Section 5.4a FUNDAMENTAL THEOREM OF CALCULUS. Deriving the Theorem Let Apply the definition of the...
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Transcript of Section 5.4a FUNDAMENTAL THEOREM OF CALCULUS. Deriving the Theorem Let Apply the definition of the...
Section 5.4a
FUNDAMENTAL THEOREM OF CALCULUS
Deriving the Theorem
x
aF x f t dtLet
Apply the definition of the derivative:
0
limh
F x h F xdF
dx h
0
lim
x h x
a a
h
f t dt f t dt
h
0
lim
x h
x
h
f t dt
h
0
1lim
x h
xhf t dt
h
Rule for Integrals!
Deriving the Theorem
This is average value of f fromx to x + h. Assuming that f iscontinuous, it takes on its averagevalue at least once in the interval…
0
1lim
x h
xhf t dt
h
1 x h
xf t dt f c
h
For some c between x and x + h
Back to the proof…
Deriving the Theorem
0
1lim
x h
xh
dFf t dt
dx h
0limh
f c
What happens to c as h goes to zero???
As x + h gets closer to x, it forces c to approach x…
Since f is continuous, this means that f(c) approaches f(x):
0
limh
f c f x
Putting it all together…
Deriving the Theorem
0
limh
F x h F xdF
dx h
0
lim
x h
x
h
f t dt
h
Definition of Derivative Rule for Integrals
0
limh
f c
f xFor some c between
x and x + h
Because f iscontinuous
The Fundamental Theorem ofCalculus, Part 1If is continuous on [a, b], then the function
x
aF x f t dt
has a derivative at every point x in [a, b], and
x
a
dF df t dt f x
dx dx
…the definite integral of a continuous function is adifferentiable function of its upper limit of integration…
f
The Fundamental Theorem ofCalculus, Part 1
x
a
df t dt f x
dx
• Every continuous function is the derivative of some other function.
• The processes of integration and differentiation are inverses of one another.
• Every continuous function has an antiderivative.
A Powerful Theorem Indeed!!!
The Fundamental Theorem ofCalculus, Part 1
cosxd
t dtdx
Evaluate each of the following, using the FTC.
cos x
20
1
1
xddt
dx t 2
1
1 x
The Fundamental Theorem ofCalculus, Part 1
2
1cos
xy t dtFind dy/dx if
1cos
uy t dt 2u x
The upper limit of integration is not x y is a composite of:
and
Apply the Chain Rule to find dy/dx:
dy dy du
dx du dx
1cos
ud dut dt
du dx
cosdu
udx
2 2cos 2 2 cosx x x x
The Fundamental Theorem ofCalculus, Part 1
53 sin
xy t t dtFind dy/dx if
53 sin
x
dt t dt
dx 53 sin
xdt t dt
dx
53 sin
xdt t dt
dx 3 sinx x
The Fundamental Theorem ofCalculus, Part 1
2
2
x t
xy e dtFind dy/dx if
2
2
x t
x
de dt
dx2 2
0 0
x xt tde dt e dt
dx
2 2 2 2x xd de x e xdx dx
ChainRule
2 22 2x xxe e
Deriving More of the Theorem
Let x
aG x f t dt
If F is any antiderivative of f, then F(x) = G(x) + C for someconstant C.
F b F a G b C G a C
Let’s evaluate F (b) – F(a):
G b G a
b a
a af t dt f t dt
b
af t dt
0
The Fundamental Theorem ofCalculus, Part 2If is continuous at every point of [a, b], and if F is anyantiderivative of on [a, b], then
b
af x dx F b F a
This part of the Fundamental Theorem is also called theIntegral Evaluation Theorem.
ff
The Fundamental Theorem ofCalculus, Part 2
b
af x dx F b F a
• Any definite integral of any continuous function can be calculated without taking limits, without calculating Riemann sums, and often without major effort all we need is an antiderivative of !!!
Another Very Powerful Theorem!!!
f
f
The Fundamental Theorem ofCalculus, Part 2
b
aF x
The usual notation for F(b) – F(a) is
A “note” on notation:
b
aF x or
depending on whether F has one or more terms…
The Fundamental Theorem ofCalculus, Part 2
Evaluate the given integral using an antiderivative.
3 3
11x dx
Antiderivative:
4
4
xx
34
14
xx
81 1
3 14 4
24How can we support thisanswer numerically???
The Fundamental Theorem ofCalculus, Part 2
Evaluate the given integral using an antiderivative.
1
23x dx
Antiderivative:
13
ln3x
267.889
3ln3
1
2
13
ln3x
1 213 3
ln 3
1 19
ln3 3
The Fundamental Theorem ofCalculus, Part 2
Evaluate the given integral using an antiderivative.
5 3 2
0x dx
Antiderivative:
5 22
5x
10 5 22.361
55 2
0
2
5x
5 225 0
5
The Fundamental Theorem ofCalculus, Part 2
Evaluate the given integral using an antiderivative.
01 cos x dx
Antiderivative:
sinx x3.142
0sinx x
sin 0 sin 0
0 0 0
The Fundamental Theorem ofCalculus, Part 2
Evaluate the given integral using an antiderivative.
3
04sec tanx x dx
Antiderivative:
sec x
4
3
04 sec x
4 sec 3 sec0
4 2 1
The Fundamental Theorem ofCalculus, Part 2
Evaluate the given integral using an antiderivative.
4
0
1 udu
u
Antiderivative:
1 22u u
4 1 2
01u du
0
41 2
02u u
2 4 4 0