areas_and_volumes_of_revolution

Scientific details and precise observational breakdown:

(a) 14 (b) 16 (c) 18 (d) 21

4) If (T,,) is an arithmetic sequence »T,
+ Tyg = 68 »
Ty + Tyg = err

(a) 68 (b) 136 (c) 34 (d) 204

5) In the arithmetic sequence (2a+5b>53a+8b 54a+11b >…)
Tyg So

(a)a+9b (b)10a+30b = (c) lla+32b (d) 12a+35b

6) If (T,,) is an arithmetic sequence in which 7 = + »
z TE seceesesnenees

@) 3 () $ (c) $ (a) 3

7) In any arithmetic sequence (T,,) » Les

(a) 2 )3 “(a (d) 5

8) An arithmetic sequence in which T
= m
T,
= » where m #n »
the common

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals · 13 / 14

Comparison aspect	Main idea	Keywords
│ └── Formula	∫ u · dv = u · v – ∫ v · du (Logarithmic Inverse trig Algebraic Trig Exponential)	LogarithmicInversetrigAlgebraic
•	└── 3. Geometric Applications of Definite Integrals	GeometricApplicationsDefiniteIntegrals
├── Plane Area between two curves	Area = ∫ [yupper – ylower] dx	yupperylower
├── Volume of Revolution about x-axis	V = π · ∫ [y²] dx
└── Volume of Revolution about y-axis	V = π · ∫ [x²] dy

Integration Method	Mathematical Condition	Fundamental Formula
Direct Form	Integrand contains function and its derivative	∫ [f(x)]ⁿ · f'(x) dx = [f(x)]ⁿ⁺¹ / (n+1) + C
Logarithmic Form	Numerator is exact derivative of denominator	∫ [f'(x) / f(x)] dx = ln \|f(x)\| + C
By Parts	Product of two distinct mathematical families	∫ u dv = u v – ∫ v du
Volume around x-axis	Revolution of y = f(x) through 360°	V = π ∫ y² dx

Curriculum

Mathematics

Unit 1

Unit 2

Unit 3

Unit 4

Unit 5

Unit 6

Unit 7

Lesson 1: Integration Techniques (Substitution and By Parts) and Definite Integrals

🌳 Concept tree

│ ├── Reminder

│ └── Rules for ex

│ ├── Rules

│ └── Geometric and physical applications

│ ├── The differential

├── 3-2 (b) Integration by substitution

│ ├── Idea

│ └── Two rules proved by substitution

├── 3-2 (c) Integration by parts

│ ├── Rule

│ └── How to choose y and dz

│ └── Finding a function from its rate of change

│ ├── Meaning

│ └── Fundamental theorem of calculus

├── 3-3 (b) Properties of the definite integral

│ ├── Rules

│ └── The modulus function

│ ├── Substitution in a definite integral

│ └── Odd and even functions

├── 3-4 (a) Area of a plane region

│ ├── Rule

│ ├── Method when the curve crosses the x-axis

└── 3-4 (b) Volumes of solids of revolution

├── Solid of revolution

└── Volume formulas proved by integration

Reminder

Rules for ex

✍️ Worked examples

Rules

✍️ Worked examples

✍️ Worked examples

The differential

✍️ Worked examples

Idea

✍️ Worked examples

✍️ Worked examples

Two rules proved by substitution

Rule

✍️ Worked examples

Finding a function from its rate of change

✍️ Worked examples

Meaning

✍️ Worked examples

Rules

✍️ Worked examples

✍️ Worked examples

Substitution in a definite integral

✍️ Worked examples

✍️ Worked examples

Rule

✍️ Worked examples

Application: cost of covering an arch

✍️ Worked examples

Solid of revolution

Rule: revolution about the x-axis

✍️ Worked examples

Volume formulas proved by integration

✍️ Worked examples

🔬 integration_areas_and_volumes_of_revolution

(a) 14 (b) 16 (c) 18 (d) 21

(a) 68 (b) 136 (c) 34 (d) 204

(a)a+9b (b)10a+30b = (c) lla+32b (d) 12a+35b

@) 3 () $ (c) $ (a) 3

7) In any arithmetic sequence (T,,) » Les

(a) 2 )3 “(a (d) 5

🔍 In-Depth Explanations, Deductions & Multi-Book Insights:

Mathematics Thanaweya Amma (2026/2027)

🌳 Conceptual Hierarchy Tree

Advanced Integration Calculus

│ └── ∫ e^f(x) · f'(x) dx = e^f(x) + C