5.1/5.2 Approximating and Computing Area &
The Definite Integral
Why might we want to compute the area under a graph? Suppose v(t) is a velocity function
where the velocity is constant.
Then computing the area under v(t) from t
1
to t
2
is the same as computing the distance
traveled from time 1 to time 2! This is true since:
Distance traveled = rate (velocity) × change in time = v∆t
But what if the velocity is not constant? Then a more accurate statement is that
Distance traveled = area under the graph of velocity over [t
1
, t
2
]
We can see that this statement extends the previous example to an example where the
velocity is 5 m/s for the first 2 seconds, 15 m/s for the next second, 10 m/s for the next
three seconds, and 5 m/s for the last two seconds:
Then the distance traveled over the time interval [0, 8] seconds is
5
m
s
· 2s + 15
m
s
· 1s + 10
m
s
· 3s + 5
m
s
· 2s = 10m + 15m + 30m + 10m = 65m
What if the function isn’t a piecewise constant function?
1
We still use what is known as a Riemann Sum, which adds up the areas of rectangles
that approximate the correct area.
To specify a Riemann sum, we need to choose
– Partition, P , of size N: a choice of points that splits [a, b] into N subintervals
P = {x
0
, x
1
, x
2
, . . . , x
N
} where a = x
0
< x
1
< x
2
< ··· < x
N
= b
Note: To specify N rectangles, we need N + 1 points (which is why we start with x
0
)
– Sample points, C: C = {c
1
, . . . , c
N
} where c
i
belongs to the subinterval [x
i−1
, x
i
] for
i = 1, . . . , N
– Width of ith subinterval: ∆x
i
= x
i
− x
i−1
Note: for a given P and C, the ith rectangle has width ∆x
i
and height f(c
i
).
Then the Riemann Sum is
R(f, P, C) = f (c
1
)∆x
1
+ f (c
2
)∆x
2
+ ···f (c
N
)∆x
N
Example 1: Calculate the Riemann Sum R(f, P, C) for f (x) = 0.5x + 3 on [−4, 8] where
P = {−4, −1, 1, 4, 8} and C = {−3, 0, 2, 5}.
-5 5
5
x
y
2
While it may be more accurate to choose a partition specific to the function, sometimes
it’s easiest to just choose subintervals of the same length. For the interval [a, b] on which we
want the approximation, choose a number of subintervals N and compute the width of each
subinterval:
∆x =
b − a
N
Then we determine how to choose the rectangles. If we want the height of the rectangle to be
the function’s value at the right end of the subinterval, then we have the Nth right-endpoint
approximation:
R
N
= f(x
1
)∆x + f(x
2
)∆x + ··· + f(x
N
)∆x
= ∆x ·[f(x
1
) + f(x
2
) + ··· + f(x
N
)]
where x
1
= a + ∆x, x
2
= a + 2∆x, . . . , x
N−1
= a + (N − 1)∆x, x
N
= a + N∆x.
For the Nth left-endpoint approximation, we have
L
N
= ∆x ·[f(x
0
) + f(x
1
) + ··· + f(x
N−1
)]
where x
0
= a, x
1
= a + ∆x, x
2
= a + 2∆x, . . . , x
N−1
= a + (N − 1)∆x.
For the Nth midpoint approximation, we have
M
N
= ∆x
f
x
0
+ x
1
2
+ f
x
1
+ x
2
2
+ ··· + f
x
N−1
+ x
N
2
where x
k
= a + k∆x (as before).
3
Example 2: Compute R
6
, L
6
and M
3
to estimate the distance traveled over [0, 6] if the
velocity at 1-second intervals is as follows:
t (s) 0 1 2 3 4 5 6
v (ft/s) 0 3 7 2 4 5 2
Example 3: Calculate L
4
for f (x) = 3x
2
over [1, 3].
4
We use summation notation to write sums in compact form. The capital Greek letter
sigma, Σ, stands for “sum,” where we start at j = m and end at j = n:
n
X
j=m
a
j
= a
m
+ a
m+1
+ ···a
n
For example,
6
X
j=2
j
2
=
2
2
+
3
2
+
4
2
+
5
2
+
6
2
=
20
2
= 10
Note that we sometimes call the summation index, j, a “dummy variable” because it
doesn’t matter what variable we use there!
5
X
k=3
(k
2
− 4) =
Important Summations (you do not need to memorize these)
•
n
X
j=m
(a
j
+ b
j
) =
n
X
j=m
a
j
+
n
X
j=m
b
j
•
n
X
j=m
Ca
j
= C
n
X
j=m
a
j
•
n
X
j=1
C = nC
•
N
X
j=1
j = 1 + 2 + ··· + N =
N(N + 1)
2
=
N
2
2
+
N
2
•
N
X
j=1
j
2
= 1
2
+ 2
2
+ ··· + N
2
=
N(N + 1)(2N + 1)
6
=
N
3
3
+
N
2
2
+
N
6
•
N
X
j=1
j
3
= 1
3
+ 2
3
+ ··· + N
3
=
N
2
(N + 1)
2
4
=
N
4
4
+
N
3
2
+
N
2
4
Note: now we can write the right-endpoint, left-endpoint, and midpoint approximations
more compactly!
R
N
= ∆x ·[f(x
1
) + f(x
2
) + ··· + f(x
N
)] = ∆x
N
X
j=1
f(x
j
)
L
N
= ∆x ·[f(x
0
) + f(x
1
) + ··· + f(x
N−1
)] = ∆x
N−1
X
j=0
f(x
j
)
M
N
= ∆x·
f
x
0
+ x
1
2
+ f
x
1
+ x
2
2
+ ··· + f
x
N−1
+ x
N
2
= ∆x
N−1
X
j=0
f
x
j
+ x
j+1
2
5
Example 4: Write the sum in summation notation:
√
1 + 1
3
+
√
2 + 2
3
+ ···
√
17 + 17
3
Example 5: Evaluate the sums:
a)
4
X
i=1
(2i + 3
i−1
)
b)
5
X
k=3
2k
2
Example 6: Express the area under the graph as a limit using the approximation indicated
(in summation notation), but do not evaluate.
L
N
, f(x) = sin x, [0, π]
6
Now that we have ways to approximate the area between a function and the x-axis, how
do we make these approximations better?
Example 7: Calculate the limit for the given function and interval. Verify your answer
using geometry.
lim
N→∞
R
N
, f(x) = 4x −2, [1, 3]
7
Note that if f is continuous on [a, b], then
L = lim
N→∞
R
N
= lim
N→∞
L
N
= lim
N→∞
M
N
= lim
kP k→0
N
X
i=1
f(c
i
) · ∆x
i
=
Z
b
a
f(x)dx
is the area between the curve and the x-axis. We call this the definite integral of f over
[a, b]. f (x) is the integrand and a and b are the limits of integration.
What if the function goes below the x-axis?
Interpretation of the Definite Integral as Signed Area:
• When f (x) ≥ 0, the definite integral is (“positive”) area under the graph (between the
graph and the x-axis).
• When f (x) ≤ 0, we consider the function to have “negative area” (between the graph
and the x-axis).
Example 8: Calculate
Z
4
0
(2 − x) dx and
Z
4
0
|2 − x| dx using geometry.
8
Properties of Definite Integrals:
•
Z
b
a
C dx = C(b − a)
•
Z
b
a
(f(x) + g(x)) dx =
Z
b
a
f(x) dx +
Z
b
a
g(x) dx
•
Z
b
a
Cf(x) dx = C
Z
b
a
f(x) dx
•
Z
b
a
f(x) dx = −
Z
a
b
f(x) dx
•
Z
c
a
f(x) dx =
Z
b
a
f(x) dx +
Z
c
b
f(x) dx
•
Z
a
a
f(x) dx = 0
•
Z
b
0
x dx =
1
2
b
2
•
Z
b
0
x
2
dx =
1
3
b
3
Example 9: Assuming
Z
5
1
f(x) dx = 15,
Z
3
0
f(x) dx = 10, and
Z
1
0
f(x) dx = 2,
calculate
Z
5
3
f(x) dx.
Example 10: Calculate
Z
5
0
(−3x + 1) dx and
Z
0
5
(−3x + 1) dx.
9
5.3 The Indefinite Integral
Now we begin talking about the inverse process of finding derivatives - finding “antideriva-
tives”! The question here is: given a derivative, can we find the original function?
Common notation:
• A function F is an antiderivative of f on an open interval (a, b) if F
0
(x) = f (x) for all
x ∈ (a, b).
•
Z
f(x) dx = F (x) + C means that F
0
(x) = f (x) where C is an arbitrary constant.
Some Examples:
• F (x) = −cos x is an antiderivative of f (x) = sin x because for all values of x,
F
0
(x) =
d
dx
(−cos x) = sin x = f(x).
• F (x) =
1
3
x
3
is an antiderivative of f(x) = x
2
because for all values of x,
F
0
(x) =
d
dx
1
3
x
3
= x
2
= f(x).
Question: What is an antiderivative of 2x?
We say an (and not the) antiderivative because antiderivatives are not unique! Since
the derivative of a constant is zero, we can always add a constant to the end and obtain a
different (but still correct) antiderivative.
10
Properties of Integrals:
i) If k is a constant,
Z
kf(x) dx = k
Z
f(x) dx
ii) If f (x) and g(x) are two funtions,
Z
(f(x) ± g(x)) dx =
Z
f(x) dx ±
Z
g(x) dx
Some Common Integrals:
i) Integral of a Constant: If k is a constant,
Z
k dx = kx + C
ii) Integral of a Power Function: For n 6= −1,
Z
x
n
dx =
x
n+1
n + 1
+ C
iii) Integral of
1
x
on the domain x > 0:
Z
1
x
dx = ln |x| + C
iv) Integral of the Exponential Function:
Z
e
x
dx = e
x
+ C and
Z
e
kx
dx =
1
k
e
kx
+ C
Trigonometric Integrals
•
Z
cos x dx = sin x + C
•
Z
sin x dx = −cos x + C
•
Z
sec
2
x dx = tan x + C
•
Z
csc x cot x dx = −csc x + C
•
Z
sec x tan x dx = sec x + C
•
Z
csc
2
x dx = −cot x + C
Note also that
Z
cos(kx) =
1
k
sin(kx) + C and
Z
sin(kx) = −
1
k
cos(kx) + C
11
Example 1: Evaluate:
a)
Z
(18t
5
− t
−3
+ 10
4
√
t) dt
b)
Z
3
x
+
4
x
2
dx
c)
Z
15 sin
θ
3
+ 14 csc (θ) cot (θ)
dθ
d)
Z
(3e
x
− x) dx
e)
Z
4e
5x−2
dx
f)
Z
x
2
− 4x + 3
√
x
dx
12
Initial Conditions: We can think of an antiderivative as a solution to a differential
equation (an equation relating an unknown function with its derivatives):
dy
dx
= f(x)
If we impose conditions on our functions, we can find a specific solution/antiderivative, rather
than a general one. These conditions are often of the form f (x
0
) = y
0
or y(x
0
) = y
0
. A
differential equation with an initial condition is called an initial value problem.
Example 3: Solve the initial value problem:
dy
dz
= sin(2z); y
π
4
= 4.
Example 4: A car traveling in a straight line with a velocity of 24 m/s begins to slow down
at time t = 0 with a deceleration of a(t) = −6 m/s
2
. Find:
a) The velocity v(t) at time t
b) The distance traveled before the car comes to a halt.
13
5.4 The Fundamental Theorem of Calculus, Part I
The Fundamental Theorem of Calculus, Part I:
Assume that f is continuous on [a, b]. If F is an antiderivative of f on [a, b], then
Z
b
a
f(x) dx = F (b) −F (a)
Proof idea:
Notation: We use F (x)|
b
a
to denote F (b)−F (a). In this notation, the FTC part I reads:
Z
b
a
f(x) dx = F (x)|
b
a
= F (b) −F (a)
Further, it doesn’t matter which antiderivative we use when evaluating a definite integral
using FTC I because the constant will cancel out.
14
Example 1: Calculate the area under the graph of each function over the given interval:
(a) f(x) = 4x
2
, [1, 5]
(b) f(x) =
2
√
x
+ 3
5
√
x
2
, [1, 4]
15
Example 2: Evaluate each definite integral.
(a)
Z
π
0
sin
x
2
dx
(b)
Z
2π
0
cos
θ
2
dθ
(c)
Z
5
3
2e
−4t
dt
16
(d)
Z
8
2
dy
y
(e)
Z
−8
−2
dx
x
(f)
Z
6
2
x +
1
x
dx
17
5.5 The Fundamental Theorem of Calculus, Part II
• Part I of the FTC says that we can use the antiderivative to compute definite integrals.
– This is a HUGE advantage over having to take limits of Riemann sums.
• Part II of the FTC says that the derivative of a certain integral is the integrand.
– That is to say that the derivative “cancels out” the action of the integral on the
original function (the integrand).
– However, the is ONLY for a very particular type of integral.
The Area Function:
A(x) =
Z
x
a
f(t) dt = signed area from a to x
Example 1: Let f(t) = t
2
.
(a) Find a formula for the area function A(x) of f(t) with lower limit a = 3.
(b) Find A(3).
(c) Find A
0
(x).
18
Example 2: Find the derivative, G
0
(x).
a) G(x) =
Z
x
−1
cos(t) dt
b) G(x) =
Z
x
4
e
−t
dt
c) G(x) =
Z
x
1
t
1/2
dt
d) What’s the pattern for taking a derivative of a function that is defined like an area
function?
e) G(x) =
Z
x
2
−1
cos(t) dt
19
The Fundamental Theorem of Calculus, Part II:
Assume that f is continuous on an open interval I and let a be a point in I. Then the
area function
A(x) =
Z
x
a
f(t) dt
is an antiderivative of f on I; that is, A
0
(x) = f (x). Equivalently,
d
dx
Z
x
a
f(t) dt = f(x).
Furthermore, A(x) satisfies the initial condition A(a) = 0.
The FTC shows that integration and differentiation are inverse operations. By FTC II,
if you start with a continuous function f and form the integral
Z
x
a
f(t) dt, then you get back
the original function by differentiating:
f(x)
Integrate
−−−−−→
Z
x
a
f(t) dt
Differentiate
−−−−−−−→
d
dx
Z
x
a
f(t) dt = f(x).
On the other hand, by FTC I, if you differentiate first and then integrate, you also recover
f(x) (but only up to a constant f (a)):
f(x)
Differentiate
−−−−−−−→ f
0
(x)
Integrate
−−−−−→
Z
x
a
f
0
(t) dt = f (x) − f(a).
Note: When the upper limit of the integral is a function of x, rather than x
itself, we will require the FTC II and the Chain Rule to differentiate the integral.
Example 3: Calculate the derivative:
d
du
Z
3u
u
e
−x
dx.
20
Increasing/Decreasing Behavior of the Area Function A(x):
• If f (x) > 0 on some interval, then the small
changes in area are also positive.
– So, ∆A > 0 and the area is increasing.
• If f (x) < 0 on some interval, then the small
changes in area are also negative.
– So, ∆A < 0 and the area is decreasing.
Example 5: Let f be the function graphed below, and A(x) =
Z
x
0
f(t) dt.
2 4 7 9
−2
−1
1
2
3
f(2, 2)
(4, 3)
(9, −2)
a) What do you know about A
0
(x)?
b) Find the intervals on which A(x) is increasing and on which A(x) is decreasing. Justify
your answer.
c) Determine the absolute extrema of A(x) on [0, 9]. Justify your answer.
21
5.6 Net Change as the Integral of a Rate of Change
Theorem: Net Change as the Integral of a Rate of Change:
The net change in s(t) over an interval [t
1
, t
2
] is given by the integral
Z
t
2
t
1
s
0
(t) dt
| {z }
Integral of the rate of change
= s(t
2
) − s(t
1
)
| {z }
Net change over [t
1
,t
2
]
Note: Remember that the integral may be approximated using Riemann Sums if neces-
sary. In these cases, we often calculate the average of the left- and right-endpoint approxi-
mations.
Example 1: A survey shows that a mayoral candidate is gaining votes at a rate of 200t+100
votes per day, where t is the number of days since she announced her candidacy.
(a) How many supporters will the candidate have after 60 days?
(b) How many additional supporters will she gain in the next 30 day period?
22
Example 2: The number of cars per hour passing an observation point along a highway is
known as the traffic flow rate (in cars per hour), call this flow rate q(t). The flow rate is
recorded at 30 minute intervals between 7 and 9 a.m. in the chart below.
(a) Which quantity is represented by
Z
t
2
t
1
q(t) dt?
(b) Estimate the number of cars using the highway during this 2 hour period.
Displacement vs. Distance Traveled:
If you travel 10 km in a straight line, then return to your start point along the same path
• Your displacement is 0 km.
• Your distance traveled is 20 km.
Theorem: The Integral of Velocity:
For an object with velocity v(t),
Displacement during [t
1
, t
2
] =
Z
t
2
t
1
v(t) dt
Distance traveled during [t
1
, t
2
] =
Z
t
2
t
1
|v(t)| dt
23
Example 3: A particle has velocity v(t) = t
3
− 10t
2
+ 24t m/s along a straight path.
Compute:
(a) Displacement over [0, 6].
(b) Total distance traveled over [0, 6].
(c) Indicate the particle’s trajectory with a motion diagram.
24
5.7 Substitution Method
In this section, we will use the Chain Rule “in reverse”.
Let’s compute the integral
Z
2x cos
x
2
dx. We can evaluate it if we remember the
Chain Rule calculation
d
dx
sin
x
2
= 2x cos
x
2
.
This tells us that sin
x
2
is an antiderivative of 2x cos
x
2
, and therefore,
Z
2x
|{z}
Derivative of
inside function
cos
x
2
|{z}
Inside
function
dx = sin
x
2
+ C.
A similar Chain Rule calculation shows that
Z
1 + 3x
2
| {z }
Derivative of
inside function
cos
x + x
3
| {z }
Inside
function
dx = sin
x + x
3
+ C.
The key in both cases is that the integrand is the product of a composite function and
the derivative of the “inner function” of the composite function.
For example, we cannot use the Chain Rule to compute
Z
cos
x + x
3
dx because the
factor (1 + 3x
2
) does not appear in the integrand.
Theorem: The Substitution Method (Often called u -substitution)
If F
0
(x) = f (x), then
Z
f(u(x))u
0
(x) dx = F (u(x)) + C
Note that u(x) is the “inner function” and u
0
(x) is the derivative of the “inner function”.
Change of Variables Formula:
We can use differentials (i.e. symbols such as dy, dx, and du) to help in clarifying the process.
Since
du
dx
= u
0
(x), we can also write du = u
0
(x)dx.
Consider the first example above:
Z
2x cos
x
2
dx. We can set u(x) = x
2
, so du = 2xdx.
Then
Z
2x cos
x
2
dx =
Z
cos(u) du.
25
Change of Variables Formula:
Z
f(u(x))
| {z }
f(u)
u
0
(x) dx
| {z }
du
=
Z
f(u) du
Example 1: Evaluate the integrals:
(a)
Z
2x
2
+ x
(4x
3
+ 3x
2
)
3
dx
(b)
Z
sec
2
θe
tan θ
dθ
(c)
Z
t
1/2
cos
t
3/2
dt
26
(d)
Z
tan θ dθ
(e)
Z
z
√
5z + 1 dz
27
The Change of Variable Formulas can be applied to definite integrals provided that the limits
of integration are changed.
Change of Variable Formula for Definite Integrals:
Z
b
a
f(u(x))u
0
(x) dx =
Z
u(b)
u(a)
f(u) du
Example 2: Calculate the integral:
(a)
Z
3
1
x
x
2
+ 1
dx
(b)
Z
π/4
0
tan
3
θ sec
2
θ dθ
28
5.8 Further Integral Formulas
Inverse Trigonometric Functions:
d
dx
sin
−1
x =
1
√
1 − x
2
,
Z
dx
√
1 − x
2
= sin
−1
x + C
d
dx
tan
−1
x =
1
x
2
+ 1
,
Z
dx
x
2
+ 1
= tan
−1
x + C
d
dx
sec
−1
x =
1
|x|
√
x
2
− 1
,
Z
dx
|x|
√
x
2
− 1
= sec
−1
x + C
In this list, we omit the integral formulas corresponding to the derivatives of y = cos
−1
x,
y = cot
−1
x, and y = csc
−1
x because the integrals differ only by a minus sign from those
already on the list. For example,
d
dx
cos
−1
x = −
1
√
1 − x
2
,
Z
dx
√
1 − x
2
= −cos
−1
x + C.
Note: The domain of inverse sine, cosine, secant, and cosecant are x ∈ [−1, 1].
Integral of Exponential Functions (b > 0, b 6= 1):
Z
e
x
dx = e
x
+ C
Z
b
x
dx =
b
x
ln b
+ C
Example 1: Evaluate the integral:
(a)
Z
1
−1
3
x
2
+ 1
dx
29
(b)
Z
tan
−1
x
1 + x
2
dx
(c)
Z
(3x + 2) dx
x
2
+ 4
30
(d)
Z
cot x ln (sin x) dx
(e)
Z
1
0
4
−2x
dx
31
