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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