BASIC
CALCULUS
6.53
Mr. John Lester C. Romero
REVIEW
TIME
7/1/20XX
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Directions. Match Column A and B with antiderivative of
the following integrands. Write the letters of your
answers in a separate sheet of paper.
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Do you recall
about the
Chain rule of
differentiation?
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This lesson will present the most
basic technique in taking the
antiderivative of a function which
is antidifferentiation by
substitution. This method is the
inverse of the chain rule in
differentiation.
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Chain Rule for Integration
The integration by substitution is justified
through the Chain Rule of Integration. Let F(u)
𝑑
be a function whose derivative is f(u), that is,
𝑑𝑥
𝐹(𝑢)=𝑓(𝑢). If u is a differentiable function of x,
say 𝑢=ℎ(𝑥) then,
∫𝑓(𝑢)=𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥
BASIC CALCULUS 4TH QUARTER –WEEK3
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Here is how to prove this rule. Given that:
𝑑
𝐹(𝑢)=𝑓(𝑢)
𝑑𝑥
𝑑 𝐹(𝑢)=𝑓(𝑢)
Integrating both sides ∫𝑑 𝐹(𝑢)=∫𝑓(𝑢) 𝑑𝑢.
𝑑𝑢
Since 𝑢=ℎ(𝑥) then =ℎ′(𝑥), by the chain rule of
𝑑𝑥
differentiation we have:
𝑑 𝐹(𝑢)
= 𝑓[ ℎ(𝑥) ] ℎ′(𝑥)
𝑑𝑥
Or
𝑑𝐹(𝑢)=𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥
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Integrating both sides
∫𝑑𝐹(𝑢) = ∫𝑓[ ℎ(𝑥) ] ℎ′(𝑥
) 𝑑𝑥
Using Transitive Property of Equality.
If ∫𝑑 𝐹(𝑢) = ∫𝑓(𝑢) 𝑑𝑢
and ∫𝑑𝐹(𝑢) = ∫𝑓[ ℎ(𝑥) ] ℎ′(𝑥
) 𝑑𝑥
then,
∫𝑓(𝑢)𝑑𝑢 = ∫𝑓[ ℎ(𝑥) ] ℎ′(𝑥)
BASIC𝑑𝑥
CALCULUS 4 QUARTER –WEEK3
TH
8
ANTIDERIVATIVE OF A FUNCTION
USING SUBSTITUTION RULE
Integration by Substitution
This technique is a great help in dealing
with integrals that cannot be evaluated
readily by direct application of the
standard integration formulas, particularly,
if the integrand is the result of a
differentiation using the chain-rule.
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Integration by substitution primarily simplifies
an integrand by setting a part of the integrand
to 𝑢 and part of integrand with 𝑑(𝑢). It is also
described as a change of variables because
we are replacing variables to obtain an
expression wherein integration rules can be
readily applied. Usually, we make a
substitution for a function whose derivative
occurs in the integrand.
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Substitution with Indefinite Integrand
Let 𝑢=ℎ(𝑥), where ℎ′(𝑥) is continuous over an interval, let
𝑓(𝑥) be continuous over corresponding range of g and
let 𝐹(𝑥) be and antiderivative of 𝑓(𝑥), then
∫𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥 = ∫𝑓(𝑢)𝑑𝑢
= 𝐹(𝑢)+𝑐
Notes : 𝑢 = ℎ(𝑥) wherein u is written instead of the
function h(x)
𝑑𝑢 = ℎ′(𝑥) while du is written instead of the
derivative
of the
function
h(x)
TH
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QUARTER –WEEK3
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Example 1. Substitution to find the antiderivative of a
function.
Given : ∫8𝑥 3 (2𝑥 4 +5)𝑑𝑥
To achieve the form ∫𝑓(𝑢)𝑑𝑢, we can rewrite the given as:
∫(2𝑥 4 +5) 8𝑥 3 𝑑𝑥
To find the antiderivative:
let 𝑢 = (2𝑥 4 +5) and 𝑑𝑢 = 8𝑥 3 𝑑𝑥
∫ 𝑢 𝑑𝑢
𝑢1+1
=
+𝐶
1+1
𝑢2
= +𝐶
2
(2𝑥 4 +5)2
=
+
2
expression)
(use power rule of integrals)
𝐶 (substitute the original
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To check :
(2𝑥 4 +5)2
ℎ(𝑥) =
2
2(2𝑥 4 +5)2−1
ℎ ′ (𝑥) =
8𝑥 3 𝑑𝑥
2
2(2𝑥 4 +5)2−1
3
ℎ ′ (𝑥) =
8𝑥 𝑑𝑥
2
4
3
ℎ ′ (𝑥) = (2𝑥 + 5) 8𝑥 𝑑𝑥
3
4
Same as ℎ ′ (𝑥) = 8𝑥 (2𝑥 + 5) 𝑑𝑥, in which
we started with.
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Example 2. Substitution with modification.
Given :
∫ ( 𝑥 𝑥 2 − 8 )𝑑𝑥
To achieve the form ∫ 𝑓(𝑢)𝑑𝑢, we can rewrite the given as:
∫ ( 𝑥 2 − 8 ) 𝑥 𝑑𝑥
Express the radical sign to fractional exponent.
1
2
∫ (𝑥 2 − 8) 𝑥 𝑑𝑥
Let 𝑢 = (𝑥 2 − 8). Then, 𝑑𝑢 = 2𝑥 𝑑𝑥
However, modification is to be done so we can apply integration by
substitution rule.
𝑑𝑢
Dividing 𝑑𝑢 = 2𝑥 𝑑𝑥 by 2, we have:
2
BASIC CALCULUS 4TH QUARTER –WEEK3
2𝑥 𝑑𝑥
=
2
𝑑𝑢
=𝑥
2
𝑑𝑥
15
1
2
Now, use 𝑑𝑢 as substitute with 𝑥 𝑑𝑥.
1
1
2
∫𝑢
𝑑𝑢
2
1
= ∫ 𝑢1 2 𝑑𝑢
2
=
1
2
=
1
2
=
=
=
1
+1
𝑢2
1
+1
2
3
𝑢2
3
2
+𝐶
(use power rule of integrals)
+𝐶
1 2
(𝑢3 2 ) + 𝐶
2 3
1
(𝑢3 2 ) +𝐶
3
1
(𝑥 2 − 8)3 2 + 𝐶
3
(substitute the original expression)
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Example 3. Substitution with integrals of exponential functions.
Given : ∫ 6𝑒 3𝑥 𝑑𝑥
Take the constant out of the integrand.
1
Let 𝑢 = 3𝑥 ; 𝑑𝑢 = 3𝑑𝑥 ; 𝑑𝑢 = 𝑑𝑥
3
∫ 6𝑒 3𝑥 𝑑𝑥
𝑢 1
= 6∫ 𝑒
𝑑𝑢
(constant rule)
=6
1
3
3
∫ 𝑒 𝑢 𝑑𝑢
= 2∫ 𝑒 𝑢 𝑑𝑢
= 2𝑒 𝑢 +𝐶
= 2𝑒 3𝑥 +𝐶
(simplify)
(used of common integral)
(substitute the value of u)
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Example 4. Antiderivative of Trigonometric Functions
using Substitution
Given :
∫ cot 𝑥 𝑑𝑥
Use trigonometric identities to replace cot x as shown.
𝑐‫𝑥 ݏ݋‬
∫
𝑑𝑥
‫𝑥 ݊݅ݏ‬
Let 𝑢 = sin 𝑥 and 𝑑𝑢 = 𝑐‫𝑥𝑑 𝑥ݏ݋‬.
Rewrite the integrand as follows.
1
∫
cos 𝑥 𝑑𝑥
‫𝑥 ݊݅ݏ‬
1
= ∫ 𝑑𝑢
𝑢
= 𝑙݊ |𝑢| + 𝐶
= 𝑙݊ |sin 𝑥| + 𝐶 (substitute value of u)
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SOLVE PROBLEMS INVOLVING
ANTIDIFFERENTIATION
In solving problems involving integrals, student’s
critical thinking skills allows them to analyze the
appropriate formula as well the applicable technique or
rules in given situations.
Since integration or antidifferentiation is a reverse of
differentiation, familiarity on the process of solving
differential calculus problems will help you in dealing
problems on integral calculus. Most of the practical
applications of integration involves how one quantity
changes with respect to another quantity. Thus, finding
their rate of change and relations.
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Rectilinear Motion
This application deals with a value which
changes with respect to a certain value of another
variable. In the study of rectilinear motion, the
velocity 𝑣 of a moving object is defined as the time
rate of change of the distance 𝑠 and the
acceleration 𝑎 as the time rate of change of the
velocity 𝑣. Symbolically, it is written as:
𝑑𝑠
𝑑𝑣
𝑣=
and 𝑎 =
𝑑𝑡
𝑑𝑡
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Illustrative Example 1.
A ball is thrown vertically
upward from the initial
height of 1 meter with an
initial velocity of 30 m/sec.
Find the maximum height
attained by the ball.
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Given from the problem : 𝑣𝑖 = 30𝑚/‫𝑐𝑒ݏ‬
𝑚
𝑚
Constant value of gravitational acceleration ( 2 ) : −9.8 ( 2 )
𝑠𝑒𝑐
𝑠𝑒𝑐
Solution :
𝑑𝑣
𝑑𝑣
Since 𝑎 = and 𝑎 is negative then, = −𝑎.
𝑑𝑡
𝑑𝑣
Therefore,
𝑑𝑡
𝑑𝑡
= −9.8. That is, 𝑑𝑣 = (−9.8)𝑑𝑡.
To solve for the integral of 𝑑𝑣 = (−9.8)𝑑𝑡, we have :
∫ (−9.8)𝑑𝑡
𝑣 = −9.8𝑡 + 𝐶 (velocity of falling object)
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However, we know that the initial velocity 𝑣 = 30𝑚/‫ 𝑐𝑒ݏ‬when 𝑡 = 0. To
determine the value of constant c, substitute the given values to the
above equation.
𝑣 = −9.8𝑡 + 𝐶
30 = −9.8(0) + 𝐶
𝐶 = 30
Then, 𝑣 = −9.8𝑡 + 𝐶 becomes 𝑣 = −9.8𝑡 + 30. This equation
represents the velocity of the ball at any given time.
𝑑𝑠
Use these two equations, 𝑣 = −9.8𝑡 + 30 and 𝑣 = to find 𝑠
𝑑𝑡
(distance) as shown.
𝑑𝑠
= −9.8𝑡 + 30
𝑑𝑡
𝑑‫( = ݏ‬−9.8𝑡 + 30)𝑑𝑡
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To find s, take the integral of 𝑑s, we have:
∫(−9.8𝑡 + 30)𝑑𝑡
9.8𝑡 2
−
+
2
2
‫=ݏ‬
30𝑡 + 𝐶
‫ = ݏ‬−4.9 𝑡 + 30𝑡 + 𝐶
From the given, when 𝑡 = 0, ‫ = ݏ‬0, and 𝐶 = 1.
The above equation becomes ‫ = ݏ‬− 4.9𝑡 2 + 30𝑡 + 1. This
equation represents the distance of the ball from the ground at
any time 𝑡.
To determine the maximum height the ball reached, we must
first calculate the time for that ball to reach its maximum.
Since 𝑣ℎ𝑚𝑎𝑥 = 0 when the ball is at its maximum, the equation
𝑣 = −9.8𝑡 + 30 becomes 0 = −9.8𝑡 + 30.
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Computing for the time, use: 0 = −9.8𝑡 + 30
−30 = −9.8𝑡 (addition property of
equality)
𝑡 ≈ 3 ‫( ݏ𝑑݊݋𝑐𝑒ݏ‬time approximated to the
nearest second, the ball reached its
max)
Substitute t = 3 seconds to determine the height.
‫ = ݏ‬−4.9𝑡 2 + 30𝑡 + 𝐶
‫ = ݏ‬−4.9(3)2 + 30(3) + 𝐶
‫ = ݏ‬45.9 𝑚 + 1𝑚
‫ = ݏ‬46.9 𝑚
The displacement of the ball from the initial point is 45.9 m. Since the
ball is thrown from 1 meter height C, the maximum height the ball
reached before going back to the ground is 46.9 m.
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Equations of the family of curves
A set of curves with the same
slope at any point (x, y) is
represented by the equation
𝑦 = 𝐹(𝑥) + 𝑐.
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Illustrative Example 2.
Find the equation of the family of the curves whose
slope is 6x - 1 with a member passing through point (1, 7)
Solution :
Since slope of the line is defined as the change in “𝑦”
Δ𝑦
𝑑𝑦
over the change of “𝑥”, then it is represented by
‫ 𝑟݋‬.
We may say that
𝑑𝑦
𝑑𝑥
Δ𝑥
𝑑𝑥
= 6𝑥 − 1. Therefore, 𝑑𝑦 = (6𝑥 − 1)𝑑𝑥
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To solve for the integral of 𝑑𝑦 = (6𝑥 − 1)𝑑𝑥, we have:
∫(6𝑥 − 1)𝑑𝑥
6𝑥 2
2
2
=
−𝑥+𝑐
(common integral)
= 3𝑥 − 𝑥 + 𝑐
(simplify)
The equation of the family of parabolas with slope (6x - 1) is 𝑦 = 3𝑥 2 − 𝑥 + 𝑐.
Now, to find the equation specific with a parabola passing to point (1, 7) we
need to determine the assigned value of c. Substitute x = 1 and y = 7 in the
equation
𝑦 = 3𝑥 2 − 𝑥 + 𝑐
7 = 3 (1)2 − 1 + 𝑐
7−3+1=𝑐
𝑐=5
Thus, the equation of the parabola passing through point (1, 7) is
𝑦 = 3𝑥 2 − 𝑥 + 5
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The family of curves
with a slope (6x - 1) and
tangent to 𝑦 = 3𝑥 2 − 𝑥 + 5
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THANK YOU
7/1/20XX
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