fundamental theorem of calculus. part i
TRANSCRIPT
Fundamental Theorem ofCalculus. Part I:
Connection between integrationand differentiation
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Motivation:
Problem of finding antiderivatives
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Definition: An antiderivative of a function f(x)is a function F (x) such that F ′(x) = f(x).
In other words, given the function f(x), you wantto tell whose derivative it is.
Example 1. Find an antiderivative of 1.
An answer: x.
Example 2. Find an antiderivative of1
1 + x2.
An answer: arctanx.
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!!..??!..!?
HOW DO YOU KNOW?
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Some antiderivatives can be found by readingdifferentiation formulas backwards:
[xα]′ = αxα−1
[cos x]′ = − sinx
[tanx]′ = 1cos2 x
[arctan x]′ = 11+x2
[arcsinx]′ = 1√1−x2
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Limitation: there aren’t formulas for
sin(x2), e(x2),√x + sin2(1 − x3)
?!?
Do these functions have antiderivatives?
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An observation
No matter what object’s velocity v(t) is, itsposition d(t) is always an antiderivative of v(t):d′(t) = v(t). This suggests that all functionshave antiderivatives.
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Antiderivative of sin(t2) exists!
Suppose the speed of my car obeys sin(t2)(do not try it on the road!). The car will moveaccordingly and the position of the car F (t)will give the antiderivative of sin(t2).
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A hypothesis
Calculating antiderivatives must be similar tocalculating position from velocity
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Fundamental Theorem of Calculus
Naive derivation
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Let, at initial time t0, position of the car onthe road is d(t0) and velocity is v(t0). After ashort period of time ∆t, the new position ofthe car is approximately
d(t1) ≈ d(t0) + v(t0)∆t, (t1 = t0 + ∆t)
4 4
4
4
4 h
hh
c
t0 t1
t2t3
v(t )0
∆t
v(t )0 ∆t
h
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After two moments of time
d(t2) ≈ d(t0) + v(t0)∆t + v(t1)∆t,
(t2 = t1 + ∆t)
4 4
4
4
4 hh c
t0 t1
t2
v(t )1
∆t ∆t
v(t )0 ∆tv(t )1 ∆t
4 4 4
hh
t3h
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After n moments of time
d(tn) ≈ d(t0) + v(t0)∆t + . . . + v(tn−1)∆t
= d(t0) +n−1∑i=0
v(ti)∆t
(tn = t0 + n∆t)
4 4
4
4
4 h
h
h
h
hc
t0 t1
t2t3 v(t )n
∆t ∆t
v(t )0 ∆tv(t )1 ∆t
v(t )n-1 ∆t
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As ∆t → 0, (∆t = (t − t0)/n).
d(t) − d(t0) = lim∆t→0
n−1∑i=0
v(ti)∆t
Compare this to∫ t
t0
v(τ )dτ = lim∆t→0
n−1∑i=0
v(ti)∆t
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Combining the two formulas
d(t) − d(t0) =∫ t
t0
v(τ )dτ
while
d′(t) = v(t)
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Assuming t0 = 0, d(t0) = 0, distance can becalculated from velocity by
d(t) =∫ t
0v(τ )dτ
Is this a function in our regular sense? Yes, foreach value of t it defines a unique number.
If d′(t) = v(t)? Yes. This constitutes theassertion of Fundamental Theorem of Calculus.
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THEOREM. Let f(x) be a continuousfunction (so, the definite integral of f(x)exists). Then the function
F (x) =∫ x
a
f(τ )dτ.
is an antiderivative of f(x), F ′(x) = f(x).
d
dx
[∫ x
a
f(τ )dτ
]= f(x)
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EXAMPLES
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 3. Find an antiderivative of
f(x) = sin(x2).
An answer: F (x) =∫ x
0sin(τ 2)dτ .
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 4. Find an antiderivative of
f(x) = e(x2).
An answer: G(x) =∫ x
0e(τ2)dτ .
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 5. Find an antiderivative of
f(x) =√
x + sin2(1 − x3).
An answer:
H(x) =∫ x
0
√τ + sin2(1 − τ 3)dτ .
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 6. Calculate F ′(x) if
F (x) =∫ x
0sin(τ 2)dτ
Answer: sin(x2).
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 7. Calculate G′(x) if
G(x) =∫ x
0e(τ2)dτ
Answer: e(x2).
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 8. Calculate H ′(x) if
H(x) =∫ x
0
√τ + sin2(1 − τ 3)dτ
Answer:√
x + sin2(1 − x3).
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 9. Calculate F ′(x) if
F (x) =∫ 0
x
sin(τ 2)dτ
Solution:d
dxF (x) =
d
dx
[∫ 0
x
sin(τ 2)dτ
]=
d
dx
[−
∫ x
0sin(τ 2)dτ
]= − sin(x2)
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 10. Calculate G′(x) if
G(x) =∫ 5x
1sin(τ 2)dτ
Trick. Introduce F (x) =∫ x
1 sin(τ 2)dτ , so,G(x) = F (5x).
Use FTC to calculate F ′(x) = sin(x2).
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Use the Chain Rule to find the derivative
d
dxG(x) =
d
dxF (5x)
= F ′(5x)[5x]′ = sin((5x)2)5.
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d
dx
[∫ x
a
f(τ )dτ
]= f(x)
Example 11. Calculate H ′(x) if
H(x) =∫ x3
π
sin(τ 2)dτ
Trick. Introduce F (x) =∫ x
π sin(τ 2)dτ , so,H(x) = F (x3).
Use FTC to calculate F ′(x) = sin(x2).
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Use the Chain Rule to find the derivative
d
dxH(x) =
d
dxF (x3)
= F ′(x3)[x3]′ = sin((x3)2)3x2.
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Example 12. Calculate G′(x) if
G(x) =∫ x3
√x
sin(τ 2)dτ
Solution: First, notice that
G(x) =∫ 0
√x
sin(τ 2)dτ +∫ x3
0sin(τ 2)dτ
Trick. Introduce F (x) =∫ x
0 sin(τ 2)dτ , so,
G(x) = −F (√
x) + F (x3).
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Use FTC to calculate F ′(x) = sin(x2).
Use the Chain Rule to find the derivative
d
dxG(x) = −
d
dxF (
√x) +
d
dxF (x3)
= −F ′(√
x)[√
x]′ + F ′(x3)[x3]′
= − sin((√
x)2)1
2√
x+ sin((x3)2)3x2.
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Control exercises
Example 13. Calculate F ′(x) if
F (x) =∫ 1+x2
1−x2sin(τ 2)dτ
Example 14. Calculate G′(x) if
G(x) =∫ x
1x2 sin(τ 2)dτ
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Fundamental Theorem of Calculus
Proof of Part I
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We will be using two facts
Interval Additive Property:∫ b
a
f(τ )dτ =∫ c
a
f(τ )dτ +∫ b
c
f(τ )dτ
Comparison Property: If m ≤ f(x) ≤ M on[a, b]
m(b − a) ≤∫ b
a
f(τ )dτ ≤ M(b − a)
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Let F (x) =∫ x
a
f(τ )dτ .
We will now show that F ′(x) = f(x).
By definition
F ′(x) = limh→0
F (x + h) − F (x)
h
= limh→0
∫ x+h
a f(τ )dτ −∫ x
a f(τ )dτ
h
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Notice, that by the Interval Additive Property∫ x+h
a
f(τ )dτ −∫ x
a
f(τ )dτ =∫ x+h
x
f(τ )dτ
Therefore,
F ′(x) = limh→0
∫ x+h
a f(τ )dτ −∫ x
a f(τ )dτ
h
= limh→0
1
h
∫ x+h
x
f(τ )dτ
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Also, by the Comparison Property, for
m = minτ∈[x,x+h]
f(τ ), M = maxτ∈[x,x+h]
f(τ )
one has
mh ≤∫ x+h
x
f(τ )dτ ≤ Mh
or, dividing by h,
m ≤1
h
∫ x+h
x
f(τ )dτ ≤ M
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Finally, by the Squeeze Theorem, expression
m ≤1
h
∫ x+h
x
f(τ )dτ ≤ M
converges to
f(x) ≤ limh→0
1
h
∫ x+h
x
f(τ )dτ ≤ f(x)
since both m = min f(τ ) and M = max f(τ )on [x, x + h] coincide with f(x) as h → 0
The theorem is proved.
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