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The definite integral of the function y = f(x) on the interval [a, b] is denoted by. ∫ b a f(x)dx and is defined by the limit.
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The definite integral of the function y = f (x) on the interval [a, b] is denoted by (^) ∫ b a
f (x) dx
and is defined by the limit
∫ (^) b
a
f (x) dx = (^) nlim→∞
∑^ n
j=
f (x∗ j ) · ∆xj
where for each n:
max(∆xj : 1 ≤ j ≤ n)
Comment: Computing definite integrals this way is usually computa- tionally intensive and frequently very difficult (if not impossible).
Happily, there is
The Fundamental Theorem of Calculus:
If F ′(x) = f (x), then ∫ (^) b
a
f (x) dx = F (b) − F (a).
Observations:
∫ (^) b
a
f (x) dx does not depend on which antiderivative of f (x) we use, because if G(x) = F (x) + C, then G(b) − G(a) = (F (b) + (^) C) − (F (a) + (^) C) = F (b) − F (a).
Example 1. Use the FTC to calculate
1
x^2 dx.
We know that
x^2 dx = x
3 3 +^ C, so ∫ (^3)
1
x^2 dx =^3
3 3 −^
Notation: We denote the difference F (b) − F (a) by
F (b) − F (a) = F (x)
b a
This makes applying the FTC a little more smooth notationally speaking.
In the previous example we can write: ∫ (^3)
1
x^2 dx = x
3 3
3
1
3 3 −^
Example 2. Compute
∫ (^2) 0
2 x^3 − 3 x^2 + 4x − 5 dx.
Using the FTC we have ∫ (^2) 0
2 x^3 − 3 x^2 + 4x − 5 dx =
( (^1) 2 x
(^4) − x (^3) + 2x (^2) − 5 x
)∣∣ ∣∣
2 0 =
( (^16) 2 −^ 8 + 8^ −^10
) − (0 − 0 + 0 − 0) = − 2
Example 3. Compute
∫ (^4) 1
x √ − 2 x dx. Using the FTC we have ∫ (^4) 1
x √ − 2 x dx^ =
∫ (^4) 1
x^1 /^2 − 2 x−^1 /^2 dx
=
( (^2) 3 x
3 / (^2) − 4 x 1 / 2
)∣∣ ∣∣
4 1 =
( 16 3 −^8
) −
( 2 3 −^4
) =^23
nlim→∞
( (^) ∑n
j=
F ′(xj− 1 )∆xj
) = F (b) − F (a)
(*) On the other hand, if F ′(x) = f (x), then from the Definition of the definite integral (using left-hand sums) we have ∫ (^) b a
f (x) dx = (^) nlim→∞
( (^) n ∑ j=
f (xj− 1 )∆xj
)
= (^) nlim→∞
( (^) ∑n
j=
F ′(xj− 1 )∆xj
) = F (b) − F (a).
Comment: This is the outline of a proof, but some important details are missing. For example, while it is true that F ′(xj− 1 )∆xj ≈ F (xj ) − F (xj− 1 ) when ∆xj is sufficiently small, there is still a small error. This means that the approximation ∑^ n j=
F ′(xj− 1 )∆xj ≈
∑^ n j=
(F (xj ) − F (xj− 1 ))
entails a sum of n small errors. The explanation of why the sum of these n errors is (very) small when n is large is a missing detail.
Substitution in a definite integral.
Example 4. Compute
0
3 x + 1
dx
Once again, using the FTC and the substitution
u = 3x + 1 =⇒ dx =^13 du,
we have ∫ (^1)
0
3 x + 1
dx =^53
?
u−^1 /^2 du =^103 u^1 /^2
?
?
(*) When making a substitution in a definite integral, the limits of integration change with the substitution. E.g., in this problem, if u = 3x + 1, then x = 0 =⇒ u = 1 and x = 1 =⇒ u = 4, so ∫ (^1)
0
3 x + 1
dx =^53
1
u−^1 /^2 du =^103 u^1 /^2
4
1
Properties of definite integrals:
These properties are easy to justify using the FTC, but they can all be justified using the definition as well. The first two are direct analogs of the same properties for indefinite integrals, while the last two do not have indefinite integral counterparts.
∫ (^) b
a
f (x) ± g(x) dx =
∫ (^) b
a
f (x) dx ±
∫ (^) b
a
g(x) dx
∫ (^) b
a
Cf (x) dx = C
∫ (^) b
a
f (x) dx
∫ (^) b
a
f (x) dx =
∫ (^) c
a
f (x) dx +
∫ (^) b
c
f (x) dx
∫ (^) a
b
f (x) dx = −
∫ (^) b
a
f (x) dx
