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- 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
- 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
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.
If we have to find an integral of the form
∫^ 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.
