http://www.ck12.org Chapter 4. Integration
We first need to divide[a,b]intonsub-intervals of length 4 x=b−na. We letx 0 =a,x 1 ,x 2 ,...,xn=bbe the endpoints
of these sub-intervals.
LetFbe any antiderivative off.
ConsiderF(b)−F(a) =F(xn)−F(x 0 ).
We will now employ a method that will express the right side of this equation as a Riemann Sum. In particular,
F(b)−F(a) =F(xn)−F(x 0 )
=F(xn)−F(xn− 1 )+F(xn− 1 )−F(xn− 2 )+F(xn− 2 )−...+F(x 1 )−F(x 0 )
=n
∑ 1 [F(xi)−F(xi− 1 )].Note thatFis continuous. Hence, by the Mean Value Theorem, there existci∈[xi−xi− 1 ,xi]
such thatF(xi)−F(xi− 1 ) =F′(ci)(xi−xi− 1 ) =f(ci) 4 x.
Hence
F(b)−F(a) =n
∑ 1 F′(ci)(xi−xi−^1 ) =n
∑ 1 f(ci)^4 x.Taking the limit of each side asn→∞we have
nlim→∞[F(b)−F(a)] =nlim→∞n
∑ 1 f(ci)^4 x.We note that the left side is a constant and the right side is our definition for∫abf(x)dx.
Hence
F(b)−F(a) =nlim→∞n
∑ 1 f(ci)^4 x=∫b
a f(x)dx.Proof of Theorem 4.2
LetF(x) =∫axf(x)dx.
By the Mean Value Theorem for derivatives, there existsc∈[a,b]such that
F′(c) =F(bb)−−Fa(a).From Theorem 4.1 we have thatFis an antiderivative off.Hence,F′(x) =f(x)and in particular,F′(c) =f(c).
Hence, by substitution we have
f(c) =F(bb)−−Fa(a).