2.4 The Fundamental Theorem of Calculus – Part II
In the previous section we discussed the first part of The Fundamental Theorem of
b
 f(x) dx by first evaluating the
Calculus. It said that we can evaluate a definite integral 
a
b
 f(x) dx. Then 
 f(x) dx = F(b) - F(a).
corresponding indefinite integral F(x) = 
a
The second part of the Fundamental Theorem says that we can construct an indefinite
integral of a function by means of the definite integral with a variable upper limit of
integration. Here is the statement of the second part.
Theorem (Fundamental Theorem of Calculus – Part II). Let f(x) be a continuous
x
a
dA(x)
dL(x)

function. If A(x) = 
f(t)
dt,
then
=
f(x).
If
L(x)
=
f(t)
dt,
then


dx
dx = - f(x).
a
x
x
Recall that 
 f(t) dt is the area A of the region R bounded by the curve y = f(t), the t-axis
a
and the lines t = a and t = x. What we are doing is freezing the lower limit t = a and
letting the upper limit t = x vary. This creates a function A(x) which is cummulative area
under the curve f(t) from t = a to t = x. Part II of the Fundamental Theorem says that the
derivative of A(x) is f(x).
y
y
y
y
F x
f t
F x
x
t
a
x
2.4 - 1
a
x
x
 f(t) dt = 
 (2t+1) dt. A(x)
Example 1. Suppose f(x) = 2x + 1 and a = 1. Then A(x) = 
a
1
1
2
is the area of a trapezoid which is (b1 + b2)h where h is the width and b1 and b2 are the
lengths of the two parallel sides. In this case h = (x-1), b1 = 3 and b2 = 2x + 1, so
x
dA(x)
 (2t+1) dt = 21 (3 + (2x+1))(x-1) = (x+2)(x-1) = x2 + x - 2. So
A(x) = 
dx = 2x + 1 = f(x)
1
which is what we expect from the Fundamental
Theorem.
y
y
2t
1
One of the main applications of the Part II of the
Fundamental Theorem is to construct indefinite
F x
integrals of functions whose indefinite integrals are
1
x
not expressable in terms of ordinary functions. A
2
good example is 
 sin(x ) dx. It turns out there is no function that can be expressed in
t
terms of familiar functions whose derivative is sin(x2). Using Part II of the Fundamental
Theorem we can construct one using definite integrals. One such function is
x
2
S(x) = 
 sin(t ) dt.
0
x
dg(x)
2
20
Example 2. Suppose g(x) = 
 (t - 1) dt. What is dx ?
1
We could solve this by multiplying (t2 - 1)20 out, integrating and then taking the
derivative. However, it is easier to use Part II of the Fundamental Theorem which says
dg(x)
2
20
dx = (x - 1) .
2
dh(x)
2
Example 3. Suppose h(x) = 
 1 + t dt. What is dx ?
x
dh(x)
By the second assertion of Part II of the Fundamental Theorem one has dx =
x
2

 sin(t ) dt
Example 4. Suppose k(x) =
1
sin x
dk(x)
. What is dx ?
Using the quotient rule one has
2.4 - 2
1 + x2.
dx
  x

2
 sin(t ) dt - 
 sin(t2) dt (cos x)
(sin x) dx
 
 

dk(x)
1
1
dx =
sin2 x
Using Part II of the Fundamental Theorem gives
x
2
(sin x) (sin(x )) - (cos x) 
 sin(t ) dt
2
dk(x)
dx = =
1
sin2 x
Proof of Part II of the Fundamental Theorem of Calculus. One has
dA(x)
A(x + h) - A(x)
=
lim
dx
h
h0
(1)
x
 f(t) dt is the area under the curve y = f(t) from t = a to t = x. Also
Note that A(x) = 
a
x+h
 f(t) dt is the area under the curve y = f(t) from t = a to t = x+h. So the
A(x+h) = 
a
difference
A(x+h) - A(x)
= Bh
is the area under the curve y = f(t) from t = x to t = x+h. Putting this into (1) we get
(2)
dA(x)
Bh
dx = hlim
0 h
Draw a horizontal line extending from the point (x, f(x)) to the point (x+h, f(x)). This
divides the area Bh into two pieces. One piece is a rectangle bounded by the lines t = x,
y = f(x), t = x+h and the x axis. It has area equal to (base)(height) = hf(x). The second
piece is bounded by the lines t = x, y = f(x), t = x+h and the curve y = f(t). Let's call this
area Ch. So Bh = hf(x) + Ch. Putting this into (2) we get
dA(x)
Ch

= lim f(x) +
dx
h 

h0
Ch
= f(x) + lim h
h0
In order to complete the proof we need to show
(3)
Ch
lim h = 0
h0
2.4 - 3
Suppose for the moment that y = f(t) is increasing between t = x and t = x+h. Then the
area Ch is contained in the rectange bounded by the lines t = x, y = f(x), t = x+h and
y = f(x+h). The area of this rectangle is (base)(height) = h[f(x+h) – f(x)]. So
0  Ch  h[f(x+h) – f(x)]
and
0 
Ch
h  f(x+h) – f(x)
So
Ch
0  lim h  lim f(x+h) – f(x)
h0
h0
Since y = f(t) is continuous, we know that
lim f(x+h) – f(x)
h0
= 0
dA(x)
dx = f(x). This completes the proof of the first assertion in Part II
of the Fundamental Theorem of Calculus. To prove the second, note that L(x) = - A(x), so
the second assertion follows from the first. //
So (3) is true and
Proof of Part I of the Fundamental Theorem of Calculus. The proof uses Part II of
x
the Fundamental Theorem of Calculus. As in Part II, let A(x) = 
 f(t) dt. Then
a
dA(x)
dF(x)
dA(x) dF(x)
=
f(x).
From
the
hypotheses
of
Part
I
we
have
=
f(x).
So
dx
dx
dx = dx . So
dA(x) dF(x)
d
dx - dx = 0. So dx [A(x) - F(x)] = 0. However, we know that if the derivative of
a function is zero, then the function is a constant. So A(x) - F(x) = C, a constant. To
evaluate C we plug in x = a. In that case A(a) = 0, so C = - F(a). So A(x) = F(x) – F(a).
b
Now we plug in x = b. This gives A(b) = F(b) – F(a). However A(b) = 
 f(t) dt. So
a
b

 f(t) dt = F(b) – F(a), which is what we wanted to prove. //
a
2.4 - 4