Integration and Antiderivatives: Fundamental Theorem of Calculus and Substitution, Study notes of Calculus

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1. Antiderivative A function G is called an antiderivative for f on [a, b] iff

(i) G is continuous on [a, b] and

(ii) G′(x) = f (x) for all x in (a, b).

Notation: If G(x) is an antiderivative for f (x) on [a, b], then G(x) + C is also an antiderivative for f (x) for any constant C. We denote by

∫

f dx, an antiderivative of f (x).

Theorem 1. If G 1 (x) and G 2 (x) are antiderivatives for f (x) on [a, b], then there exists a constant C such that G 1 (x) − G 2 (x) = C for all x in [a, b].

Example: Let n be a positive integer. Then

d dx

[

n + 1

xn+

]

= xn,

and so G(x) = (^) n^1 +1 xn+1^ is an antiderivative of the function f (x) = xn. Therefore,

∫

xndx = (^) n+1^1 xn+1^ + C.

Example: Since

d dx

[ex] = ex, ∫ exdx = ex^ + C.

Example: We will prove later that

∫ 1 x

dx = ln|x| + C.

1

2. Connection Between Derivative and Integral : Fundamental Theorem of Calculus Let f (t) be a continuous function [a, b]. For each x in the interval [a, b], let’s define the function F (x) by

F (x) =

∫^ x

a

f (t)dt.

Then, F (x) is continuous on x on [a, b], and for all x in the open interval (a, b),

F ′(x) = f (x).

Thus, F (x) =

∫^ x a

f (t)dt is an antiderivative of f on [a, b].

Corollary 1. Let f (t) be a continuous function [a, b]. If G is an antiderivative of f on [a, b], then ∫b

a

f (t)dt = [G(x)]ba = G(b) − G(a).

Example: Let n be a positive integer. Then, Since G(x) = (^) n+1^1 xn+1^ is an antiderivative of the function f (x) = xn, we have that

∫^2

1

xndx = [

n + 1

xn+1]^21 =

n + 1

2 n+1^ −

n + 1

1 n+1^ =

n + 1

(2n+1^ − 1).

Example: Since G(x) = ex^ is an antiderivative of the function f (x) = ex, we have that ∫^2

1

exdx = [ex]^21 = e^2 − e.

Example: Since G(x) = 2

√

t is an antiderivative of the function f (t) = √^1 t , we have that ∫^9

4

√

t

dt = [

√

t]^94 = 2

√

9 − 2

√

4 = 2.

Definition 1.

∫^ b a

f (t)dt = −

∫^ a b

f (t)dt.

Some Properties of Integral:

Let f and g be continuous functions on [a, b]. Let a < c < b. Also, let α be a constant.

(1)

∫^ b a

[αf (t)]dt = α

∫^ b a

f (t)dt.

(2)

∫^ b a

[f (t) + g(t)]dt =

∫^ b a

f (t)dt +

∫^ b a

g(t)dt.

(3)

∫^ b a

[f (t) − g(t)]dt =

∫^ b a

f (t)dt +

∫^ b a

g(t)dt.

(4)

∫^ b a

f (t)dt =

∫^ c a

f (t)dt +

∫^ b c

f (t)dt.

Example: Evaluate

∫^1

0

(2x − 6 x^4 + 5)dx.

Since G(x) = x^2 − 65 x^5 + 5x is an antiderivative of the function f (x) = 2x − 6 x^4 + 5, we have that

∫^1

0

(2x − 6 x^4 + 5)dx = [x^2 −

x^5 + 5x]^10 = (1^2 −

15 + (5)(1)) − (0^2 −

05 + (5)(0)) =

.

3. Substitution Let g be a continuous function on [a, b] such that the derivative g′^ is also continuous on [a, b]. Let f be a function which is continuous on g([a, b]). Then,

(3.1)

∫^ b

a

f (g(x))g′(x)dx =

∫^ g(b)

g(a)

f (u)du,

Example: Evaluate

∫^2

1

2 x

√

x^2 − 1 dx.

We apply the formula (3.1). Let g(x) = x^2 − 1, and let f (x) =

√

x. Then we have that ∫^2

1

2 x

√

x^2 − 1 dx =

∫^2

1

f (g(x))g′(x)dx.

Also, F (x) = 23 x (^32) is an antiderivative for f (x) =

√

x. Note that, g(1) = 0, g(2) = 3, and g′(x) = 2x So by applying the formula (3.1), we get ∫^2

1

f (g(x))g′(x)dx =

∫^3

0

f (u)du =

∫^3

0

u

1 (^2) du =

[

u

3 2

] 3

0

=

[

3 (^2) − 0 3 2

]

=

3 (^2).

If we have to find an integral of the form

(3.2)

∫^ b

a

f (g(x))g′(x)dx =

∫^ g(b)

g(a)

f (u)du,

then we follow the following steps. We let u = g(x), and then du = g′(x)dx, and we change the limits of integration a and b by g(a) and G(b).

Example: Evaluate

∫^2

x=

2 xex 2 dx

Let u = x^2. Then du = 2xdx, and ∫^2

x=

2 xex^2 dx =

∫^4

u=

eudu = [eu]^41 = e^4 − e^1 = e^4 − e.