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FTCMVT

# FTCMVT - Math 173 Kouba MVT and F TC’s If f is a...

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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 Minimum-Value 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 =WW-Q*WW*Q =mm—mm b = F(1:) ...
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