Applications of Differentiation in economics, optimization, kinematics, tangent and normal lines, related rates, linear approximation or small changes

1. SMA 2101: CALCULUS I
c
⃝Francis O. Ochieng
francokech@gmail.com
YouTube: Prof. Francis Oketch∗
Department of Pure and Applied Mathematics
Jomo Kenyatta University of Agriculture and Technology
Course content
• Functions: definition, domain, range, codomain, composition (or composite), inverse.
• Limits, continuity and differentiability of a function.
• Differentiation by first principle and by rule for xn (integral and fractional n).
• Other techniques of differentiation, i.e., sums, products, quotients, chain rule; their applications
to algebraic, trigonometric, logarithmic, exponential, and inverse trigonometric functions all of
a single variable.
• Implicit and parametric differentiation.
• Applications of differentiation to: rates of change, small changes, stationary points, equations of
tangents and normal lines, kinematics, and economics and financial models (cost, revenue and
profit).
• Introduction to integration and its applications to area and volume.
References
[1] Calculus: Early Transcendentals (8th Edition) by James Stewart
[2] Calculus with Analytic Geometry by Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards;
5th edition
[3] Calculus and Analytical Geometry (9th edition) by George B. Thomas and Ross L. Finney
[4] Advanced Engineering Mathematics (10th ed.) by Erwin Kreyszig
[5] Calculus by Larson Hostellem
1 Applications of differentiation
1.1 Equation of a tangent line and normal line to a curve
Consider the diagram below
∗
https://www.youtube.com/channel/UC7wd3x6 08GNoUbLxmaC4vA
1

2. 1.1 Equation of a tangent line and normal line to a curve c
⃝Francis Oketch
A tangent line to a curve is a line that touches the
curve at one point, say (x0, y0), while a normal
line to a curve is a line perpendicular to the
tangent line and passes through the point (x0, y0).
1. The rate of change of y with respect to x, i.e.,
dy
dx
= y′(x), gives the gradient function to the
curve y = f(x).
2. If y′(x) (or
dy
dx
) is evaluated at point x = x0, the result is the gradient of the tangent line at the
point x = x0.
3. Since the normal line is perpendicular to the tangent line, the product of their gradients must
be equal to −1, i.e.,
.
.
m1 × m2 = −1 ,
where m1 and m2 represents the gradient of the tangent line and normal line, respectively.
4. To find the equation of a straight line, we require a known point (x0, y0), a general point (x, y),
which must all lie on the line, and the gradient, m =
∆y
∆x
, of the line. Thus, the equation of a
straight line is given by
.
.
y − y0
x − x0
= m
Example(s):
1. Find the equation of the tangent line and normal line to the curve x2 + 2xy + 3y2 = 17 at point
(1, 2).
Solution
Clearly, the point (1, 2) lies on the given curve. Now, differentiating the given curve implicitly
with respect to x yields 2x + 2x
dy
dx
+ 2y + 6y
dy
dx
= 0. Therefore,
dy
dx
=
−(x + y)
x + 3y
The gradient of the tangent line at point (1, 2) is
dy
dx
=
−(1 + 2)
1 + 3(2)
= −
3
7
= m1. Thus, the
equation of the tangent line at point (1, 2) is
y − 2
x − 1
= −
3
7
⇒ y = −
3
7
x +
17
7
The gradient of the normal line at (1, 2) is m2 = −
1
m1
=
7
3
. Thus, the equation of the normal
line at (1, 2) is
y − 2
x − 1
=
7
3
⇒ y =
7
3
x −
1
3
2. Find the equation of tangent line and normal line to the following curves at the indicated points.
2

3. 1.1 Equation of a tangent line and normal line to a curve c
⃝Francis Oketch
(a) 2e−x + ey = 3ex−y at point (0,0).
Solution
Clearly, the point (0, 0) lies on the given curve. Now, differentiating the given curve
implicitly with respect to x yields −2e−x + eyy′ = 3(1 − y′)ex−y. Therefore,
y′
=
2e−x + 3ex−y
ey + 3ex−y
The gradient of the tangent line at point (0, 0) is
dy
dx
=
2e−0 + 3e0
e0 + 3e0
=
2 + 3
1 + 3
=
5
4
= m1.
Thus, the equation of the tangent line at point (0, 0) is
y − 0
x − 0
=
5
4
⇒ y =
5
4
x
The gradient of the normal line at (0, 0) is m2 = −
1
m1
= −
4
5
. Thus, the equation of the
normal line at (0, 0) is
y − 0
x − 0
= −
4
5
⇒ y = −
4
5
x
(b) xy = 6e2x−3y at point (3,2). [hint: m1 = 10/21]
(c) x =
t2
1 + t
, y =
t3
1 − t
at t = 2. [hint: m1 = −9/2]
(d) y = a cos3 t, x = a sin3 t at t =
π
4
. [hint: m1 = −1]
3. The parametric equations of a curve are x = 3 (2θ − sin 2θ), y = 3 (1 − cos 2θ). The tangent
and normal to the curve at the point P where θ =
π
4
meet the y−axis at L and M, respectively.
Show that the area of triangle PLM is
9
4
(π − 2)2
.
Solution
dx
dθ
= 3 (2 − 2 cos 2θ) and
dy
dθ
= 6 sin 2θ. Therefore,
dy
dx
=
dy/dθ
dx/dθ
=
6 sin 2θ
3 (2 − 2 cos 2θ)
=
sin 2θ
1 − cos 2θ
.
When θ =
π
4
,
dy
dx
=
sin
(π
2
)
1 − cos
(π
2
) =
1
1 − 0
= 1. Hence,
Gradient of the tangent at P is 1. Now, the value of x when θ =
π
4
is x = 3
[π
2 − sin
(π
2
)]
=
3π − 6
2
. The value of y when θ =
π
4
is y = 3
[
1 − cos
(π
2
)]
= 3. Thus, the co-ordinate of
point P is
(
3π − 6
2
, 3
)
.
Equation of the tangent at P is
y − 3
x −
3π − 6
2
= 1 ⇒ y = x −
3π − 6
2
+ 3 ⇒ y = x +
12 − 3π
2
Thus, the y−intercept is y =
12 − 3π
2
. Hence, the co-ordinate of point L is
(
0,
12 − 3π
2
)
.
Gradient of the normal at P is −1. Thus, equation of the normal at P is
y − 3
x −
3π − 6
2
= −1 ⇒ y = −x +
3π − 6
2
+ 3 ⇒ y = −x +
3π
2
Thus, the y−intercept is y =
3π
2
. Hence, the co-ordinate of point M is
(
0,
3π
2
)
.
3

4. 1.2 Linear approximation/small changes c
⃝Francis Oketch
Height of triangle PLM is h =
3π − 6
2
and the
base length is LM =
3π
2
−
12 − 3π
2
= 3π − 6.
Thus, area of triangle PLM
=
1
2
(3π − 6)
(
3π − 6
2
)
=
9
4
(π − 2)2
Exercise:
1. Find the equation of tangent line and normal line to the following curves
(a) 12(x2 + y2) = 25xy at point (3,4)
(b) x2y = x + 2 at point (2,1)
(c) xy = ln
(
x
y
)
at point (1,3)
(d) xy = 6e2x−3y at point (3,2)
(e)
1
x + 1
+
1
y + 1
= 1 at point (1,1)
(f) x =
t2 + 1
t + 3t2
, y =
t3 − 7
t + 3t2
at t = 1
2. Find the equations of the tangent lines and normal lines to the curve y = x2 − 4x + 3 at the
points where the curve cuts the x-axis.
3. (a) Show that the equation of the tangent to x2 + xy + y = 0 at the point (x1, y1) is (2x1 +
y1)x + (x1 + 1)y + y1 = 0.
(b) Show that the equation of the tangent at (x1, y1) to the curve ax2 + by2 + cxy + dx = 0 is
ax1x + by1y +
1
2
c(y1x + x1y) +
1
2
d(x1 + x) = 0.
Lecture 10
1.2 Linear approximation/small changes
Linear approximation is a technique used to estimate values of some functions close to some known
results. The equation of tangent line at the point (x0, y0) can be used to approximate the function
y(x) close to this point. Now, the gradient of tangent line to the curve y(x) at point (x0, y0) is denoted
by
dy
dx (x0,y0)
= y′
(x0)
Thus, the equation of the tangent line at this point is
∆y
∆x
= gradient ⇒
y − y0
x − x0
= y′(x0).
Therefore, for x close to x0 and denoting y0 = y(x0), we have the following approximation of y(x):
.
.
y(x) ≈ y(x0) + y′
(x0)(x − x0).
→ Notes:
(1) This formula comes from the slope
dy
dx
≈
y − y0
x − x0
for x close to x0.
(2) This is also equivalent to taking
∆y
∆x
≈
dy
dx
, i.e. the differential as an approximation of the
increment.
4

5. 1.2 Linear approximation/small changes c
⃝Francis Oketch
Example(s):
(a) Use linear approximation to estimate (i)
√
26 and (ii) 4
√
80.
Solution
i) Note that
√
26 = 26
1
2 . Let y(x) = x
1
2 ⇒ y′(x) =
1
2
x−1
2 . Take x0 = 25 (a value close
to 26 and has exact square root), we have y(x0) = (25)
1
2 = 5 and y′(x0) =
1
2
(25)−1
2 =
1
10
.
By linear approximation, we have
y(x) ≈ y(x0) + y′
(x0)(x − x0) = 5 +
1
10
(x − 25)
Plugging in x = 26 yields
y(26) ≈ 5 +
(26 − 25)
10
= 5 +
1
10
= 5.1
Therefore,
√
26 ≈ 5.1.
ii) Note that 4
√
80 = 80
1
4 . Let y(x) = x
1
4 ⇒ y′(x) =
1
4
x−3
4 . Take x0 = 81 (a value close
to 80 and has exact fourth root), we have y(x0) = (81)
1
4 = 3 and y′(x0) =
1
4
(81)− 3
4 =
1
108
.
By linear approximation, we have
y(x) ≈ y(x0) + y′
(x0)(x − x0) = 3 +
1
108
(x − 81)
Plugging in x = 80 yields
y(80) ≈ 3 +
(80 − 81)
108
= 3 −
1
108
=
324 − 1
108
=
323
108
≈ 2.9907
Therefore, 4
√
80 ≈ 2.9907.
(b) Use differentials to approximate (i) cos(44o) and (ii) sin(60o1′).
Solution
i) Let y = cos x ⇒
dy
dx
= − sin x. Take x0 = 45o =
π
4
(a value close to 44o and its cosine
can be obtained without using SMP table or a calculator), we have y(x0) = cos(45o) =
1
√
2
and y′(x0) = − sin(45o) = −
1
√
2
. By linear approximation, we have
y(x) ≈ y(x0) + y′
(x0)(x − x0) =
1
√
2
−
1
√
2
(
x −
π
4
)
Plugging in x = 44o ≡
11π
45
radians yields
y
(
11π
45
)
≈
1
√
2
−
1
√
2
(
11π
45
−
π
4
)
= 0.7194
Therefore, sin(44o) ≈ 0.7194.
ii) Let y = sin x ⇒
dy
dx
= cos x. Take x0 = 60o =
π
3
(a value close to 60o1′ and its sine
can be obtained without using SMP table or a calculator), we have y(x0) = sin(60o) =
√
3
2
and y′(x0) = cos(60o) =
1
2
. By linear approximation, we have
y(x) ≈ y(x0) + y′
(x0)(x − x0) =
√
3
2
+
1
2
(
x −
π
3
)
5

6. 1.3 Related rates c
⃝Francis Oketch
Plugging in x = 60o1′ ≡
(
π
3
+ 0.0003
)
radians yields
y
(
π
3
+ 0.0003
)
≈
√
3
2
+
1
2
(
π
3
+ 0.0003 −
π
3
)
= 0.86618
Therefore, sin(60o1′) ≈ 0.86618.
Exercise:
(a) Estimate 5
√
30 by linear approximation. [ans: 5
√
30 ≈ 1.975]
(b) Find the approximate value of 80
3
4 using linear approximation. [ans: 80
3
4 ≈ 26.75]
(c) Find the cube root of 24 without using a calculator.
(d) Find the linearization of the function y =
√
x + 3 at x0 = 1 and use it to approximate the
numbers
√
3.98 and
√
4.05. Are these approximations overestimates or underestimates?
(e) Use linear approximation to estimate ln(1.1). You must make an appropriate choice of where
to center your approximation. Draw a picture illustrating your approximation and write and
explanation of why you chose to base your approximation where you did (In other words, explain
your choice of x0).
(f) Use linear approximation to estimate y(4.1) given that y(4) = 2 and
dy
dx
=
√
x2 + 20.
1.3 Related rates
If a variable x is a function of time t, then the time rate of change of x is given by
dx
dt
. When two
or more variables, all functions of t, are related by an equation, the relation between their rates of
change may be obtained by differentiating the equation with respect to t. For example, if z2 = x2 +y2
then differentiating the equation implicitly with respect to t yields 2z
dz
dt
= 2x
dx
dt
+ 2y
dy
dt
. To solve
related rates problems, follow these steps:
i) Assign symbols to all quantities given and their respective rates of change. Use a sketch where
necessary.
ii) Write an equation relating all the variables whose rates of change are given or are to be
determined.
iii) Apply chain rule of differentiation to differentiate implicitly both sides of the equation with
respect to time t
iv) Substitute into the resulting equation all the known variables and their rates of change and then
solve for the required rate of change.
Example(s):
1. Sand is being emptied from a hopper at a rate of 10 cubic feets per second. The sand forms
a conical pile whose height is always twice its radius. At what rate is the radius of the pile
increasing when its height is 5 feets.
Solution
Step I: Let V be the volume of the conical pile, h the height and r the radius. Given
dV
dt
= 10 ft3/s and h = 2r. We are required to find
dr
dt
when h = 5 ft.
6

7. 1.3 Related rates c
⃝Francis Oketch
Step II: At time t, the conical pile has the volume V =
1
3
πr2h. Putting h = 2r yields
V =
2
3
πr3.
Step III: Differentiating the equation implicitly with respect to t, we obtain
dV
dt
= 2πr2 dr
dt
⇒
dr
dt
=
1
2πr2
dV
dt
(∗)
Step IV: Substituting
dV
dt
= 10 and r =
h
2
=
5
2
into equation (∗) yields
dr
dt
=
1
2π(5
2 )2
(10) =
4
5π
≈ 0.254648 ft/s
Therefore, the radius of the conical pile is increasing at the rate of 4/(5π) feets per second
when its height is 5 feets.
2. Gas is escaping from a spherical balloon at the rate of 900 cm3/s. How fast is the surface area
shrinking when the radius is 360 cm.
Solution
Step I: Let V be the volume of the sphere, S the surface area and r the radius. Given
dV
dt
= −900 m3/min. We are required to find
dS
dt
when r = 360m.
Step II: At time t, the sphere has the volume V =
4
3
πr3 and surface area S = 4πr2.
Step III: Differentiating implicitly with respect to t, we obtain
dV
dt
= 4πr2 dr
dt
and
dS
dt
= 8πr
dr
dt
Thus,
dS
dt
= 8πr
(
1
4πr2
dV
dt
)
⇒
dS
dt
=
2
r
dV
dt
(∗).
Step IV: Substituting
dV
dt
= −900 and r = 360 into equation (∗) yields
dS
dt
=
2
360
(−900) = −5
Therefore, the surface area of the balloon is decreasing at the rate of 5cm2 per second when
its radius is 360 cm.
3. Car A is traveling west at 50km/h and car B is traveling north at 60km/h. Both are headed for
the intersection of the two roads. At what rate are the cars approaching each other when car A
is 0.3km and car B is 0.4km from the intersection?
Solution
Step I: Let x and y be the distance of car A and
B from the junction C at any time t, respectively.
Let z be the distance between car A and B at
any time t. Given
dx
dt
= −50km/h and
dy
dt
=
−60km/h. We are required to find
dz
dt
when x =
0.3km and y = 0.4km.
Step II: At time t the distance between A and B
is z2 = x2 + y2. (use Pythagoras theorem)
7

8. 1.3 Related rates c
⃝Francis Oketch
Step III: Differentiating the equation implicitly with respect to t, we obtain
2z
dz
dt
= 2x
dx
dt
+ 2y
dy
dt
⇒
dz
dt
=
1
z
(
x
dx
dt
+ y
dy
dt
)
(∗)
Step IV: When x = 0.3km and y = 0.4km, we have
z =
√
x2 + y2 =
√
(0.3)2 + (0.4)2 = 0.5km
Substituting x = 0.3km, y = 0.4km, z = 0.5km,
dx
dt
= −50km/h and
dy
dt
= −60km/h into
equation (∗) yields
dz
dt
=
1
0.5
[(0.3)(−50) + (0.4)(−60)] = −78. Therefore, the two cars are
approaching each other at a rate of 78km/h when car A is 0.3km and car B is 0.4km from
the intersection.
4. Water is running out of a conical funnel at the rate of 5cm3/s. If the radius of the base of the
funnel is 10cm and the altitude is 20cm, find the rate at which the water level is dropping when
it is 5cm from the top.
Solution
Step I: Let r be the radius, h the height of
the surface of the water, and V the volume of
water in the cone at time t. Given
dV
dt
= −5
cm3/min, r = 10cm and h = 20cm. We are
required to find
dh
dt
when h = 5cm from the
top.
Step II: The volume of the cone at time t is given by the equation V =
1
3
πr2h. But by
similar triangles, we have
r
10
=
h
20
⇒ r =
h
2
. Substituting into the above equation
yields
V =
1
3
π
(
h
2
)2
h ⇒ V =
π
12
h3
Step III: Differentiating the equation implicitly with respect to t, we obtain
dV
dt
=
π
4
h2 dh
dt
⇒
dh
dt
=
4
πh2
dV
dt
(∗)
Step IV: Substituting
dV
dt
= −5cm3/s and h = 20 − 5 = 15cm into equation (∗), we get
dh
dt
=
4
π(15)2
(−5) = −
4
45π
Therefore, the water level is dropping at the rate of
4
45π
cm/s when it is 5cm from the top.
Exercise:
1. Sands falling from a chute form a conical pile whose altitude is equal to
4
3
of the radius of the
base.
(a) How fast is the volume increasing when the radius of the base is 0.3m and is increasing at
the rate of 0.025m/s. [ans:
dV
dt
= 0.003π m3/s]
8

9. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
(b) How fast is the radius increasing when it is 0.6m and the volume is increasing at the rate
of 0.024m3/s. [ans:
dr
dt
=
1
20π
m/s]
2. Water is running out of a conical funnel at the rate of 1000m3/s. If the radius of the base of
the funnel is 40mm and the altitude is 80mm, find the rate at which the water level is dropping
when it is 20mm from the top. [ans:
dh
dt
= −
10
9π
m/s]
3. (a) Gas is escaping from a spherical balloon at the rate of 0.02 m3/s. How fast is the surface
area shrinking when the radius is 4 meters. [ans:
dS
dt
= −0.01 m2/s]
(b) A spherical balloon is blown up so that its volume increases at a constant rate of 2cm3/s.
Find the rate of increase of the radius when the volume of the balloon is 50cm3. [ans:
dr
dt
= 0.030472 cm/s]
4. Ink is dropped on to a blotting paper forming a circular stain which increases in area at the rate
of 5cm2/s. Find the rate of change of the radius when the area is 30cm2. [ans:
dr
dt
=
5
2
√
30π
cm/s]
5. The length of a thin rectangular metal sheet is twice its width. If the sheet is subjected to a
constant heating, find the rate of change of its perimeter when its width is 1m and its area is
changing at the rate of 18cm2/s. (Assume that the expansion is uniform and that the thickness
of the metal sheet is negligible) [ans:
dP
dt
= 27 m/s]
6. A machine produces two products and the number of each product is related by y =
10(20 + x)
50 − x
,
where x and y denote the number of each product produced. How fast is y changing if x is
changing at the rate of 2 units per hour when x = 40. [ans:
dy
dt
= 14 units per hour]
7. An air traffic controller spots two airplanes at the same altitude converging to a point as they
fly at right angles to each other. One airplane is 150 miles from the point and has a speed of 300
miles per hour. The other is 200 miles from the point and has a speed of 400 miles per hour. At
what rate is the distance z between the planes changing? [ans:
dz
dt
= −500 miles per hour]
8. A company is increasing the production of a product at the rate of 20 units per week. The demand
and cost functions for the product are given by p = 60 − 0.01x and C = 4000 + 40x − 0.03x2,
where x is the number of units produced per week. Find the rate of change of the profit with
respect to time (in dollars per week) when the weekly sales are x = 700 units. [ans:
dP
dt
= 960
dollars per week]
9. A ladder of length 5 metres rests against a vertical wall with the bottom end on horizontal
ground. If the foot of the ladder is 3m from the wall and moving away at 20m/s, find the rate
at which the top end is sliding down the wall. [hint: x2 + y2 = 25 at x = 3, y = 4,
dx
dt
= 20, ans:
dy
dt
= −15m/s]
10. Car A is traveling due West at 50km/h while car B is traveling due North at 60km/h. Both
cars are heading for an intersection of two roads. Determine the rate at which the cars are
approaching each other when car A is 0.3 km and car B is 0.4 km from the intersection. [hint:
z2 = x2 + y2 at x = 0.3, y = 0.4,
dx
dt
= 50,
dy
dt
= 60, ans:
dz
dt
= 78 km/h]
1.4 Turning points/Stationary points/Critical points/Extrema
The points at which the slope (gradient) of a curve is zero are called stationary points (or turning
points). For example, consider the following curve y = f(x):
9

10. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
A- a maximum turning point
B- a minimum turning point
C- a point of inflection
To classify the stationary values, consider the points A1 and A2, B1 and B2, C1 and C2 which are left
and right of A, B, and C, respectively, and close to them.
(1) First derivative test (or sign test)
This test relies on the sign of
dy
dx
just to the LHS and just to the RHS of the turning point.
Consider the behaviour of the gradient
dy
dx
at points A, B and C.
For A (a maximum point)
at point A1,
dy
dx
is positive (+ve)
at point A,
dy
dx
is zero (0)
at point A2,
dy
dx
is negative (-ve)
For B (a minimum point)
at point B1,
dy
dx
is negative (-ve)
at point B,
dy
dx
is zero (0)
at point B2,
dy
dx
is positive (+ve)
For C (a point of inflection)
at point C1,
dy
dx
is positive (+ve)
at point C,
dy
dx
is zero (0)
at point C2,
dy
dx
is positive (+ve)
Summary:
Maximum Minimum Inflection
Sign of
dy
dx
when + 0 − − 0 + + 0 + OR − 0 −
moving through a stationary value
OR
(2) Second derivative test: first compute
dy
dx
and
d2y
dx2
. Now, when passing through point A,
dy
dx
changes from +ve to -ve i.e.,
dy
dx
decreases as x increases
d2y
dx2
is negative (i.e.,
d2y
dx2
0)
Similarly, when passing through point B,
dy
dx
changes from -ve to +ve i.e.,
dy
dx
increases as x
increases
d2y
dx2
is positive (i.e.,
d2y
dx2
0)
Summary:
10

11. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
Maximum Minimum Inflection
Sign of
d2y
dx2
at negative positive zero
a turning point i.e.,
(
d2y
dx2
0
)
i.e.,
(
d2y
dx2
0
)
i.e.,
(
d2y
dx2
= 0
)
In summary, the steps for finding critical points are?
i) Find the first derivative
ii) Equate the first derivative to zero
iii) Solve the equation to get the turning points
iv) Use the second derivative test to check whether the points you found are maxima or minima or
points of inflection.
Example(s):
1. Find the stationary points on the following curves and classify them.
(a) y = x4 + 4x3 − 6.
Solution
Differentiating the given curve, we obtain
dy
dx
= 4x3
+ 12x2
and
d2y
dx2
= 12x2
+ 24x
At a stationary point,
dy
dx
= 0, i.e., 4x3 + 12x2 = 0. Solving yields x = 0 or x = −3. The
value of y at x = 0 is y = (0)4 + 4(0)3 − 6 = −6. Similarly, the value of y at x = −3 is
y = (−3)4 + 4(−3)3 − 6 = −33. So the turning points are (0, −6) and (−3, −33). Using the
second derivative test, we classify the points as follows:
When x = −3, we have
d2y
dx2
= 12(−3)2 + 24(−3) = 36 0. Therefore, the point
(−3, −33) is a minimum point.
When x = 0, we have
d2y
dx2
= 12(0)2 +24(0) = 0. Therefore, the point (0, −6) is a point
of inflection.
(b) y = x2(x + 1).
Solution
Differentiating the given curve, we obtain
dy
dx
= 3x2
+ 2x and
d2y
dx2
= 6x + 2
At a stationary point, 3x2+2x = 0. Solving yields x = 0 or x = −
2
3
. The value of y at x = 0
is y = (0)2(0 + 1) = 0. Also, The value of y at x = −
2
3
is y =
(
−
2
3
)2 (
−
2
3
+ 1
)
=
4
27
. So
the turning points are (0, 0) and
(
−
2
3
,
4
27
)
. Using the second derivative test, we classify
the points as follows:
When x = 0, we have
d2y
dx2
= 6(0) + 2 = 2 0. Therefore, (0, 0) is a minimum point.
When x = −
2
3
, we have
d2y
dx2
= 6
(
−
2
3
)
+ 2 = −2 0. Therefore,
(
−
2
3
,
4
27
)
is a
maximum point.
11

12. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
Exercise:
1. Find the maximum and minimum values of the function y = 2 sin θ + cos 2θ. [ans: max point
(
π
6
,
3
2
), min point (
π
2
, 1)]
2. Find the turning points and point of inflection on the curve y = x5 − 5x4 + 5x3 − 1. [ans: max
point (1, 0), min point (3, −28), point of inflection (0,-1)]
3. Find and classify all the points on the curve y = 3x4 − 8x3 − 24x2 + 96x at which the tangent
to the curve is parallel to the x−axis.
4. Find and classify the stationary points of the following functions.
i) y = 3x5 + 6x4 − 4x3 + 1.
ii) f(x) = 2x3 − 3x2 + 1.
iii) f(x) =
1
4
x4 −
1
3
x3 − 6x2. [ans: stationary points at x = −3, 0, 4]
5. Show that the curve y = a sec θ − b tan θ has a minimum value if
a2
√
a2 − b2
0.
6. Find the inflection points in the interval [0, 2π] of the function f(x) = 2 sin x − sin x cos x. [ans:
(0, 0), (π
6 , 0.57), (π, 0), (11π
6 , −0.57)]
Areas where the concept of maxima and minima is applied
1.4.1 Optimization
Example(s):
1. A box with a square base and an open top is to have volume 62.5 in3. Neglect the thickness of
the material used to make the box, and find the dimensions that will minimize the amount of
material used.
Solution
Let the base width be x in. Thus, the height is
62.5
x2
in. The surface area is given by
A = x2
+ 4x
(
62.5
x2
)
= x2
+
250
x
Thus,
dA
dx
= 2x −
250
x2
and
d2A
dx2
= 2 +
500
x3
.
For maximum or minimum area,
dA
dx
= 0. That is, 2x −
250
x2
= 0. Solving yields x = 5. At
x = 5,
d2A
dx2
= 2 +
500
(5)3
= 6 0. Therefore, the box has a minimum area when its base is 5 in
by 5 in and height is
62.5
(5)2
= 2.5 in.
2. Find the height of the right circular cylinder of greatest volume which can be cut from a sphere
of radius a.
Solution
12

13. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
Let the radius of the cylinder be r and its height
be h. From triangle OBX, Pythagoras theorem
yields
r2
+
h2
4
= a2
Thus, r2 = a2 −
h2
4
. The volume of the cylinder
is given by
V = πr2
h = π
(
a2
−
h2
4
)
h = π
(
a2
h −
h3
4
)
Thus,
dV
dh
= π
(
a2 −
3
4
h2
)
and
d2V
dh2
= −
3
2
πh. For maximum or minimum volume,
dV
dh
= 0.
That is, π
(
a2 −
3
4
h2
)
= 0. Solving yields h =
2a
√
3
. At h =
2a
√
3
,
d2V
dh2
= −
3πa
√
3
0. Therefore,
the cylinder has a maximum volume when its height is
2a
√
3
.
3. Find the values of x and y that will maximize the function f(x, y) = xy subject to the constraint
4x + 2y = 40.
Solution
Given 4x + 2y = 40, we have y = 20 − 2x. Substituting into f(x, y) = xy, we get
f(x) = x(20 − 2x) = 20x − 2x2
Differentiating with respect to x, we have f′(x) = 20 − 4x. At maxima or minima, f′(x) = 0,
i.e., 20 − 4x = 0 ⇒ x = 5. Substituting into the given constraint, we get y = 10. To classify
the optimal value, we use second derivative test. Now, f′′(x) = −4 0. Therefore, x = 5 and
y = 10 will maximize xy.
Exercise:
1. Find the height of the right circular cone of maximum volume, given that the sum of the height
and radius of the base is 0.12m [ans: h = 0.04m]
2. Find the dimensions of the rectangle of greatest area which can be inscribed in a circle of radius
r. [ans: a square of side
√
2r]
3. (a) A manufacturer wants to design an open box having square base and surface area of 108
square meters. Find the dimensions of the box that will give maximum volume. [hint:
x2 + 4xh = 108, V = 27x −
x3
4
, ans: length x= 6m and height h= 3m]
(b) An open box is to be made from a twelve-inch by twelve-inch square piece of material by
cutting equal squares from the corners and turning up the sides. Find the volume of the
largest box that can be made. [ans: V = 128in3]
4. ABCD is a square ploughed field of side 132m, with a path along its perimeter. A man can walk
at 8 km/h along the path, but only at 5 km/h across the field. He starts from A along AB,
leaves AB at a point P, and walks straight from P to C. Find the distance of P from A, if the
time taken is the least possible.
5. (a) Choose x and y to maximize xy subject to the constraint 3x + y = 60. (Note: you do not
need to confirm that your solution is a maximum).
(b) Choose x and y to maximize xy2 subject to the constraint x + y = 200. [ans: x = 200/3
and y = 400/3]
13

14. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
6. Coughing forces the trachea (windpipe) to contract, which in turn affects the velocity of the air
through the trachea. The velocity of the air during coughing can be modeled by
v = k(R − r)r2
, 0 ≤ r R
where k is a positive constant, R is the normal radius of the trachea, and r is the radius during
coughing. What radius r will produce the maximum air velocity? [ans: r =
2
3
R]
7. A rectangular solid with a square base has a surface area of 121.5 square centimeters. (Let w
represent the width of the sides of the square base and let h represent the height of the solid.)
(a) Determine the dimensions (in cm) that yield the maximum volume. [ans: w = 4.5cm and
h = 4.5cm]
(b) Find the maximum volume (in cm3). [ans: V = 91.125cm3]
8. A rectangular solid with a square base has a volume of 1,728 cubic inches. (Let w represent the
length of the sides of the square base and let h represent the height of the solid.)
(a) Determine the dimensions (in inches) that yield the minimum surface area. [ans: w = 12in
and h = 12in]
(b) Find the minimum surface area (in in.2). [ans: S = 864in2]
9. A campground owner plans to enclose a rectangular field adjacent to a river. The owner wants the
field to contain 125,000 square meters. No fencing is required along the river. What dimensions
(in meters) will use the least amount of fencing? [ans: l = 500m and w = 250m]
10. A storage box with a square base must have a volume of 128 cubic centimeters. The top and
bottom cost $0.40 per square centimeter and the sides cost $0.20 per square centimeter. Find
the dimensions that will minimize cost. (Let x represent the length of the sides of the square
base and let y represent the height.) [ans: x = 4cm and y = 8cm]
1.4.2 Economics: cost, revenue and profit
The cost function, C(x), is the total cost incurred in producing x units of a commodity. It is
given by the sum of the fixed costs and variable cost.
→ Note: The fixed cost is determined when zero units are produced.
Marginal cost (MC) is the rate of change of the total cost function with respect to the number
of units produced, i.e., MC =
dC
dx
. It represents the extra cost incurred in producing one extra
unit of a commodity when the level of production is already at x.
Average cost (AC or C(x)) is given by C(x) =
C(x)
x
.
Revenue function, R(x), is the (total) revenue received when x units of a given commodity are
produced and sold at a unit price p(x) (or the demand function). Thus, R(x) = x · p(x), where
x is the number of units produced and sold. Note that the supply function is given by x = f(p).
Marginal revenue, MR =
dR
dx
.
Average revenue (AR or R(x)) is given by R(x) =
R(x)
x
.
Profit function, P(x) or Π(x), is given by total revenue minus total cost, i.e., P(x) = R(x)−C(x).
Marginal profit, MP =
dP
dx
.
Average profit (AP or P(x)) is given by P(x) =
P(x)
x
.
14

15. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
At maxima or minima, we have MC =
dC
dx
= 0, or MR =
dR
dx
= 0 or MP =
dP
dx
= 0. The values of
x which satisfy these equations are the critical points (or the optimizing level of production).
→ Note:
Break-even point refers to the point in which total cost incurred and total revenue earned are
equal, i.e., the value(s) of x for which C(x) = R(x).
Point of diminishing returns refers to the point (x, P(x)) in which
d2P
dx2
= 0.
Example(s):
1. In marketing a certain commodity, a business has discovered that the demand for the commodity
is represented by p(x) =
50
√
x
. The cost of producing x units of the commodity is given by
C(x) = 0.5x + 500. Find the price per unit that will yield maximum profit. (Note: p is in
dollars)
Solution
The profit function is given by
P(x) = xp(x) − C(x) = x
50
√
x
− (0.5x + 500) = 50
√
x − 0.5x − 500
The marginal profit (MP) is given by MP =
dP
dx
=
25
√
x
− 0.5. At maxima or mimima,
dP
dx
= 0.
That is,
25
√
x
− 0.5 = 0 ⇒ x = 2500
We need to test if this value of x will lead to maximum profit. Now,
d2P
dx2
= −12.5x−3/2. When
x = 2500, P′′(x) = −12.5(2500)−3/2 = −0.0001 0. Hence, the business will realize maximum
profit if 2500 units of the commodity are produced. The optimal price per unit (demand) is
p(2500) =
50
√
2500
=
50
50
= 1 dollar.
2. A certain company faces market demand given by p = 48 − 3x. This company has total cost
given by C(x) = 2x2 − 12x + 100. Find:
(a) Price when revenue is maximized. [hint: R(x) = xp = 48x − 3x2, ans: p = 24]
(b) Revenue when cost is maximized. [ans: R(3) = 117]
(c) Maximum possible profit. [hint: P(x) = −5x2 + 60x − 100, ans: P(6) = 80]
(d) How many items should be produced and sold in order to break-even. [hint: R(x) = C(x),
ans: x = 2 and x = 10]
3. A monopolist faces the demand function p = 200 − x. The total cost is C = 100 − 40x + 5x2.
(a) Write down the monopolist’s profit as a function of the quantity produced x.
(b) Find the profit-maximizing level of production and confirm that your solution is a
maximum.
(c) How does the profit-maximizing level of production change if the government imposes a
lump sum tax of L dollars per each unit sold?
Solution
15

16. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
(a) Given x = 200 − p, we have p = 200 − x. The profit function is given by
P(x) = xp(x) − C(x) = x(200 − x) −
(
100 − 40x + 5x2
)
= −6x2
+ 240x − 100
(b) The marginal profit (MP) is given by MP =
dP
dx
= −12x + 240. At maxima or minima,
dP
dx
= 0. That is,
−12x + 240 = 0 ⇒ x = 20
We need to test if this value of x will lead to maximum profit. Now,
d2P
dx2
= −12 0.
Hence, the business will realize maximum profit if 20 units of the commodity are produced.
(c) Since the government imposes a lump tax of L per each unit sold, the total tax bill is given
by
T = Lx
The profit function, given a lump sum tax L, is given by P = −6x2 + 240x − 100 − T, i.e.,
P = −6x2
+ 240x − 100 − Lx
At maximum or minimum,
dP
dx
= 0. That is, −12x + 240 − L = 0.
⇒ x = 20 −
L
12
This is the optimal level of production. The government raises a total tax bill of
T = Lx = L
(
20 −
L
12
)
⇒ T = 20L −
L2
12
Differentiating with respect to L, we have
dT
dL
= 20 −
L
6
. To maximize the government’s
tax revenue, then
dT
dL
= 0. That is,
20 −
L
6
= 0 ⇒ L = 120
Thus, L = 120 maximizes the government’s tax revenue since
d2T
dL2
= −
1
6
0. Therefore,
the new profit-maximizing level of production is x = 20 −
L
12
= 20 −
120
12
= 10 units.
Exercise:
1. For a production level of x units of a commodity, the cost function is C(x) = 100 + 30x and the
demand function is p(x) = 90 − x. What price p will maximize profit?
Solution
The profit function is given by
P(x) = xp(x) − C(x) = x(90 − x) − (100 + 30x) = 90x − x2
− 100 − 30x = 60x − x2
− 100
The marginal profit (MP) is given by MP =
dP
dx
= 60 − 2x. At maxima or minima,
dP
dx
= 0.
That is,
60 − 2x = 0 ⇒ x = 30
Now,
d2P
dx2
= −2. When x = 30, P′′(x) = −2 0. Hence, the profit-maximizing level of
production x = 30 units. Therefore, the optimal unit price is p(30) = 90 − 30 = 60 dollars.
16

17. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
2. A monopolist faces the demand function given by x = 10 − 0.5p, where x is the number of units
produced and sold at price p dollars per unit. The total cost incurred consists of a fixed overhead
of 28 dollars plus production cost of 2 dollars per unit of x. [hint: C(x) = 2x + 28]
(a) Write down the monopolist’s profit as a function of the quantity produced x.
(b) Find the profit-maximizing level of production and confirm that your solution is a
maximum.
(c) Find the break-even points. What is the slope of the profit function at each of the break-
even points?
(d) Assume now that the government imposes a fixed tax of t dollars per each unit sold. What
t maximises the government’s tax revenue?
3. A manufacturer estimates that if x units of a particular commodity are produced, the total cost
will be C(x) dollars, where C(x) = x3 − 24x2 + 350x + 338.
(a) At what level of production will the marginal cost be minimized?
(b) At what level of production will the average cost be minimized?
4. A manufacturer estimates that if q units of a particular commodity are produced, the total
cost incurred will be C(q) dollars and the total revenue earned will be R(q) dollars, where
C(q) = q3 − 6q2 + 140q + 750 and R(q) = 1400q − 7.5q2.
(a) What will be the maximum profit? [ans: Π(20) = 15850 dollars]
(b) What will be the quantity demanded when the market price is set at 275 dollars per unit
of the commodity? [hint: demand p =
R(q)
q
= 1400 − 7.5q, ans: q = 150 units]
5. A manufacturer can produce digital recorders at a cost of 70 dollars each. It is estimated that if
the recorders are sold for p dollars each, consumers will buy x = 120 − p recorders each month.
(a) Express the manufacturer’s profit P as a function of x. [ans: P(x) = 50x − x2]
(b) What is the average rate of change in profit obtained as the level of production increases
from x = 0 to x = 20 recorders. [ans: =
P(20) − P(0)
20 − 0
= 30]
(c) At what rate is profit changing when x = 20 recorders are produced? Is the profit increasing
or decreasing at this level of production? [ans: P′(20) = 10, increasing]
6. Consider the two functions f(x) and g(x) where 0 ≤ x ≤ 12 represents the number of items
(in thousands) sold by a certain small business and both f and g are in thousands of dollars.
The graphs of f (curve in red) and g (line in blue) are given below. One of these two functions
represents the total costs, and the other the total revenue. All we know so far is that the fixed
costs are nonzero.
17

18. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
(a) How much is the fixed costs? How much is each item being sold for? Explain.
(b) Approximate f′(4) and interpret its meaning. [hint: f′(4) = f(5) − f(4)]
(c) How many items (approximately) should the company produce and sell in order to break-
even? Explain briefly.
(d) Using the graph, approximate the maximum profit. Briefly explain.
(e) Copy the functions and sketch a rough graph of the profit function.
7. Student Government at a University is chartering a plane for Spring Break. The plane can seat
150 passengers. The airline will charge $120 per passenger and added to this a surcharge of $15
per passenger for each unsold seat. Let x represent the number of unsold seats.
(a) Find the revenue function R(x). [ans: R(x) = (150 − x)(120 + 15x)]
(b) How many seats should be unused to maximize the airline’s revenue.
(c) What price would each passenger pay if the airline maximized its revenue?
(d) Is this a good deal for the Student Government?
8. You have a business that sells phone chargers. You know that the cost of producing x units
is C = 200 + 40x. In order to sell x units, you also know that you need to set the price at
p = 90 − 2x dollars. Find the price for which the profit is maximized. [ans: p = 65 dollars]
9. The weekly demand function for x units of a product sold by a firm is p = 400 −
1
2
x dollars, and
the average cost of production is AC = 100 + 2x dollars.
(a) Find the quantity that will maximize profit. [ans: x = 30]
(b) Find the selling price at this optimal quantity. [ans: p = 370 dollars per unit]
(c) What is the maximum profit? [ans: P(30) = 9000 dollars]
10. When a wholesaler sold a product at $40 per unit, sales were 180 units per week. After a price
increase of $5 per unit, however, the average number of units sold dropped to 165 per week.
Assuming that the demand function is linear, what price per unit will yield a maximum total
revenue? [hint: demand function p = −
1
3
x + 100, ans: p(150) = 50]
1.4.3 Kinematics
The motion of a particle P along a straight line is completely described by the equation S = f(t),
where t 0 is time and S is the distance of P from a fixed point O in its path. The velocity of P at
time t is V =
dS
dt
.
If V 0, P is moving in the direction of increasing S.
If V 0, P is moving in the direction of decreasing S.
If V = 0, P is instantaneously at rest.
The acceleration of P at time t is a =
dV
dt
=
d2S
dt2
.
If a 0, V is increasing.
If a 0, V is decreasing.
If V and a have the same sign, the speed of P is increasing.
If V and a have the opposite signs, the speed of P is decreasing.
Example(s):
18

19. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
1. A body moves along a straight line according to the law S = t3 − 6t2 + 9t + 4. Find
(a) S and a when V = 0
(b) S and V when a = 0
(c) when is S increasing?
Solution
(a)
dS
dt
= V = 3t2 − 12t + 9 = 3(t − 1)(t − 3) and a =
dV
dt
= 6t − 12 = 6(t − 2). When V = 0,
t = 1 or t = 3.
When t = 1, S = (1)3 − 6(1)2 + 9(1) + 4 = 8 and a = 6(1 − 2) = −6.
When t = 3, S = (3)3 − 6(3)2 + 9(3) + 4 = 4 and a = 6(3 − 2) = 6.
(b) When a = 0, we have 6(t−2) = 0 ⇒ t = 2. When t = 2, S = (2)3 −6(2)2 +9(2)+4 = 6
and V = 3(2 − 1)(2 − 3) = −3
(c) S is increasing when V 0 i.e., when t 1 and t 3.
2. A body moves in a straight line so that the distance moved S metres is given in terms of the
time t seconds by S = t3 − t2. Find an expression for the acceleration of the body at time t and
find the times at which the body is at rest.
Solution
dS
dt
= V = 3t2 − 2t and
d2S
dt2
=
dV
dt
= a = 6t − 2. The body is at rest when V = 0, i.e.,
3t2 − 2t = 0 ⇒ t = 0 or t =
2
3
seconds.
3. The distance S moved in a straight line by a particle in time t is given by S = bt2 +ct+d, where
b, c and d are constants. If V is the velocity of the particle at time t, show that 4b(S−d) = V 2−c2.
Proof. V =
dS
dt
= 2bt + c. Now,
4b(S − d) = 4b(bt2
+ ct + d − d) = 4b2
t2
+ 4bct) = (2bt + c)2
− b2
= V 2
− c2
Exercise:
1. The displacement S at time t of a moving particle is given by S = b sin 2t + c cos 2t, where b and
c are constants. If V is the speed at time t, prove that V = 2
√
b2 + c2 − S2.
2. A particle moves in a horizontal line according to the law S = t4 − 6t3 + 12t2 − 10t + 3. Find
(a) the velocity and acceleration
(b) the time when the particle is at rest
3. If the velocity of a body varies inversely as the square root of the distance, prove that the
acceleration varies as the fourth power of the velocity.
4. A body moves in a straight line so that its distance S meters from a fixed point O at time t
seconds is given by S = (t − 2)2(2t − 7). Find the time when the body passes through point O
and the acceleration each time it passes through point O.
5. The velocity V m/s of a particle which has traveled a distance S metres from a fixed point is
given by V 2 = 16S. Find the acceleration of the particle.
19

20. 1.4 Turning points/Stationary points/Critical points/Extrema c
⃝Francis Oketch
6. A ball is thrown vertically upwards so that its height S metres after t seconds is given by
S =
1
27
t2 + 4
√
t. Find its:
i) velocity at any time t.
ii) acceleration when t = 1.
iii) maximum height reached.
7. The distance traveled by a projectile at any time t seconds is given by S = 4t3 + 3t + 2 feet.
Find the velocity and acceleration after 1 second. [ans: velocity = 15 ft/s, acceleration = 24
ft/s2]
20

21. c
⃝Francis Oketch
Lecture 11
2 Introduction to integration
Definition 2.1 (Integration). Integration is the reverse process of differentiation.
Suppose
dy
dx
= f(x). To obtain y, we integrate the function f(x) with respect to the independent
variable x. This is put in notation form as
.
.
y =
∫
f(x)dx + C ,
where C is a constant of integration and f(x) is the integrand. For example,
If
dy
dx
= cos x, then y =
∫
cos xdx = sin x + C.
The following are some important results of integration, based on the definition of integration as the
reverse process of differentiation.
(1)
∫
1dx = x + C (2)
∫
xn
dx =
xn+1
n + 1
+ C, for n ̸= −1
(3)
∫
1
x
dx = ln |x| + C (4)
∫
ex
dx = ex
+ C
(5)
∫
cos(kx)dx =
sin(kx)
k
+ C, for k ̸= 0 (6)
∫
sin(kx)dx = −
cos(kx)
k
+ C, for k ̸= 0
(7)
∫
sec2
xdx = tan x + C (8)
∫
cosec2
xdx = − cot x + C
(9)
∫
sec x tan xdx = sec x + C (10)
∫
cosec x cot xdx = −cosec x + C
These types of integrals are called indefinite since they lack limits of integration.
2.1 Techniques of integration
2.1.1 Power rule of integration
.
.
∫
xn
dx =
xn+1
n + 1
+ C, for n ̸= −1 ,
where C is a constant of integration.
Example(s):
(a)
∫
x3
dx =
x3+1
3 + 1
+ C =
x4
4
+ C.
(b)
∫
x−7
dx =
x−7+1
−7 + 1
+ C = −
x−6
6
+ C.
(c)
∫
√
xdx =
∫
x1/2
dx =
x
1
2
+1
1
2 + 1
+ C =
2
3
x3/2
+ C.
(d)
∫
x(1 − 3x)dx =
∫
(x − 3x2
)dx =
x2
2
− 3 ·
x3
3
=
x2
2
− x3
+ C.
(e)
∫
(2x − 3)2
dx =
∫
(4x2
− 12x + 9)dx =
4
3
x3
− 6x2
+ 9x + C.
(f)
∫ √
x + 3
√
x + 6
x2
dx =
∫ (
x1/2
x2
+
x1/3
x2
+
6
x2
)
dx =
∫ (
x− 3
2 + x−5
3 + 6x−2
)
dx
= −2x−1
2 −
3
2
x− 2
3 − 6x−1
+ C
21

22. 2.1 Techniques of integration c
⃝Francis Oketch
(g)
∫
x3
x + 1
dx.
Solution
Since the degree of the polynomial in the numerator is greater than that in the denominator,
long division yields
x2 − x + 1
x + 1
)
x3
− x3 − x2
− x2
x2 + x
x
− x − 1
− 1
Therefore,
∫
x3
x + 1
dx =
∫ (
x2
− x + 1 −
1
x + 1
)
dx =
x3
3
−
x2
2
+ x − ln |x + 1| + C
2.1.2 The u-substitution method
This technique requires that a new variable, say u, be introduced in the integrand to reduce the
problem to a form in which the power rule of integration can be applied.
Example(s):
(a) Evaluate
∫
(2x + 1)1/3
dx.
Solution
Let u = 2x + 1. Differentiating with respect to x yields
du
dx
= 2 ⇒ dx =
du
2
. Substituting
into the given integral, we get
∫
(2x + 1)1/3
dx =
∫
u1/3 du
2
=
1
2
∫
u1/3
du =
1
2
[
3
4
u4/3
]
+ C =
3
8
(2x + 1)4/3
+ C
(b) Evaluate
∫
x
(1 − x2)3
dx.
Solution
Let u = 1 − x2. Differentiating with respect to x yields
du
dx
= −2x ⇒ dx = −
du
2x
.
Substituting into the given integral, we get
∫
x
(1 − x2)3
dx =
∫
x
u3
(
−
du
2x
)
= −
1
2
∫
u−3
du = −
1
2
[
1
−2
u−2
]
+ C =
1
4
(1 − x2
)−2
+ C
(c) Evaluate
∫
5x(1 + 3x)5
dx.
Solution
Let u = 1 + 3x. Differentiating with respect to x yields
du
dx
= 3. Therefore, dx =
du
3
and
x =
1
3
(u − 1). Substituting into the given integral, we get
∫
5x(1 + 3x)3
dx =
∫
5
3
(u − 1) u5
(
du
3
)
=
5
9
∫ (
u6
− u5
)
du =
5
9
[
1
7
u7
−
1
6
u6
]
+ C
=
5
9
[
1
7
(1 + 3x)7
−
1
6
(1 + 3x)6
]
+ C
22

23. 2.1 Techniques of integration c
⃝Francis Oketch
(d) Evaluate
∫
sec2
(5x + 1)dx.
Solution
Let u = 5x + 1. Differentiating with respect to x yields
du
dx
= 5 ⇒ dx =
du
5
. Substituting
into the given integral, we get
∫
sec2
(5x + 1)dx =
∫
sec2
(u)
(
du
5
)
=
1
5
∫
sec2
(u)du =
1
5
tan u + C =
1
5
tan(5x + 1) + C
Exercise:
1. Evaluate the following integrals
(a)
∫
(7x − 2)3
dx. [hint: put u = 7x − 2, ans: =
1
28
(7x − 2)4 + C]
(b)
∫
3x
√
1 + 2x2dx. [hint: put u = 1 + 2x2, ans: =
1
2
(1 + 2x2)3/2 + C]
(c)
∫
1
(x + 1)2
dx. [hint: put u = x + 1, ans: =
−1
x + 1
+ C]
(d)
∫
(3x + 5)−2
dx. [hint: put u = 3x + 5, ans: = −
1
3
(3x + 5)−1 + C]
(e)
∫
cosec2
(
x − 1
3
)
dx. [hint: put u =
x − 1
3
, ans: = −3 cot
(
x−1
3
)
+ C]
(f)
∫
(ln x)2
x
dx. [hint: put u = ln x, ans: =
1
3
(ln x)3
+ C]
(g)
∫
x(3 − 5x2
)4
dx. [hint: put u = 3 − 5x2, ans: = −
1
50
(3 − 5x2)5 + C]
(h)
∫
x + 1
5
√
x2 + 2x + 7
dx. [hint: put u = x2 + 2x + 7, ans: =
5
8
(x2 + 2x + 7)4/5 + C]
(i)
∫
x(2 + 3x)4
dx. [hint: put u = 2 + 3x, ans: =
1
3
(
(2 + 3x)6
18
−
2(2 + 3x)5
15
)
+ C]
(j)
∫
x2
(1 + 4x3
)3
dx. [hint: put u = 1 + 4x3, ans: =
1
48
(
1 + 4x3
)4
+ C]
(k)
∫
1
√
x (1 +
√
x)
4 dx. [hint: put u = 1 +
√
x, ans: = −
2
3
(1 +
√
x)
−3
+ C]
2. If the marginal revenue function for output x is given by MR =
6
(x + 2)2
+ 5, find the total
revenue function and the demand equation. [hint: R(0) = 0, ans: revenue R(x) = −
6
x + 2
+ 5x,
demand p =
R(x)
x
]
23