MATH 12002 - CALCULUS I 4.3: The Evaluation Theorem Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.3: The Evaluation Theorem

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 9

Evaluation Theorem

Evaluation Theorem

If y = f (x) is a continuous function on [a, b], then b

a

f (x) dx = F(b) − F(a), where F is any antiderivative of f .

D.L. White (Kent State University) 2 / 9

Evaluation Theorem

Notes: The Evaluation Theorem is actually Part 2 of the Fundamental Theorem of Calculus (see text, page 237). This generalizes the solution to the distance problem: b

a v(t) dt = displacement = s(b) − s(a).

See pages 224–225 of the text for a proof of the Evaluation Theorem. If G is another antiderivative of F, then we know G(x) = F(x) + C for a constant C. Hence G(b) − G(a) = (F(b) + C) − (F(a) + C) = F(b) + C − F(a) − C = F(b) − F(a), and so it does not matter what antiderivative we choose. For convenience, we will use the notation F(x)|b

a = F(b) − F(a).

D.L. White (Kent State University) 3 / 9

Evaluation Theorem

Because of this relationship between the antiderivative and the definite integral, we use the notation

• f (x) dx

for the general antiderivative of f , and we call

• f (x) dx the indefinite integral of f .

So for example,

• x2 dx = 1

3x3 + C

and

• 2x + cos x dx = x2 + sin x + C.

Note that b

a f (x) dx is a NUMBER,

whereas

• f (x) dx is a family of FUNCTIONS.

D.L. White (Kent State University) 4 / 9

Evaluation Theorem

We can translate our properties of antiderivatives into properties of indefinite integrals:

• xn dx

= 1 n + 1xn+1 + C (n = −1)

• sin x dx

= − cos x + C

• cos x dx

= sin x + C

• kf (x) dx

= k

• f (x) dx
• [f (x) ± g(x)] dx

=

• f (x) dx ±
• g(x) dx

D.L. White (Kent State University) 5 / 9

Examples

1 Evaluate

3

−1(6x2 + 5) dx.

3

−1

(6x2 + 5) dx = (2x3 + 5x)

• 3

−1

= [2(33) + 5(3)] − [2(−1)3 + 5(−1)] = (54 + 15) − (−2 − 5) = 69 − (−7) = 76.

2 Evaluate

π

0 sin x dx.

π sin x dx = − cos x|π = (− cos π) − (− cos 0) = −(−1) − (−1) = 1 + 1 = 2.

D.L. White (Kent State University) 6 / 9

Examples

3 Evaluate

4

1 (x−2 + 2√x − 4x) dx.

4

1

(x−2 + 2√x − 4x) dx = 4

1

(x−2 + 2x1/2 − 4x) dx =

• −x−1 + 4

3x3/2 − 2x2

• 4

1

= [−4−1 + 4

3(43/2) − 2(42)] −

[−1−1 + 4

3(13/2) − 2(12)]

= [−1

4 + 4 3(8) − 32] − [−1 + 4 3 − 2]

= −1

4 + 32 3 − 32 + 1 − 4 3 + 2

= −239

12 ≈ −19.9.

D.L. White (Kent State University) 7 / 9

Examples

4 Evaluate

1

0 (x2 + 1)(x3 + 5) dx.

1 (x2 + 1)(x3 + 5) dx = 1 (x5 + x3 + 5x2 + 5) dx = 1

6x6 + 1 4x4 + 5 3x3 + 5x

• 1

= 1

6(16) + 1 4(14) + 5 3(13) + 5(1)

• −

1

6(06) + 1 4(04) + 5 3(03) + 5(0)

• =

1

6 + 1 4 + 5 3 + 5

• − (0)

=

85 12 ≈ 7.08.

D.L. White (Kent State University) 8 / 9

Examples

Note that the Evaluation Theorem is valid only if f is continuous on [a, b]:

5 Consider

1

−1 1 x4 dx; f (x) = 1 x4 is not continuous at x = 0.

Since

1 x4 is always positive where it is defined,

if 1

−1 1 x4 dx exists, it has to be positive.

However,

1 x4 = x−4 has antiderivative −1 3x−3 = − 1 3x3 , and

• − 1

3x3

• 1

−1

=

• −

1 3(13)

• −
• −

1 3(−1)3

• =
• −1

3

• −
• − 1

−3

• =
• −1

3

• −

1 3

• = −2

3, and so the formula from the Evaluation Theorem fails.

D.L. White (Kent State University) 9 / 9