第二节·微积分基本定理

Fundamentals of Advanced Mathematics (I)
Xueli Wang School of Science, BUPT Tel: 6228-2117 E-Mail: wangxlpku@
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Section 5.2
The Fundamental Theorems of Calculus
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Newton-Leibniz Formula
Definition. (Primitive function) If F ′( x ) = f ( x ) , x ∈ I , then F ( x ) is called an antiderivative of the function f ( x ) on I .
For instance, sin x is an antiderivative of cos x and ln x 1 . is an antiderivative of x The evaluation of a definite integral is closely related to the antiderivative of the integrand function.
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Newton-Leibniz Formula
Example: Suppose that a particle moves along a straight line from t = a to t = b . If the velocity v = v ( t ) is known, then by the definition of definite integral we know that
s = ∫ v ( t )dt
a b
if the displacement function, s = s( t ) , is known, then
s = s( b ) − s( a )
Hence, we have
∫
b a
v ( t )dt = s(b ) − s(a )
It is well known that s′( t ) = v ( t ) or s( t ) is an antiderivative of v ( t ) , then, by the last equation, we can establish the following theorem.
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Newton-Leibniz Formula
Theorem: (Newton-Leibniz formula) If antiderivative of f ( x ) on [a , b], then
f ∈ R[a , b] , and F ( x ) is an
b
∫
b a
f ( x )dx = F (b ) − F (a ) = F ( x ) a = [ F ( x )]a
b
d
(
)
Newton, Isaac (1642-1727) English physicist and mathematician
Leibniz, Gottfried (1646-1716) German philosopher, physicist, and mathematician
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Newton-Leibniz Formula
Proof: Arbitrarily insert n − 1 points of partition between a and b such that Then,
a = x0 < x1 < " < xn−1 < xn = b
F (b ) − F (a ) = [ F ( xn ) − F ( xn−1 )] + [ F ( xn−1 ) − F ( xn− 2 )] + " + [ F ( x2 ) − F ( x1 )] + [ F ( x1 ) − F ( x0 )] = ∑ [ F ( xk ) − F ( xk −1 )]
k =1 n
Since F ′( x ) = f ( x ) , so F ( x ) is continuous. By the mean value theorem of differential calculus gives
F ( xk ) − F ( xk −1 ) = F ′(ξ k )( xk − xk −1 ) = f ( ξ k ) Δxk
where xk −1 ≤ ξ k ≤ xk and k = 1, 2,", n .
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Newton-Leibniz Formula
Proof (continued) Since the antiderivative F ( x ) of f ( x ) exists and F ′( x ) = f ( x ) , then
F ( x ) is continuous, hence we may choose the
ξ k ∈ [ x k −1 , x k ]
in the Riemann sum as the one obtained by the mean value theorem. Then
F ( b ) − F (a ) = ∑ f (ξ k )Δxk
Taking limit on both sides we have
n d →0 k =1 k =1
n
F (b ) − F (a ) = lim ∑ f (ξ k )Δxk = ∫ f ( x )dx
b a
Finish.
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Newton-Leibniz Formula
Example Find the following definite integrals:
(1) (2)
∫
1 0
x 2 dx
∫
π
2 0
sin xdx
Solution:
1 1 3 2 = x dx = x (1) ∫0 3 0 3
1
1
(2)
∫
π
2 0
sin xdx = − cos x 02 = 0 − ( −1) = 1
π
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