MTH 1410 Blank Notes
§5.3 Fundamental Theorem of Calculus
Page 1 of 9
Trying to compute every definite integral via the definition would be a really horrible experience! In this section, we
will discuss the “shortcut” to computing definite integrals – the Fundamental Theorem of Calculus.
Definition: Area Function
Let f be a continuous function for t ≥ a. The area function for f with left endpoint a is
Z x
A(x) =
f (t) dt
a
where x ≥ a. The area function gives the net area of the region bounded by the graph of f and the
t-axis on the interval [a, x].
x
Z
Z
8
2
for f . Evaluate each of the following expressions.
Z
x
f (t) dt be two area functions
f (t) dt and F (x) =
Example 5.3.1: The graph of f is shown below. Let A(x) =
2
(a) A(2) =
Z
8
(b) A(8) =
f (t) dt = 0
2
f (t) dt = 22
2
18
Z
(c) A(18) =
Z
f (t) dt
2
14
f (t) dt = −28
(d) F (14) =
8
= 22 + (−28) + 11
= 5
Now let’s jump into the Fundamental Theorem of Calculus! The first part shows the connection between differentiation and integration. The second gives us a fast method for computing definite integrals.
Theorem: The Fundamental Theorem of Calculus (Part 1)
Z x
If f is continuous on [a, b], then the area function A(x) =
f (t) dt, for a ≤ t ≤ b, is continuous on [a, b] and
a
differentiable on (a, b). The area function satisfies A′ (x) = f (x); or equivalently,
Z x
d
′
f (t) dt = f (x)
A (x) =
dx a
which means that the area function of f is an antiderivative of f on [a, b].
! This is easily shown to be true with the second part of the Fundamental Theorem of Calculus.
△
Example 5.3.2: Simplify each expression.
d
(a)
dx
Z
d
dx
Z
x
e
−3t
2
2
sin (t + 1) dt
0
0
x
e−3t sin2 (t2 + 1) dt = e−3x sin2 (x2 + 1)
d
(b)
dx
Z
d
dx
Z
x2 p
t4 + 1 dt (need to use the Chain Rule)
1
1
x2 p
t4 + 1 dt =
p
p
(x2 )4 + 1 (2x) = 2x x8 + 1
§5.3 Fundamental Theorem of Calculus
MTH 1410 Blank Notes
Page 2 of 9
Theorem: The Fundamental Theorem of Calculus (Part 2)
If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then
Z
b
b
= F (b) − F (a)
f (x) dx = F (x)
a
a
! This theorem is extremely important! A proof of it (as well as a justification for Part 1) will be in the annotated
△
notes!
! When using the Fundamental Theorem of Calculus, there is no need to write the +C! Just remember, if your
△
integral has numbers on it, your answer is a number (for now)!
b
Z
f (x) dx = F (x) + C
b
a
a
= F (b) + C − F (a) + C = F (b) + C − F (a) − C = F (b) − F (a) ✓
Example 5.3.3: Evaluate each of the following definite integrals.
Z
2
3x2 − 4x + 5 dx
(a)
0
Z
2
3x2 − 4x + 5 dx = x3 − 2x2 + 5x
2
0
0
3
2
= (2) − 2(2) + 5(2) − (0)3 − 2(0)2 + 5(0)
= 10 − 0
= 10
Z
π/6
(b)
sec(2x) tan(2x) dx
0
Z
π/6
π/6
1
1
2π
sec(2x)
=
sec
− sec(0)
2
2
6
0




1
1 1
1
1
1  11
−
=  −  = (2 − 1) =
= 
2 cos π
cos(0)
2 1
1
2
2
3
2
sec(2x) tan(2x) dx =
0
Z
(c)
4
9
√
Z 9
Z 9
3x − x
3x
x1/2
1/2
√
dx =
−
dx
=
3x
−
1
dx
x
x1/2
x1/2
4
4
Z 9
4
3x
1/2
9
2 3/2
− 1 dx = 3
x
−x
3
4
3/2
= 2(9) − 9 − 2(4)3/2 − 4
= 45 − 12
= 33
§5.3 Fundamental Theorem of Calculus
MTH 1410 Blank Notes
Page 3 of 9
Review: Even and Odd Functions
ˆ A function f (x) is even if and only if f (−x) = f (x). Even functions are symmetric with respect to
the y-axis. Examples: f (x) = x2 , f (x) = cos x, f (x) = x2/3 , f (x) = |x|.
– “E” for even, “e” for negatives get eaten, “e” for y-axis.
ˆ A function f (x) is odd if and only if f (−x) = −f (x). Odd functions are symmetric with respect to
the origin. Examples: f (x) = x3 , f (x) = sin x, f (x) = tan x, f (x) = x1/3 .
– “O” for odd, “o” for negatives factor out, “o” for origin.
Theorem: Integrals of Even and Odd Functions
Let a be a positive real number and suppose f is integrable on the interval [−a, a].
Z a
Z a
Z a
If f is even:
f (x) dx = 2
f (x) dx
If f is odd:
f (x) dx = 0
−a
−a
0
Example 5.3.4: Evaluate each integral.
Z π/2 14 17/15 13 7/9
(a)
x
− x + 4 sin x dx
8
−π/2 19
Z
π/2
−π/2
Z
14 17/15 1 7/9
x
− x + 4 sin x
19
8
(each power of x is an odd function and sine is also odd)
Z
Z
Z π/2
14 π/2 17/15
1 π/2 7/9
x
dx −
x dx + 4
sin x dx
19 −π/2
8 −π/2
−π/2
1
14
(0) − (0) + 4(0)
=
19
8
= 0
dx =
3
(3 − |x|) dx
(b)
(f (x) = 3 − |x| is even)
−3
Z
3
Z
(3 − |x|) dx = 2
−3
3
(3 − x) dx
0
3
1 2
= 2 3x − x
2
0
1 2
= 2 3(3) − (3) − 0
2
= 9
To find the average (mean) of n values x1 , x2 , x3 , . . . , xn we would add up the values, and then divide by n:
n
P
x1 + x2 + x3 + · · · + xn
x=
=
n
xk
k=1
n
=
“add up the things”
“divide by the number of things”
This idea of “average” generalizes to functions! Over an interval, on average, how high (or low) does a function go
on the y-axis?
§5.3 Fundamental Theorem of Calculus
MTH 1410 Blank Notes
Page 4 of 9
Definition: Average Value of a Function
1
Z
b
f (x) dx.
b−a a
Add up the things (the integral) and divide by the number of things (the length of the interval).
The average value of an integrable function f on the interval [a, b] is f =
The average value of a function f on an interval [a, b] has a clear geometrical interpretation.
The rectangle constructed with base (b − a) and height f will have the same
area as the area under f over the interval [a, b].
Z b
1
f (x) dx
b−a a
Z b
(b − a)f =
f (x) dx
f=
(multiply the (b − a) to the left side)
a
area of rectangle = area under curve
Example 5.3.5: Find the average value of f (x) = 30x(20 − x) on the interval [0, 20].
f=
1
b−a
Z
b
f (x) dx
a
Z 20
1
600x − 30x2 dx
20 − 0 0
i20
1h
300x2 − 10x3
=
20
0
1
2
=
300(20) − 10(20)3 − 0
20
= 2000
=
Theorem: Mean Value Theorem for Integrals
Let f be continuous on the interval [a, b]. There exists a point c in (a, b) such that
f (c) = f =
1
b−a
Z
b
f (x) dx
a
! Given an interval, we can find at least one x-value for which a function attains its average value on that interval.
△
MTH 1410 Blank Notes
§5.3 Fundamental Theorem of Calculus
Page 5 of 9
Example 5.3.6: Find the point(s) at which the function f (x) = 4x2 equals its average value on the interval [0, 3].
Z 3
1
4x2 dx
3−0 0
1 4 3
= · x3
3 3 0
4
= (3)3 − 0
9
f = 12
f (x) = f
f=
4x2 = 12
x2 = 3
√
x=± 3
√
x= 3
√
− 3 is not in (0, 3)
x
Z
Your Turn!
Z
1. The graph of f is shown below. Let A(x) =
x
f (t) dt and F (x) =
0
Evaluate each of the following expressions.
f (t) dt be two area functions for f .
5
(a) A(5)
(b) F (5)
(c) A(15)
(d) F (15)
2. Simplify the following expressions.
Z 5x
2
d
(a)
et dt
dx 2
(b)
d
dx
Z
π
sin(t3 ) dt
0
3. Evaluate each definite integral.
−1
Z
(2x − 1) dx
(a)
Z
0
Z
(d)
1
Z 3
π
(2eu + cos u) du
2
(b)
Z
(f)
Z
9
dx
2+9
x
−3
5
(|x| − 3) dx
(b)
−π
sec2 θ dθ
π/4
Z 3
1
4. Use symmetry to evaluate each integral.
Z π
(a)
sin3 x dx
π/3
(c)
(2z − 1)2 dz
(e)
0
5t6 + t − 1
dt
t2
−5
5. Find the point(s) at which the function f (x) = x2 − 3x + 1 equals its average value on the interval [1, 13].
4a. 0, 4b. −5,
1a. −7, 1b. 0, 1c. 4, 1d. 11,
2a. 5e25x , 2b. 0,
2
√
61
62
3π
+ ln 2, 3c. 3 − 1, 3d. 2eπ − 2, 3e.
, 3f.
2
3
2
5. x = 8
3a. 2, 3b.
§5.3 Fundamental Theorem of Calculus
MTH 1410 Blank Notes
Page 6 of 9
Proof of Part 2 of the Fundamental Theorem of Calculus
Z
b
f (x) dx = F (b) − F (a).
If f is continuous on [a, b] and F is any antiderivative of f on [a, b], then
a
Proof.
Let’s start by partitioning [a, b] as a = x0 < x1 < x2 < · · · < xn−1 < xn = b. Let each subinterval
have equal length ∆x, in particular, the kth subinterval, [xk−1 , xk ], will have length ∆x = xk − xk−1 . Then for any
antiderivative F of a continuous function f , we have the following.
F (b) − F (a) = F (xn ) − F (x0 )
= F (xn )+0 + 0 + · · · + 0 − F (x0 )
= F (xn )+ − F (xn−1 ) + F (xn−1 ) + − F (xn−2 ) + F (xn−2 ) + · · · + − F (x1 ) + F (x1 ) − F (x0 )
= F (xn ) − F (xn−1 ) + F (xn−1 ) − F (xn−2 ) + F (xn−2 ) − F (xn−3 ) + · · · + F (x1 ) − F (x0 )
F (b) − F (a) =
n
X
F (xk ) − F (xk−1 )
k=1
Because F ′ (x) = f (x), F is differentiable, which tells us that F is continuous. Hence, we can apply the Mean Value
Theorem to F on the interval [xk−1 , xk ] to find a x∗k in (xk−1 , xk ) such that
F (xk ) − F (xk−1 )
xk − xk−1
F
(x
k ) − F (xk−1 )
f (x∗k ) =
∆x
∗
F (xk ) − F (xk−1 ) = f (xk )∆x
F ′ (x∗k ) =
Therefore, after substituting f (x∗k )∆x for F (xk ) − F (xk−1 ) and taking the limit as n → ∞ of both sides, we arrive
at the desired result!
F (b) − F (a) =
F (b) − F (a) =
n
X
k=1
n
X
F (xk ) − F (xk−1 )
f (x∗k )∆x
k=1
n
X
lim F (b) − F (a) = lim
f (x∗k )∆x
n→∞
n→∞
Z
F (b) − F (a) =
(the left hand side is constant with respect to n)
k=1
b
f (x) dx
■
a
Justification for Part 1 of the Fundamental Theorem of Calculus
Z x
d
If f (x) is continuous on [a, b], then
f (t) dt = f ′ (x), for a ≤ t ≤ b.
dx a
Justification.
Since f (x) is continuous on [a, b], it is integrable. Let F be any antiderivative of of f on [a, b]; in
′
other words, F (t) = f (t) on [a, b]. Then by Part 2 of the Fundamental Theorem of Calculus, we have
Z x
xi
d
d h
d f (t) dt =
=
F (t)
F (x) − F (a)] = F ′ (x) − 0 = f (x)
dx a
dx
dx
a
§5.3 Fundamental Theorem of Calculus
MTH 1410 Blank Notes
Page 7 of 9
Your Turn Solutions
Z x
Z x
1. The graph of f is shown below. Let A(x) =
f (t) dt and F (x) =
f (t) dt be two area functions for f .
Evaluate each of the following expressions.
(a) A(5)
0
5
(b) F (5)
Z
5
5
Z
f (t) dt = −7
A(5) =
F (t) =
0
f (t) dt = 0
5
(c) A(15)
(d) F (15)
Z
A(15) =
15
Z
15
F (15) =
f (t) dt
0
f (t) dt
5
= −7 + 20 + (−9)
= 20 + (−9)
= 4
= 11
2. Simplify the following expressions. Don’t forget the Chain Rule!
Z 5x
2
d
(a)
et dt
dx 2
d
dx
Z
d
(b)
dx
Z
d
dx
Z
5x
2
2
et dt = e(5x) (5) = 5e25x
2
2
π
sin t3 dt
0
0
π
sin t3 dt = 0 because
Z
0
π
sin t3 dt will be a constant, and the derivative of a constant is zero!
§5.3 Fundamental Theorem of Calculus
MTH 1410 Blank Notes
Page 8 of 9
3. Evaluate each definite integral.
Z −1
(a)
(2x − 1) dx
We don’t care if the integral is “backwards”. The FTC will always work itself out!
0
Z
−1
h
i−1
(2x − 1) dx = x2 − x
= (−1)2 − (−1) − (0)2 − (0) = 2 − 0 = 2
0
0
Z
2
(b)
1
Z
2
5t6 + t − 1
dt =
t2
1
Z
1
5t + − t−2
t
4
Z 2
1
5t6
t
1
+ 2− 2
t2
t
t
2
Z
dt =
5t4 +
1
1
− t−2
t
dt
i2
h
0 1
61
*
+ ln 2
dt = t5 + ln |t| + t−1 = (2)5 + ln 2 + − (1)2 + ln 1+1 =
2
2
1
π/3
sec2 θ dθ
(c)
π/4
Z
π/3
sec2 θ dθ = tan θ
Z
= tan
π/4
π/4
(d)
π/3
π
3
− tan
π
4
=
√
3−1
π
(2eu + cos u) du
0
Z
1
h
iπ
0
:−0 2e
:
0
(2eu + cos u) du = 2eu + sin u = 2eπ +
sinπ
+
sin
0 = 2eπ − 2
π
0
0
Z
3
2
Z
(2z − 1) dz =
(e)
1
Z
Z
4z 2 − 4z + 1 dz
1
3
4 3
4z − 4z + 1 dz =
z − 2z 2 + z
3
2
1
(f)
3
3
−3 x
2
3
1
4
= (3)3 − 2(3)2 + 3 −
3
9
dx
+9
Z 3
x 3
9
1
1
dx = 9 2
dx = 18
arctan
2
2
2
3
3 0
−3 x + 9
0 x +3
π
3π
= 6 arctan(1) − arctan(0) = 6
−0 =
4
2
Z
3
4 3
(1) − 2(1)2 + 1
3
= 21 −
1
62
=
3
3
MTH 1410 Blank Notes
§5.3 Fundamental Theorem of Calculus
Page 9 of 9
4. Use symmetry to evaluate each integral.
Z π
(a)
sin3 x dx
−π
f (x) = sin3 x = (sin x)3 is the composition of two odd functions, so it is also odd. Therefore,
Z π
sin3 x dx = 0
−π
Z
5
(|x| − 3) dx
(b)
−5
Z
5
Z
(|x| − 3) dx = 2
5
(x − 3) dx
−5
0
5
1 2
= 2 x − 3x
2
0
1 2
=2
(5) − 3(5) − 0
2
= −5
5. Find the point(s) at which the function f (x) = x2 − 3x + 1 equals its average value on the interval [1, 13].
f=
=
=
=
Z 13
1
x2 − 3x + 1 dx
13 − 1 1
13
1 1 3 3 2
x − x +x
12 3
2
1
1 1
3
1 3 3 2
(13)3 − (13)2 + 13 −
(1) − (1) + 1
12 3
2
3
2
1 2951 1
+
12
6
6
f = 41
f (x) = f
x2 − 3x + 1 = 41
x2 − 3x − 40 = 0
(x − 8)(x + 5) = 0
X
x=8, X
x=
−5
X
X