Warmup 1) 2). 5.4: Fundamental Theorem of Calculus

warmup

1)

2)

5.4: Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus, Part 1

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

x

aF x f t dt

has a derivative at every point in , and ,a b

x

a

dF df t dt f x

dx dx

x

a

df t dt f x

dx

First Fundamental Theorem:

1. Derivative of an integral.

a

xdf t dt

xf x

d

2. Derivative matches upper limit of integration.



a

xdf t dt f x

dx



3. Lower limit of integration is a constant.


x

a

df t dt f x

dx




New variable.


cos xd

t dtdx cos x 1. Derivative of an integral.



sinxdt

dx

sin sind

xdx

0

sind

xdx

cos x

The long way:First Fundamental Theorem:

20

1

1+t

xddt

dx 2

1

1 x




2

0cos

xdt dt

dx

2 2cosd

x xdx

2cos 2x x

22 cosx x

The upper limit of integration does not match the derivative, but we could use the chain rule.

53 sin

x

dt t dt

dxThe lower limit of integration is not a constant, but the upper limit is.

53 sin xdt t dt

dx

3 sinx x

We can change the sign of the integral and reverse the limits.

3cos3x (E) 3sin3x (D)

cos3x (C)3sin3x - (B) 3sin)(

)cos( )23

0

xA

dttdx

d x

theseof none (E) 13x (D)

1)-11)((t3

2 (C)

3

1 (B) 1)(

1 )1

32

33

2

33

0

3

x

tt

ttA

dxxdt

d t

Group Problem:

points critical no has f (e)

5- at x max local a has f (d)

5- at x min local a has f (c)

-5 xifonly increases f (b)

xallfor increases f (a)

thatfollowsIt .5

5)( suppose

0 2

dtt

txf

x

xdttg

a)(f(x)given

The graph above is g(t)

answer?your justify

minimum? a have f(x) does

2

2

1

2

x

tx

ddt

dx eNeither limit of integration is a constant.

2 0

0 2

1 1

2 2

x

t tx

ddt dt

dx e e

It does not matter what constant we use!

2 2

0 0

1 1

2 2

x x

t t

ddt dt

dx e e

2 2

1 12 2

22xx

xee

(Limits are reversed.)

(Chain rule is used.)2 2

2 2

22xx

x

ee

We split the integral into two parts.

The Fundamental Theorem of Calculus, Part 2

If f is continuous at every point of , and if

F is any antiderivative of f on , then

,a b

b

af x dx F b F a

,a b

(Also called the Integral Evaluation Theorem)

tiveantiderivaan using dx 1 1

4

2 xevaluate

xx

3

3

is a general antiderivativeso…

3

80

3

76

3

4

)4(3

)4(1

3

1

F(4)-F(1)F(a)-F(b) 3

33

1

4

3

xx

Remember, the definite integral gives us the net area

Net area counts area below the x-axis as negative

The net area, or if this werea definite integral, would=5-3+4=6

The area, or “total area”,or area to the x-axis,would be 5+3+4=12

b

anetdxxf ent)(Displacem area )(

Group Work

a) Find g(-5)

b) Find all values of x on the open interval (-5,4) where g is decreasing. Justify your answer.

c) Write an equation for the line tangent to the graph of g at x = -1

d) Find the minimum value of g on the closed interval [-5,4]. Justify your answer.

xdtthlet

1)(g(x)by definedfunction thebe g

a) Find g(-5)

Find all values of x on the open interval (-5,4) where g is decreasing. Justify your answer.

c) Write an equation for the line tangent to the graph of g at x = -1

d) Find the minimum value of g on the closed interval [-5,4]. Justify your answer.

Solution

1

5

5

1)()()5( dtthdtthg

(-2,1)on decreasing is g (-2,1),on 0(x)g since ).()( xhxg

1)-2(x0-y

is line tangent the(-1,0)point theand 2,- m Using

graph) the(from -2h(-1) h(x).(x)g since h(-1) of value theis slope The

throughgoes tangent that the(-1,0)point thehave we,0)(g(-1) since1

1

th

.-g(-5) is g of valueminumum The

-2.h(t)dt g(1) and before) (-g(-5) left. thefrom 4 approachesit

as increasing its since 4 xbet can'It 4. and 1, 5,- xare candidatesonly The

sign changes (x)g or whereendpoint an at occur must g of valueminumum The

1

1-

from

Group Problem

right the tographed

[-1,6]domain ith function w

theis f where,)(H(x) x

1-

continuous

dttfLet

(1)H (2),H and , H(2) find(f)

Explain value?maximum its achieve H does where(e)

Explain negative?or positive H(6) is (d)

Explain up? concave H is intervalon what (c)

Explain ?increasing H is intervalon what (b)

H(-1) )(

Finda

Group Problem

t

0meters. )(s is axis coordinate a along

moving particle a of (sec) t at timeposition The

shown. isgraph hosefunction w abledifferenti theis f

dxxf

6? tat time lie particle thedoesorigin theof sideon what (g)

away? origin? the towardsmoving particle theis When (f)

zero?on accelerati theisely when Approximat (e)

origin? he through tpass particle thedoes When (d)

3? t ? 1 t at timeposition sparticle' theis whay (c)

negative? of positive 1 tat time particle theofon accelerati theIs (b)

1? tat time velocity sparticle' theis what )(

a

Using FTC with an initial condition:

x

aaFxFdttf )()()(

x

axFdttfaF )()()(

IF the initial condition is given, it accumulates normallyand then adds the initial condition.

Ex. If oil fills a tank at a rate modeled by

and the tanker has 2,500 gallons to start. min

barrelsin ,250 )022.1( xey

How much oil is in the tank after 50 minutes pass?

50

0

)022.1(250)0()( dxefxF x

f(a) a is the lower limit

Ex.

f(5) ,4)1( and ,58)( 3 findfxxf Given

xdxxFF

1

3 58)1()5(

5

1

3 58)1()5( dxxFF

1264

12684)5(

F

1)

2)

tt

rr

drrxtx

t

t

12

35

12

35

13)0()(

2

0

2

0

1834

)2(3)2(4344

344

38)2()(

2

22

22

2

tt

tt

ww

dwwxtx

t

t

2) Where does is the particle at t=5 ?

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Warmup 1) 2). 5.4: Fundamental Theorem of Calculus