Drill: Find dy/dx

1. y = sin(x2)

2. y = sec2x – tan2 x 3. y = 2x

4. y’ = 2cos(x2 )5. Let u = sec x and v = tan x

du = secxtanx and dv = sec2 x

y = u2 – v2

y’ = 2u du - 2v dvy’ = 2(sec x)(secxtanx) - 2(tanx)

(sec2x)= 0

3. y’ =2x ln 2

Fundamental Theorem of Calculus

Lesson 5.4Day 1 Homework:

p. 302/3: 1-25 ODD

Objectives

• Students will be able to– apply the Fundamental Theorem of Calculus.– understand the relationship between the derivative

and definite integral as expressed in both parts of the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus, Part 1

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

has a derivative at every point x in [a, b], and

x

a

dttfxF

.xfdttfdxd

dxdF x

a

Example 1 Applying the Fundamental Theorem

Find

x

x

dttdxd

dttdx

d

sin

42

42 4

22

x

xsin

Note: It does not matter what a is, as long as b is ‘x.’

.xfdttfdxd

dxdF x

a

Example 2 The Fundamental Theorem with the Chain Rule

Find .cos if 2

1x

dttydxdy 2Let xu

u

tdty1

cos xdxdu 2

dxdu

dudy

dxdy

ududy cos xu

dxdy 2cos

xxdxdy 2sin 2

Example 3 The Fundamental Theorem with the Chain Rule

.cos1sin1 if 2

2

x

dttty

dxdy

xu Let

.cos1sin1

2

2

u

dttty

1dxdu

dxdu

dudy

dxdy

uu

dudy

2

2

cos1sin1

)1(cos1sin1

2

2

uu

dxdy

)(cos1)(sin1

2

2

xx

dxdy

Example 4 The Fundamental Theorem with the Chain Rule

Find . if 3

1

xx

t dtteydxdy xxu 3Let

u

t dttey1

dxdu

dudy

dxdy

13 2 xdxdu

13 23 3

xexx xx

uuedudy

)13( 2 xuedxdy u

Example 5 Variable Lower Limits of Integration

• Find dy/dx

• Need to use the rules of integrals. Remember:

• So,

• dy/dx = -3xsin(x)

• Find dy/dx5

sin3x

tdtty

a

b

b

a

dxxfdxxf )()(

x

x

tdtttdtty5

5

sin3sin3

2

2 21x

xt dt

ey

ab

a

b

dxxfdxxfdxxf00

)()()(

x

t

x

t dte

dte

y2

00 21

21

2

22

122

122

xx e

xedx

dy

xx eex

dxdy

222

22

2

Example 5 Constructing a Function with Given Derivative and Value

• Find a function y = f(x) with derivative dy/dx = tanx that satisfies the condition f(3) = 5.

• To construct a function with derivative of tanx

(always let the lower bound be the given value of x.)

• Remember that ,

• So, if x = 3, then

• Therefore, we would only need to add 5 to construct a function whose derivative is tan(x) and where f(3) = 5

x

tdty3

tan

0)( a

a

dxxf

0tan3

3

tdty

5tan)(3

x

tdtxf

Drill: Construct a function of the form that satisfies the given conditions.

x

a

Cdttfy )(

8;0;tan xyxedxdy x

3;4;cos3 xyxdxdy

x

t tdtey8

0tan

x

dtty3

4cos3

The Fundamental Theorem of Calculus, Part 2

If f is continuous on [a, b], and F is any antiderivative of f on the interval, then

This part is also called the Integral Evaluation Theorem.

.aFbFdxxfb

a

Example 6 Evaluating an Integral

Evaluate using an antiderivative.

5

3

3 564 dxxx

5

3

245

3

3 53564

xxxdxxx

3533355535 2424

536

4

1ln2 2

3

xx

Example 6 Evaluating an Integral

Evaluate using an antiderivative.

4

1

13 dxx

x

4

1

14

1

21

313 dxxxdxx

x

1ln124ln42 23

23

024ln82

4ln14

How to Find Area Analytically

To find the area between the graph of y = f(x) and the x-axis over the interval [a,b] analytically,

1) Graph the equation.2) Partition [a,b] with the zeros of f.3) Integrate f over each subinterval4) Add the absolute values of the integrals

Find the area of the region between the curve y = 4- x2, [0, 3], and the x-axis

1) Graph the equation.2) Partition [a,b] with the zeros of

f.3) Integrate f over each

subinterval4) Add the absolute values of the

integrals

• The first region, where the graph is positive, is from [0, 2]. The next region, where the graph is negative, is from [2, 3]

3

160)388(

344

2

0

32

0

2

xxdxx

37

316)912(

344

3

2

33

2

2

xxdxx

323|

37||

316|

How to find Total Area on the Calculator

• To find the area between the graph of y = f(x) and the x-axis over the interval [a, b] numerically, evaluate fnInt(|f(x)|, x, a, b) on the calculator.

• Try the previous example: fnInt(|4-x2|, x, 0, 3)• 7.66666667

Homework

• Page 303: 27-47 (ODD)