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Unformatted text preview: Math 173
Kouba MVT and F TC’s If f is a continuous function on the closed interval 3’ 41—
etc) Mean Value Theorem for Integrals :
[a,b], then there is at least one number c , a S c _<_ b , so that b
f(C)(b—a)=/ f(a:)dz.
a a o b Proof : Since f is a continuous function on the closed interval [a,b] , by the Maximum and
MinimumValue Theorems, f has a maximum vaule M and a minimum value m on [0, b], i.e., m S f (1:) S M on [a,b] . By property l'\. of deﬁnite integrals,
7 1c M b
m(b—a) 3/ 1(2) dz: 5 M(b — a), so that .yo
1 6
ms ,M/a imideM,
‘T """—“ X
call this number y,
By the Intermediate Value Theorem there is at least one number c , a S c g b , so
that
1 b
rw=ymuqﬂo=b '/fbwm
so that b
new: — a) = f f(z)d:r. First Fundamental Theorem of Calculus (FTCI) : Assume that f is a continuous function 3
on the closed interval [a,b] and that F(z) = / f(t)dt . Then F’ (2:) = f(:r) .
G 2
Proof : Consider F(z) = f f (t) dt as the area under the graph of f above the interval [0, 1:] .
Then F(z+h) is the area urider the graph of f above the interval [a, z+h] and F(z+h) — F(z) z+h
is the area of the “ thin strip ” from a: to z+ h, i.e., F(.r + h) — F(z) = / f(t) dt . By the
Mean Value Theorem for integrals there is at least one number c , x S c :5 :r + h , so that 4: The derivative of F(z) can now be computed as m) = lim h+0 h.
z+h
/ f(t) dt
zﬂtx h
__ . h ‘E& h x Ca x+k
= lim f(c) (Recall that a: g c S a: + h.)
h+0
=ﬂﬂ Second Fundamental Theorem of Calculus (FTC2) : Let f be a continuous function on the closed interval [a,bI. Assume that F(:r) is an antiderivative of f(x), i.e., assume that F’ (2:) = f(x). Then :=Fm—me [5 f(x) dz = M) Proof : Let A(x) = [3 f(t) dt . Then A(a) = 0, A(b) = /b f(t) dt, and A’(z) = f(z) by FTCI. But F’(x) = flz) . By Corollary to the“ Mean Value Theorem F(a:) =
A(z) + C for any constant C', or A(x) = F(:c) — C .
Then 6 b
[a f(z)dx= / f(t)dt
=Am =Mm—Mn
=WWQ*WW*Q
=mm—mm b
= F(1:) ...
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 Winter '07
 Kouba

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