2008 W -C L Calculus(I)

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Calculus (I)

W^EN-C^HING L^IEN

Department of Mathematics National Cheng Kung University

2008

WEN-CHINGLIEN Calculus (I)

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5.3 The Fundamental Theorem of Calculus.

Theorem (I)

If f is continuous on[a,b],then the function F defined by

F(x) = Z x

a

f(u)du, a≤x ≤b

is continuous on [a,b]and differentiable on(a,b),and d

dxF(x) =f(x)

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pf of Theorem I:

d

dx F ( x ) = lim

h→0

1 h

R

x+h

x

f ( u ) du

For u ∈ [ x , x + h] , m ≤ f (u ) ≤ M

⇒ m ≤ 1 h

R

x+h

x

f (u)du ≤ M

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pf of Theorem I:

d

dx F ( x ) = lim

h→0

1 h

R

x+h

x

f ( u ) du

For u ∈ [ x , x + h] , m ≤ f (u ) ≤ M

⇒ m ≤ 1 h

R

x+h

x

f (u)du ≤ M

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pf of Theorem I:

d

dx F ( x ) = lim

h→0

1 h

R

x+h

x

f ( u ) du

For u ∈ [ x , x + h] , m ≤ f (u ) ≤ M

⇒ m ≤ 1 h

R

x+h

x

f (u)du ≤ M

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⇒ I.V.T. ∃Ch∈ [x,x+h]

such that f(Ch) = Ih, lim

h→0Ch=x

∵f is continuous, lim

h→0f(Ch) =f(x)

⇒ d

dxF(x) =f(x). 2

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⇒ I.V.T. ∃Ch∈ [x,x+h]

h→0Ch=x

h→0f(Ch) =f(x)

⇒ d

dxF(x) =f(x). 2

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⇒ I.V.T. ∃Ch∈ [x,x+h]

h→0Ch=x

h→0f(Ch) =f(x)

⇒ d

dxF(x) =f(x). 2

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Remark:

(1)F(x) =Rx

a f(u)du

⇒ F^′(x) =f(x)

⇒ F(x)is an anti-derivative of f(x)

We call R

f(x)dx an indefinite integral.

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Remark:

(1)F(x) =Rx

a f(u)du

⇒ F^′(x) =f(x)

We call R

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Remark:

(1)F(x) =Rx

a f(u)du

⇒ F^′(x) =f(x)

We call R

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Remark:

(1)F(x) =Rx

a f(u)du

⇒ F^′(x) =f(x)

We call R

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Exercises:

1

R ( sinx + x

³

) dx

2

R 1

x dx

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Exercises:

1

R ( sinx + x

³

) dx

2

R 1

x dx

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Theorem (Leibniz^′s Rule) d

dx Z h(x)

g(x)

f(u)du =f(h(x))h^′(x) −f(g(x))g^′(x) Here f is continuous ,g and h are differentiable

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Exercises:

1 d

dx Rx

0(e^u²−u)du

2 d

dx Rsinx

0 (x²−2x)dx

3 d

dx Rx³

x² udu

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Exercises:

1 d

dx Rx

0(e^u²−u)du

2 d

dx Rsinx

0 (x²−2x)dx

3 d

dx Rx³

x² udu

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Exercises:

1 d

dx Rx

0(e^u²−u)du

2 d

dx Rsinx

0 (x²−2x)dx

3 d

dx Rx³

x² udu

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Theorem (II)

Assume that f is continuous on [a,b],them Z b

a

f(x)dx =F(b) −F(a), where F(x)is an anti-derivative of f(x)

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pf of Theorem II:

G(x ) ≡ R

x

a

f (u)du

⇒ G(x) = F (x ) + C

⇒ G(a) = 0, G(a) = F (a) + C

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pf of Theorem II:

G(x ) ≡ R

x

a

f (u)du

⇒ G(x) = F (x ) + C

⇒ G(a) = 0, G(a) = F (a) + C

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pf of Theorem II:

G(x ) ≡ R

x

a

f (u)du

⇒ G(x) = F (x ) + C

⇒ G(a) = 0, G(a) = F (a) + C

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⇒ C= −F(a)

⇒ G(x) = F(x) −F(a)

⇒ Rx

a f(u)du =F(x) −F(a) 2

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⇒ C= −F(a)

⇒ G(x) = F(x) −F(a)

⇒ Rx

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⇒ C= −F(a)

⇒ G(x) = F(x) −F(a)

⇒ Rx

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Exercises:

1 R2

1 x²− 2 xdx

2 R^π

0 cosx dx

3 R1

0 xe⁻²^{x 2}dx

4 Re

1

1 xdx

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Exercises:

1 R2

1 x²− 2 xdx

2 R^π

0 cosx dx

3 R1

0 xe⁻²^{x 2}dx

4 Re

1

1 xdx

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Exercises:

1 R2

1 x²− 2 xdx

2 R^π

0 cosx dx

3 R1

0 xe⁻²^{x 2}dx

4 Re

1

1 xdx

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Exercises:

1 R2

1 x²− 2 xdx

2 R^π

0 cosx dx

3 R1

0 xe⁻²^{x 2}dx

4 Re

1

1 xdx

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Example:

Compute the following integrals

R (x −1)² R cos3x dx R1

0(x²−√ x)dx

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Example:

Compute the following integrals

0(x²−√ x)dx

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Example:

Compute the following integrals

0(x²−√ x)dx

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Exercises:

R

2

0

2te

^t²

dt

h

lim

→0

1 h

R

h 0

e

^x

dx

Suppose that R

x

0

f ( t ) dt = 2x

²

, find f ( x )

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Exercises:

R

2

0

2te

^t²

dt

h

lim

→0

1 h

R

h 0

e

^x

dx Suppose that R

x

0

f ( t ) dt = 2x

²

, find f ( x )

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Exercises:

R

2

0

2te

^t²

dt

h

lim

→0

1 h

R

h 0

e

^x

dx Suppose that R

x

0

f ( t ) dt = 2x

²

, find f ( x )

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Example:

f(x) =

0 , if x is irrational,

1 , if x is rational, for 0≤ x ≤1.

DoesR1

0 f(x)dx exist ?

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