CSUN MATH 150A: Practice Problem Set on the Cumulative
Effect of a Function (Integration)
Section I: Conceptual Understanding and Riemann Sums
This section focuses on approximating the cumulative effect using geometry. Show
your work, including the calculation for
Δ x
and the summation formula.
Problem 1: Approximating Area with
Rn
and
Ln
Consider the function
f(x)=x2+1
on the interval
[0,6]
.
1. Calculate
Δ x
for
n=3
subintervals.
2. Estimate the area under the curve using the Right Endpoint Riemann Sum
(
R3
).
3. Estimate the area under the curve using the Left Endpoint Riemann Sum (
L3
).
4. Is
R3
an overestimate or an underestimate? Justify your answer based on
the functions behavior (increasing/decreasing).
Problem 2: Midpoint Riemann Sum and Summation Notation
Consider the function
g(x)=4x − x2
on the interval
[1,5]
.
1. Calculate
Δ x
for
n=4
subintervals.
2. Determine the midpoint
xi
∗
for each of the four subintervals.
3. Estimate the area under the curve using the Midpoint Riemann Sum (
M4
).
4. Write the expression for the general Right Endpoint Riemann Sum (
Rn
) for
g(x)
over
[1,5]
using sigma notation.
Problem 3: Interpreting the Definite Integral
A particles velocity, in meters per second, is modeled by
v(t)=3t2−6t
.
1. Write a definite integral expression that represents the net change in
position (displacement) of the particle from
t=0
to
t=4
seconds.
2. Write a definite integral expression that represents the total distance
traveled by the particle from
t=0
to
t=4
seconds.
3. Explain the difference between the physical meaning of
∫
1
3
v(t)dt
and
∫
1
3
¿v(t)∨dt
.
Section II: Fundamental Theorem of Calculus Part II (Computational)
Use the FTC Part II (
∫
a
b
f(x)dx=F(b)− F (a)
) to evaluate the following definite
integrals.
Problem 4: Polynomial and Power Rule
Evaluate:
∫
1
3
(¿6x2−4x+5)dx ¿
Problem 5: Rational and Logarithmic Functions
Evaluate:
∫
e
e2
(
1
x−1
x2
)
dx
Problem 6: Trigonometric Functions
Evaluate:
∫
0
π/4
(¿3 sin(x)+sec2(x))dx ¿
Problem 7: Piecewise Functions
Evaluate the definite integral for the piecewise function
f(x)
:
f(x)=
{
4x3if x<1
5− x if x ≥ 1∫
0
2
f(x)dx
Problem 8: Exponential Functions
Evaluate:
∫
−1
1
(¿e2x+2)dx ¿
Section III: Fundamental Theorem of Calculus Part I (Theoretical)
This section requires applying the FTC Part I (
d
dx ∫
a
x
f(t)dt=f(x)
), often combined
with the Chain Rule.
Problem 9: Simple Application
Find the derivative
dy
dx
of the function
y
:
y=∫
3
x
√
t3+5t+1dt
Problem 10: Chain Rule with the Upper Limit
Find the derivative
dy
dx
of the function
y
:
y=∫
0
x2
cos(t3)dt
Problem 11: Chain Rule with the Lower Limit
Find the derivative
dy
dx
of the function
y
. (Hint: Use the order of integration
property.)
y=∫
tan(x)
5
ln(t)dt
Problem 12: Chain Rule with Both Limits
Find the derivative
dy
dx
of the function
y
. (Hint: Use the additivity property to split
the integral.)
y=∫
√
x
x3
et2dt
Section IV: Application and Net Change Theorem
Apply the FTC and the concept of accumulation to solve real-world problems.
Problem 13: Total Production
The rate of production of a factory (in units per hour) is given by
P(t)=15 t2+10 t
,
where
t
is the number of hours since the factory started its shift.
1. Find the total number of units produced during the first 4 hours of the
shift.
2. Find the average rate of production over the first 4 hours. (Hint: Average
rate =
1
b −a ∫
a
b
P(t)dt
).
Problem 14: Population Change
The rate of change of a towns population, in people per year, is modeled by
r(t)=100
√
t+50
, where
t
is the time in years since the year 2000 (
t=0
). The
population in 2000 was 5,000 people.
1. Write the integral expression for the net change in population between
2004 (
t=4
) and 2009 (
t=9
).
2. Calculate the net change in population over this period.
3. Determine the population of the town in the year 2009. (Hint: Final
Population = Initial Population + Net Change).
Problem 15: Total Distance vs. Displacement
A particle moves along a line with velocity
v(t)=t2−3t+2
(in
cm/sec
).
1. Calculate the displacement (net change in position) from
t=0
to
t=3
.
2. Calculate the total distance traveled from
t=0
to
t=3
. (Hint:
∫
a
b
¿v(t)∨dt
).
Answer Key and Detailed Solutions
Section I: Conceptual Understanding and Riemann Sums
Problem 1: Approximating Area with
Rn
and
Ln
Δ x=b − a
n=6−0
3=2
Subintervals:
[0,2],[2,4],[4,6]
.
1. Right Endpoint Sum (
R3
): Sample points are
x1
∗=2, x2
∗=4, x3
∗=6
.
R3=Δ x[f(2)+f(4)+f(6)]f(2)=22+1=5f(4)=42+1=17 f(6)=62+1=37 R3=2[5+17+37 ]=2[59]=118 units2
2. Left Endpoint Sum (
L3
): Sample points are
x1
∗=0, x2
∗=2, x3
∗=4
.
L3=Δ x[f(0)+f(2)+f(4)]f(0)=02+1=1f(2)=5f(4)=17 L3=2[1+5+17 ]=2[23]=46 units2
3. Overestimate/Underestimate: The function
f(x)=x2+1
is always
increasing on
[0,6]
(since
f′(x)=2x>0
for
x>0
). For an increasing function,
the Right Endpoint Sum (
R3
) is an overestimate because the top-right
corner of each rectangle lies above the curve. Conversely, the Left Endpoint
Sum (
L3
) is an underestimate.
Problem 2: Midpoint Riemann Sum and Summation Notation
Δ x=5−1
4=1
Subintervals:
[1,2],[2,3],[3,4],[4,5]
.
1. Midpoints
xi
∗
:
o
[1,2]:x1
∗=1.5
o
[2,3]:x2
∗=2.5
o
[3,4]:x3
∗=3.5
o
[4,5]:x4
∗=4.5
2. Midpoint Sum (
M4
):
M4=Δ x[g(1.5)+g(2.5)+g(3.5)+g(4.5)] g(x)=4x − x2g(1.5)=4(1.5)−¿
3. General Right Endpoint Riemann Sum (
Rn
):
The right endpoint of the
i
-th subinterval is
xi=a+i⋅Δ x
. Here,
a=1
and
Δ x=4/n
.
xi=1+i
(
4
n
)
Rn=∑
i=1
n
g(xi)⋅Δ x=∑
i=1
n
[
4
(
1+4i
n
)
−
(
1+4i
n
)
2
]
(
4
n
)
Problem 3: Interpreting the Definite Integral
1. Net Change in Position (Displacement): The integral of velocity gives the
net change in position.
∫
0
4
(¿3t2−6t)dt ¿
2. Total Distance Traveled: Total distance is the integral of the speed
(absolute value of velocity).
∫
0
4
¿3t2−6t∨dt
3. Physical Difference:
o
∫
1
3
v(t)dt
(Displacement): This represents the net change in the
particles position from
t=1
to
t=3
. If the particle moves forward and
then backward, the area counted below the axis (negative velocity)
subtracts from the area above the axis (positive velocity).
o
∫
1
3
¿v(t)∨dt
(Total Distance): This represents the total path length
the particle traveled from
t=1
to
t=3
. All movement (forward or
backward) is counted as positive distance. The area below the
t
-axis is
reflected above, ensuring that all accumulated movement is positive.
Section II: Fundamental Theorem of Calculus Part II (Computational)
Problem 4: Polynomial and Power Rule
∫
1
3
(¿6x2−4x+5)dx=
[
6x3
3−4x2
2+5x
]
1
3
=
[
2x3−2x2+5x
]
1
3=¿¿
Problem 5: Rational and Logarithmic Functions
∫
e
e2
(
1
x− x−2
)
dx=
[
ln∨x∨−x−1
−1
]
e
e2
=
[
ln(x)+ 1
x
]
e
e2
=
[
ln(e2)+ 1
e2
]
−
[
ln(e)+ 1
e
]
=
[
2+1
e2
]
−
[
1+1
e
]
=2+1
e2−1−1
e=1+1
e2−1
e
Problem 6: Trigonometric Functions
∫
0
π/4
(¿3 sin(x)+sec2(x))dx=
[
−3 cos(x)+tan(x)
]
0
π/4=
[
−3 cos(π/4)+tan (π/4)
]
−
[
−3 cos (0)+tan (0)
]
=
[
−3
(
√
2
2
)
+1
]
−
[
−3(1)+0
]
=−3
√
2
2+1−(−3)=4−3
√
2
2¿
Problem 7: Piecewise Functions
We must use the additivity property to split the integral at the transition point
x=1
:
∫
0
2
f(x)dx=∫
0
1
4x3dx+∫
1
2
(¿5− x)dx ¿
1. First Integral:
∫
0
1
4x3dx=
[
4x4
4
]
0
1
=
[
x4
]
0
1=14−04=1
2. Second Integral:
∫
1
2
(¿5− x)dx=
[
5x − x2
2
]
1
2
=
[
5(2)−22
2
]
−
[
5(1)−12
2
]
=[10−2]−[5−0.5 ]=8−4.5=3.5=7
2¿
3. Total Accumulation:
∫
0
2
f(x)dx=1+3.5=4.5or 9
2
Problem 8: Exponential Functions
∫
−1
1
(¿e2x+2)dx ¿
(Using
u
-substitution for
e2x
:
u=2x , du=2dx ⇒dx=1
2du
)
∫ e2xdx=1
2e2x∫
−1
1
(¿e2x+2)dx=
[
1
2e2x+2x
]
−1
1
=
[
1
2e2(1)+2(1)
]
−
[
1
2e2(−1)+2(−1)
]
=
(
e2
2+2
)
−
(
e−2
2−2
)
=e2
2−e−2
2+4¿
Section III: Fundamental Theorem of Calculus Part I (Theoretical)
Problem 9: Simple Application
y=∫
3
x
√
t3+5t+1dt
Applying FTC I directly:
dy
dx =
√
x3+5x+1
Problem 10: Chain Rule with the Upper Limit
y=∫
0
x2
cos(t3)dt
Let
u=x2
. Then
du
dx =2x
.
dy
dx =cos ¿
Problem 11: Chain Rule with the Lower Limit
y=∫
tan(x)
5
ln(t)dt
First, use the order of integration property:
∫
a
b
¿−∫
b
a
❑
.
y=−∫
5
tan (x)
ln(t)dt
Let
u=tan(x)
. Then
du
dx =sec2(x)
.
dy
dx =−
[
ln(tan (x))⋅sec2(x)
]
dy
dx =− sec2(x)ln(tan(x))
Problem 12: Chain Rule with Both Limits
y=∫
√
x
x3
et2dt
Split the integral at a constant (e.g.,
c=0
):
y=∫
√
x
0
et2dt+∫
0
x3
et2dty=−∫
0
√
x
et2dt+∫
0
x3
et2dt
1. Derivative of the First Term:
u=
√
x=x1/2
,
du
dx =1
2x−1/2=1
2
√
x
.
−¿
2. Derivative of the Second Term:
v=x3
,
dv
dx =3x2
.
e¿¿
3. Total Derivative:
dy
dx =3x2ex6
−ex
2
√
x
Section IV: Application and Net Change Theorem
Problem 13: Total Production
1. Total Units: The total accumulation is
∫
0
4
P(t)dt
.
∫
0
4
(¿15 t2+10 t)dt=
[
15 t3
3+10 t2
2
]
0
4
=
[
5t3+5t2
]
0
4=¿¿
2. Average Rate of Production:
Average Rate=1
4−0∫
0
4
P(t)dt=1
4(400)=100 units per hour
Problem 14: Population Change
r(t)=100
√
t+50=100t1/2+50
1. Net Change Expression:
∫
4
9
(¿100
√
t+50)dt ¿
2. Calculate Net Change:
∫
4
9
(¿100 t1/2+50)dt=
[
100 t3/2
3/2+50 t
]
4
9
=
[
200
3t3/2+50 t
]
4
9
=¿¿
3. Population in 2009 (
t=9
):
Pop(9)=Pop(0)+∫
0
9
r(t)dt
We need
∫
0
9
r(t)dt
:
∫
0
9
r(t)dt=
[
200
3t3/2+50t
]
0
9
=
[
200
3(27)+450
]
−[0]=1800+450=2250 peoplePop(9)=5000+2250=7250 people
Problem 15: Total Distance vs. Displacement
v(t)=t2−3t+2=(t − 1)(t −2)
Velocity is zero at
t=1
and
t=2
.
1. Displacement (Net Change):
∫
0
3
(¿t2−3t+2)dt=
[
t3
3−3t2
2+2t
]
0
3
=
[
33
3−3(32)
2+2(3)
]
−[0]=9−27
2+6=15 −13.5=1.5 cm ¿
2. Total Distance Traveled: We must split the integral where velocity changes
sign.
o
v(t)>0
on
¿
and
¿
(positive)
o
v(t)<0
on
(1,2)
(negative)
Total Distance=∫
0
1
v(t)dt −∫
1
2
v(t)dt +∫
2
3
v(t)dt
From step 1,
F(t)= t3
3−3t2
2+2t
.
Distance1=F(1)− F(0)=
(
1
3−3
2+2
)
−0=2−9+12
6=5
6Distance2=F(2)− F (1)=
(
8
3−12
2+4
)
−5
6=
(
8
3−6+4
)
−5
6=8
3−2−5
6=16 −12−5
6=−1
6Distance3=F(3)− F (2)=1.5 −(−1/6)=3/2+1/6=9/6+1/6=10 /6=5/3 Total Distance=
|
5
6
|
+
|
−1
6
|
+
|
5
3
|
=5
6+1
6+10
6=16
6=8
3cmTotal Distance ≈2.67 cm
