warmup 1) 2). 5.4: fundamental theorem of calculus
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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
a
xdf t dt
xf x
d
2. Derivative matches upper limit of integration.
First Fundamental Theorem:
1. Derivative of an integral.
a
xdf t dt f x
dx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
First Fundamental Theorem:
x
a
df t dt f x
dx
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
New variable.
First Fundamental Theorem:
cos xd
t dtdx cos x 1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
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
1. Derivative of an integral.
2. Derivative matches upper limit of integration.
3. Lower limit of integration is a constant.
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
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 )(
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
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