basic integration in Calculus 1

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Notation Slices along x
(with respect to x)

The integration symbol “න " is an extended S
for "Summation" (the idea of summing slices)

Integral symbol
Integrand

(Function we want to integrate)

• After the Integral Symbol we put the function we want to integral of (called the Integrand),
and then finish with dx to mean the slices go in the x direction.

• Here is how we write the answer :

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Plus C.. Constant of integration

We wrote the answer as + . But… why + C ?
It is the “constant of Integration”. It is there because all the function whose derivative is 2x :

• The derivative of + is , and the derivative of − is also , and so on! Because the
derivative of a constant is zero.

• So, when we reverse the operation (to find the integral) we only know , but there could have
been a constant of any value.

• So we wrap up the idea just writing + C at the end.

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Throughout this table, a and b are given constants, independent of x and C is a constant of integration

Common Differentiation Rules Common Indefinite Integration Rules

1. 1. න 0 =
= 0

2. Derivatives of any 2. න = +
= 0 constants equals

to zero

3. 3. න 1 = +
1 = 0

4. 2
= 1 4. න = 2 +

5. 2 = 2 5. න 2 = 3 +
3

1 1 −12 1 3
= 2 2 1 2 23
6. 2 = = 6. න = න 2 = 3 + = 3 2 + C

2

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Throughout this table, a and b are given constants, independent of x and C is a constant of integration

Common Differentiation Rules Common Indefinite Integration Rules

7. = −1 For any values of n 7. න = +1 + ≠ −1
+ 1 (any value except -1)

8. 1 ‫ ׬‬ =?? 8. න −1 = න 1 = + = −1
= (only)
(use integration by
parts, Calculus 2

9. = 9. න = +


10. = 10. න = 1 +


11. = ∙ 11. න = 1 ∙ +


12. 12. න = + = 2.718 …
= 0 (Euler number)

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Differentiation and Integration of Trigonometric Functions

Common Differentiation Rules Common Indefinite Integration Rules

13. 13. න = +
=

14. 14. න = − +
= − 15. න 2 = +
16. න 2 = − +
15. = 2 17. න = +


16. = − 2


17.
=

18. 18. න = − +
= −

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Throughout this table, a and b are given constants, independent of x and C is a constant of integration

Common Differentiation Rules Common Indefinite Integration Rules

19. 1
+ = + 19. න + = + +

20. 1
+ = − + 20. න + = − + +

21. = 2 + 21. න 2 1 +
+ + = +

22. = − 2 + 22. න 2 + = 1 + +
+ −

23. 1
+ = + + 23. න + + = + +

24. 1
+ = − + + 24. න + + = − + +

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Throughout this table, a and b are given constants, independent of x and C is a constant of integration

Common Differentiation Rules Common Indefinite Integration Rules

25. = 2 Change to the basic 25. න = − + = +
trigonometric function
( and ). Then,

use u-substitution

26. = − 2 26. න = + = − +

27. 27. න = + +
=
Can move o2u8t . Calculus 2
28.
= − the constants න = − +

29. Never move out29. න = න Multiplication by
= ′ constant
the variables

30. Sum and 30. න ± = න ± න
± = ′ ± ′
difference rule

(integrate one by one)

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Examples for multiplication function (that possible to expand)

expand

Upper limit

1 3 21 23
1.a) න 1 + 3 − 2 = න 3 − 2 + − 2 1.b) න 1 + 3 − 2 = න 3 − 2 + − 2

1 1

3 Lower limit 2 3
= න 1 − 2 + 1 − 2 +
=න

1

1 Integrate =න 21
= න 1 − 2 + 3 ∙ one by one 1 − 2 + 3 ∙
1

2 2 2
= − 2 ∙ 2 + 3 ∙ + = − 2 ∙ 2 + 3 ∙ 1

= − 2 + 3 + # = − 2 + 3 2
1

0

= 2 − 22 + 3 2 − 1 − 12 + 3 1

= 0.07944 #

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Examples for multiplication function (that possible to expand)

Square – possible to expand! Expand bracket square

2) 1 1 (a-b)2 = a2 - 2ab + b2
− 1 2= 2 − 2 + 1
න 3 − 1 2 = න 3 2 − 2 + 1
Integrate one by one
am•an = am+n am•an = am+n 7 41 (with respect to t)

1 ∙ 1 = 13+1 = 4 1 ∙ 2 = 31+2 = 7 = න 3 − 2 3 + 3

3 3 3 3

10 7 4

3 3 3
-1) = 10 − 2 7 + 4 +

3 33

3 10 37 34
= 10 3 − 2 ∙ 7 3 + 4 3 +

3 10 6 7 3 4
= 10 3 − 7 3 + 4 3 + #

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Examples for division function (that possible to separate – with single term in denominator)

+ 3 2 1 Expand bracket square
3) න 3 2
+ 6 2 + 9
= න 3 2

= − 1 = − 21 Separate (because single
term in denominator)
6 2 3 9
= න 3 2 + 3 2 + 3 2 Simplify (using the
properties of indices)
1 1 1 2 −32
2 2 3
2 = 12−2 = −32 = −2 =න + + 3 −2

=න 1 ∙ 1 + 2 −32 + 3 −2 Integrate one by one
3 (with respect to x)

1 −21 −1 (a+b)2 = a2 + 2ab + b2
= 3 ∙ + 2 ∙ + 3 ∙ −1 +
-1) − 1 + 3 2= 2 + 6 + 9
2
1
1 2 −12 3 + 3 2= + 6 2 + 9
= 3 + 2 ∙ − 1 − +

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= 3 − − + #

Examples for division function (that possible to separate – with single term in denominator)

1 1 − 2 2 =න 1 1 − 2 2 + 4 Expand bracket square
4) න 2 2
0 =
0

1 =න 1 1 2 2 4 Separate (because single 2 2= 2 2
= − 2 − 2 + 2 term in denominator) = 4
0
Simplify (using the ∙ = +
1 = −2 1 properties of indices)
2 2 2= 2 ∙ 2
= න −2 − 2 + 2 Integrate one by one = 2 +2
(with respect to x) = 4
0

= 1 −2 − 2 + 1 2 1

−2 2 0

= − 1 −2 1 − 2 1 + 1 2 1 − 1 −2 0 1 0 1 2 0 1

− −2 0 +
2 22 2
= −

4 = 4 −2 = 2 = − 1 −2 − 2 + 1 2 − − 1 + 1 (a-b)2 = a2 - 2ab + b2
2 1 − 2 2= 12 − 2 2 + 2 2
2 2 22

= 1.627 # 1 − 2 2= 1 − 2 2 + 4

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"Integration by Substitution" (also called "u-Substitution" or "The Reverse Chain Rule") is a method to
find an integral, but only when it can be set up in a special way.

The first and most vital step is to be able to write our integral in this form:

′ =


′ = ∙ =


Note that we have u and its derivative u’ in general, integration by
So, we can write u-substitution is a

combination of derivatives
and rules of integration.

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Like in this example:

Here = , and we have = 2 and its derivative 2 . This integral is good to go!
When our integral is set up like that, we can do this substitution :

Then we can integrate f(u) with respect to u and finish by putting g(x) back as u. © Amirah Hana
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Throughout this table, a is the given constants and C is a constant of integration

Rules of Integration (with respect to x) Rules of Integration (with respect to u)

1. න 0 = 1. න 0 =

2. න = + 2. න = +

3. න 1 = + 3. න 1 = +

2 2
4. න = 2 + 4. න = 2 +

5. න 2 = 3 + 5. න 2 = 3 +
3 3

3 3
1 2 23 1 2 23
6. න = න 2 = 3 + = 3 2 + C 6. න = න 2 = 3 + = 3 2 + C
2 2

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Throughout this table, a and b are the given constants and C is a constant of integration

Rules of Integration (with respect to x) Rules of Integration (with respect to u)

7. න = +1 + 7. න = +1 + ≠ −1
+ 1 + 1 (any value except -1)

8. න −1 = න 1 = + 8. න −1 = න 1 = + = −1
(only)

9. න = + 9. න = +

For instant, if is a linear function, +

න + = 1 ∙ + +1 ≠ −1
+ 1 + (any value except -1)

න + = 1 ∙ + + = −1
(only)

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Integration of Trigonometric Functions Rules of Integration (with respect to u)

Rules of Integration (with respect to x) 13. න = +

13. න = +

14. න = − + 14. න = − +

15. න 2 = + 15. න 2 = +

16. න 2 = − + 16. න 2 = − +

17. න = + 17. න = +

18. න = − + 18. න = − +
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3 In this case, we can use u-substitution
Evaluate න 2 + 1 . (since the degree of x in numerator less 1
from the degree of x in denominator, so they

can be related with derivative)

න 3 1 = න 3 ∙ Re-write in terms = 2 + 1 if we make the right choice of
2 + 2 of u and du u, we can simplify (or eliminate)
all the variables x from the
No Quotient rule 3 = 2 given question
in integration ! = න ∙ 2
© Amirah Hana
31 = 2
= 2 න
Integrate wrt u

3
= 2 +

= 3 2 + 1 + # Replace back,
2 = 2 + 1

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3
Evaluate න 2 + 1 .

Re-write in terms = 3
of u and du

න 3 1 = න 1 ∙ = 3
2 + 2 + 3

= 3

if we make the wrong choice of
u, we can’t simplify (or
eliminate) all the variables x
from the given question

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Evaluate න + 1 4 . In this case, we can use u-substitution
(since the functions sinx and cosx can be

related with derivative)

Cannot expand, because has ( )4 Re-write in terms = + 1 if we make the right choice of
of u and du u, we can simplify (or eliminate)
න + 1 4 = න ∙ 4 − = − all the variables x from the
given question

Integrate wrt u = − න 4 = −

5 ≠ −1
= − 5 + (any value except -1)

1 + 1 5 + # No Product, Quotient &
= −5 Generalized Power Rules

Replace back, in integration !
= + 1
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Evaluate න + 1 4 .

Re-write in terms =
of u and du
= −
න + 1 4 = න ∙ + 1 4 ∙
−
= −
= − න + 1 4

We can simplify sinx, but are we
capable of expanding + 1 4 or
use substitution again ? So, the
choice of = is incorrect

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Evaluate න + 1 4 .

Re-write in terms =
of u and du
=
න + 1 4 = න +1 4
∙
=

if we make the wrong choice of
u, we can’t simplify (or
eliminate) all the variables x
from the given question

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Evaluate න 3 + 5 2 . We can use u-substitution
(since the functions and 2 can be

related with derivative)

Re-write in terms =
of u and du

= 2
2
න 3 + 5 2 = න 3 + 5 2 ∙
= 2

Integrate wrt u = න 3 + 5 if we make the right choice of u,
4 6 we can simplify (or eliminate) all
the variables x from the given
= 4 + 6 + question

≠ −1 = 1 4 + 1 6 + # Replace back,
(any value except -1) 4 6 =

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Evaluate න 3 + 5 2 .

Re-write in terms
of u and du

න 3 + 5 2 = න 3 + 5 2 ∙ =


=


= න 2 + 4 ∙ =

= න 2 + 4 ∙ if we make the wrong choice of
u, we can’t simplify (or
eliminate) all the variables x
from the given question

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Evaluate න 3 2 1 + 4 . We can use u-substitution
(since the functions 1 + 4 and 3 can be related

with derivative). We also can integrate 2…

Re-write in terms = 1 + 4

of u and du

= 4 3
∙ 4 3
න 3 2 1 + 4 = න 3 2
= 4 3
Integrate wrt u
= න 2 ∙
4
if we make the right choice of u, we

can simplify (or eliminate) all the

= 1 න 2 variables x from the given question and
4 its easily can solve the integrals

1 Replace back,
= 4 + = 1 + 4

= 1 1 + 4 + #
4

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Evaluate න 3 2 1 + 4 .

Re-write in terms
of u and du

= 2 1 + 4 = 1 + 4 2
∙ 8 3 2 1 + 4 1 + 4
න 3 2 1 + 4 = න 3 2 1 + 4

= 2 1 + 4 ∙ 1 + 4 1 + 4 ∙ 4 3


= න 8 1 + 4 = 8 3 2 1 + 4 1 + 4

if we make the wrong choice of u, we
can’t simplify (or eliminate) all the
variables x from the given question

and its so hard to solve the integrals

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Evaluate න 4 − 3 . We can use u-substitution
(since the functions and can be related

with derivative).

Re-write in terms =
of u and du
=
න 4 − 3 = න 4 − 3
∙
=

Integrate wrt u = න 4 − 3 if we make the right choice of u, we
4 can simplify (or eliminate) all the

= 4 − 4 + variables  from the given question and
its easily can solve the integrals

≠ −1 = 4 − 1 4 + #
(any value except -1) 4
Replace back,
=

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Evaluate න 4 − 3 .

Re-write in terms
of u and du

න 4 − 3 = න 4 − 3 =
∙ −
= −


= −

if we make the wrong choice of u, we
can’t simplify (or eliminate) all the

variables  from the given question and
its so hard to solve the integrals

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Evaluate න 4 − 3 .

Re-write in terms
of u and du

= 4 − 3 = 4 − 3

න 4 − 3
= න ∙ ∙ −3 2
= −3 2 = −3 2


= න ∙ −3 2 = −3 2

if we make the wrong choice of u, we
can’t simplify (or eliminate) all the

variables  from the given question and
its so hard to solve the integrals

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5 Re-write in terms ≠ −1
Evaluate න + 3 4 . of u and du (any value except -1)

Integrate wrt u

න 5 5 = + 3 = − 3 5 −2 −3 +
+ 3 4 = න 4 ∙ න + 3 4 = 5 −2 − 3 ∙ −3
= 1
Can move out = How to eliminate 11
constant from = 5 න 4 ∙ variable x = 5 − 2 2 + 3 +
integration sign.

= 5 න − 3 ∙ Single term in 55
4 the denominator = − 2 + 3 2 + + 3 3 + #
But… be careful!
never move out 3 Replace back,
= 5 න 4 − 4 = + 3
variable!

Separating the
fraction…

= 5 න −3 − 3 −4

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3 We can use u-substitution
Evaluate න 3 . (since the functions 3 and 3 can be related

with derivative).

3 = න 3 − 3 Re-write in terms = − 3
න 3
of u and du = − −3 3
Integrate wrt u
= න 3 ∙ 3
3 = 3 3

= න ∙
3

= 1 න if we make the right choice of u, we
3 can simplify (or eliminate) all the

= 1 + variables x from the given question and
3 its easily can solve the integrals

= 1 − 3 + # Replace back,
3 = − 3

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3
Evaluate න 3 .

Re-write in terms
of u and du

3 = 3
න 3 = න 3 − 3

= න − 3 ∙ = 3 3
3 3

Can’t simplify ! = 3 3

if we make the wrong choice of u, we
can’t simplify (or eliminate) all the
variables x from the given question

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Evaluate න 1 2 Definite Integral using
. u-substitution
0 3 2 + 1

Step 1: 2 1 2 1
Solve Indefinite Integral න = 3 ∙ 1 ∙ 2 +
3 2 + 1

න 2 = න 2 3 2 + 1 −21 = 3 2 + 1 Replace back, 2
= 3 2 + 1 = 3 +
= 6
3 2 + 1

Re-write in terms = න 2 ∙ −21 ∙ = 6 2 3 2 + 1 +
of u and du 6 =3
≠ −1
3 (any value except -1)

= 1 න −12
3
Integrate wrt u
1
1 2
=3∙ +
1
2

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Evaluate න 1 2 Definite Integral using
. u-substitution
0 3 2 + 1

Step 1: න 2 2 3 2 + 1 +
Solve Indefinite Integral =3

3 2 + 1

Step 2: 1 2 2 1
Solve Definite Integral =3
න 3 2 + 1
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0 0

2 3 1 2+1− 3 0 2+1
=3
Substitute Substitute
Lower limit
2 upper limit
=3 2−1

2 © Amirah Hana
=3 #

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