MAT186H1F - Calculus I
Fall 2025
Pre-class Essentials: E4-1 Modeling with Riemann Sums
Introduction: In the next four PCEs, we will explore applications of definite integrals to physical quantities, and focus on their formulation and justification through Riemann sums. This PCE will focus on two
specific quantities: total displacement and area between curves.
Regardless of the physical situation, you will see that the same pattern emerges in every application. The
goal is to learn this pattern, so you are able to model new physical quantities with definite integrals without
referencing a textbook - or the internet - for a formula. The idea is to:
1. divide the problem into small parts.
2. estimate the physical quantity in each part.
3. add up the estimations to approximate the physical quantity in its entirety.
4. take a limit as the number of divisions goes to ∞.
Learning Objectives: By the end of this PCE you should be able to
• derive a definite integral that represents total displacement.
• derive a definite integral that represents area between two curves.
• explain why the definite integral formula you come up with makes sense using Riemann sums.
Total Displacement:
Here, we justify why an object moving along a straight line with velocity v(t) has total displacement from
time t = a to time t = b given by
Z b
v(t) dt.
a
1. We divide the problem into small parts by dividing the time interval [a, b] into n equal subintervals of
b−a
(units of time).
length ∆t =
n
Our n subintervals are [t0 , t1 ], [t2 , t2 ], . . . , [tn−1 , tn ] where t0 = a, tn = b, and tk − tk−1 = ∆t.
2. On each subinterval [tk−1 , tk ], k = 1, 2, . . . , n we can approximate the total displacement of the object
by choosing a representative time t∗k ∈ [tk−1 , tk ] and taking the object’s velocity as v(t∗k ) over the
entire time interval [tk−1 , tk ]. This is only an approximation of the object’s velocity on the time
interval [tk−1 , tk ] since the object’s velocity is not necessarily constant on [tk−1 , tk ]. However, if our
time intervals [tk−1 , tk ] are small enough, the approximation will be reasonable since the velocity of
the object will not vary much. Therefore, the total displacement of the object on [tk−1 , tk ] can be
approximated by
v(t∗k )∆t
metres
(Notice the units here are in units
of time · units of time = metres)
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3. Adding up across all subintervals, we can approximate the total displacement of the object on [a, b] by
n
X
v(t∗k )∆t
k=1
4. As the number of subintervals n → ∞, the approximations become better and approaches the total
displacement of the object from time t = a to t = b. That is
Total Displacement = lim
n→∞
n
X
v(t∗k )∆t
k=1
Z b
=
v(t) dt
a
by definition of the definite integral.
Area Between Two Curves:
x2
x
Example: Find the area of the region bounded by the curves f (x) = , g(x) = x − , between x = 5 and
6
30
x = 10. This region is illustrated in the left figure below.
Area Between Curves
Approximation of Area Between Curves
1. We divide the region between x = 5 and x = 10 into n subintervals of equal length ∆x =
10 − 5
5
= .
n
n
Our n subintervals are [x0 , x1 ], [x1 , x2 ], . . . , [xn−1 , xn ] where x0 = 5, xn = 10, and xk − xk−1 = ∆x.
2. On each subinterval [xk−1 , xk ], k = 1, 2, . . . , n we can approximate the area between f and g using
rectangles by choosing
point x∗k ∈ [xk−1 , xk ] and taking the height of the rectangle to
a representative
∗ 2
∗
(x )
x
be g(x∗k ) − f (x∗k ) = x∗k − k
− k , and its width to be ∆x. In other words, the area between f
30
6
and g on [xk−1 , xk ] can be approximated by
(x∗k )2
x∗k 5
∗
xk −
−
30
6 n
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3. Adding up across all subintervals, we can approximate the area between f and g on [5, 10] by
n X
x∗ 5
(x∗ )2
x∗k − k − k
30
6 n
k=1
4. As the number of subintervals n → ∞, the approximations become better and approaches the area
between f and g on [5, 10]. That is
n X
x∗k 5
(x∗k )2
∗
−
Area = lim
xk −
n→∞
30
6 n
k=1
Z 10 x2
x
=
x−
−
dx
30
6
5
by definition of the definite integral.
Example: Find the area of the region bounded by the curves x = y, and x = 6 − y 2 . This region is
illustrated in the left figure below.
Area Between Curves
Approximation of Area Between Curves
Note that the two curves intersect at (−3, −3) and (2, 2). We essentially mimic our approach in the previous
example by dividing our region along the y-axis as opposed to the x-axis.
1. We divide the region between y = −3 and y = 2 into n subintervals of equal length ∆y =
2 − (−3)
5
= .
n
n
Our n subintervals are [y0 , y1 ], [y1 , y2 ], . . . , [yn−1 , yn ] where y0 = −3, yn = 2, and yk − yk−1 = ∆y.
2. On each subinterval [yk−1 , yk ], k = 1, 2, . . . , n we can approximate the area between the curves using
rectangles by choosing a representative point yk∗ ∈ [yk−1 , yk ] and taking the height of the rectangle to
be (6 − (yk∗ )2 ) − yk∗ , and its width to be ∆y. In other words, the area between the curves on [yk−1 , yk ]
can be approximated by
5
6 − (yk∗ )2 − yk∗
n
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3. Adding up across all subintervals, we can approximate the area between two curves on [−3, 2] (on the
y-axis) by
n
X
5
6 − (yk∗ )2 − yk∗
n
k=1
4. As the number of subintervals n → ∞, the approximations become better and approaches the area
between the two curves. That is
Area = lim
n→∞
Z 2
=
n
X
6 − (yk∗ )2 − yk∗
k=1
5
n
6 − y 2 − y dy
−3
by definition of the definite integral.
Supplementary Exercises and References
For the following problems, set up the Riemann sum
n
X
f (x∗k )∆x, and find the corresponding definite integral.
k=1
We are not evaluating these integrals yet. To check if your model provides the correct area, you can check
using an online calculator.
Textbook Section 6.2:
Fundamental Problems: 9, 11, 13, 15, 21, 23, 27, 31, 43, 47, 63.
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