MATH 12002 - CALCULUS I 4.4: Fundamental Theorem of Calculus - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.4: Fundamental Theorem of Calculus

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 8

The Fundamental Theorem

Fundamental Theorem of Calculus

Let f be continuous on [a, b].

1 If we define

g(x) = x

a

f (t) dt, then g′(x) = f (x) for all x in (a, b).

2 If F is any antiderivative for f , then

b

a

f (x) dx = F(b) − F(a).

D.L. White (Kent State University) 2 / 8

The Fundamental Theorem

Notes: Part 1 says that the function g is an antiderivative of f . In particular, every continuous function has an antiderivative. Part 1 also says that if we integrate f (i.e., to get x

a f (t) dt)

and then differentiate, we get f back; that is, d dx x

a

f (t) dt = f (x). Part 2 says that if we differentiate F (i.e., to get f ) and then integrate, we get F back (except for a constant); that is, x

a

F ′(t) dt = F(x) − F(a). Together, Parts 1 & 2 say that differentiation and integration are inverse processes. That Fundamental Theorem really ties the branches of calculus together, does it not?

D.L. White (Kent State University) 3 / 8

Proof of FTC

Assume f is continuous on [a, b] and define g(x) = x

a f (t) dt.

For Part(1), we need to show that g′(x) = f (x) for a < x < b; that is, lim

h→0

g(x + h) − g(x) h = f (x). We first informally outline the idea of the proof of Part (1). Note that g(x + h) = x+h

a

f (t) dt and g(x) = x

a f (t) dt.

For a positive function f and positive h, we have the following picture:

✲ ✻

f

a b x

q

f (x) x + h

q

f (x + h)

g(x + h) − g(x) = R x+h

a

f (t) dt − R x

a f (t) dt

= area under graph on [x, x + h] ≈ area of rectangle

h f (x) = h · f (x)

Hence for small h,

g(x+h)−g(x) h

≈ f (x) and so f (x) = lim

h→0 g(x+h)−g(x) h

= g′(x).

D.L. White (Kent State University) 4 / 8

Proof of FTC

We now prove Part (1) more rigorously. By one of our properties of definite integrals, g(x + h) = x+h

a

f (t) dt = x

a f (t) dt +

x+h

x

f (t) dt = g(x) + x+h

x

f (t) dt, and so g(x + h) − g(x) = x+h

x

f (t) dt. Therefore, g(x + h) − g(x) h = 1 h x+h

x

f (t) dt.

D.L. White (Kent State University) 5 / 8

Proof of FTC

Since f is continuous on the closed interval [x, x + h], it has an absolute maximum and an absolute minimum on the interval. Thus there is a number u, x u x + h, such that f (u) is the absolute minimum value of f on the interval, and there is a number v, x v x + h, such that f (v) is the absolute maximum value of f on the interval. Hence on the interval [x, x + h], we have f (u) f (t) f (v). By a comparison property of integrals, we have f (u)((x + h) − x) x+h

x

f (t) dt f (v)((x + h) − x); that is, f (u)h x+h

x

f (t) dt f (v)h, and finally f (u) 1 h x+h

x

f (t) dt f (v).

D.L. White (Kent State University) 6 / 8

Proof of FTC

Recalling that

1 h

x+h

x

f (t) dt = g(x+h)−g(x)

h

and f (u) 1

h

x+h

x

f (t) dt f (v), we now have f (u) g(x + h) − g(x) h f (v). Since x u x + h & x v x + h, we have u → x & v → x as h → 0. Since f is continuous, this implies lim

h→0 f (u) = lim u→x f (u) = f (x) and lim h→0 f (v) = lim v→x f (v) = f (x).

Therefore, f (x) = lim

h→0 f (u) lim h→0

g(x + h) − g(x) h lim

h→0 f (v) = f (x),

and so by the Squeeze Theorem g′(x) = lim

h→0

g(x + h) − g(x) h = f (x).

D.L. White (Kent State University) 7 / 8

Proof of FTC

We now have that Part (1) of the Fundamental Theorem holds, and so g(x) = x

a f (t) dt is an antiderivative of f .

If F is any antiderivative of f , then F(x) = g(x) + C for some constant C. In particular, F(b) = g(b) + C = b

a

f (t) dt + C, and so b

a f (t) dt = F(b) − C.

But also, F(a) = g(a) + C = a

a

f (t) dt + C = 0 + C = C, and so C = F(a). Therefore b

a

f (t) dt = F(b) − F(a) and Part (2) of the Fundamental Theorem holds.

D.L. White (Kent State University) 8 / 8