Fundamental Theorem of Calculus. Part I:

Fundamental Theorem of

Calculus. Part I:

Connection between integration

and differentiation

Motivation:

Definition: An antiderivative of a function f (x) is a function F (x) such that F 0(x) = f (x).

In other words, given the function f (x), you want to tell whose derivative it is.

Example 1. Find an antiderivative of 1.

An answer: x.

Example 2. Find an antiderivative of 1

1 + x2.

!!..??!..!?

Some antiderivatives can be found by reading differentiation formulas backwards:

[xα]0 = αxα−1 [cos x]0 = − sin x [tan x]0 = _cos1_{2 x} [arctan x]0 = _1+x1 ₂ [arcsin x]0 = √ 1 1−x2

Limitation: there aren’t formulas for

sin(x2), e(x2), q

x + sin2(1 − x3)

?!?

An observation

No matter what object’s velocity v(t) is, its position d(t) is always an antiderivative of v(t): d0(t) = v(t). This suggests that all functions have antiderivatives.

Antiderivative of sin(t2) exists!

Suppose the speed of my car obeys sin(t2) (do not try it on the road!). The car will move accordingly and the position of the car F (t) will give the antiderivative of sin(t2).

A hypothesis

Calculating antiderivatives must be similar to calculating position from velocity

Fundamental Theorem of Calculus

Naive derivation

Let, at initial time t₀, position of the car on the road is d(t₀) and velocity is v(t₀). After a short period of time ∆t, the new position of the car is approximately

d(t₁) ≈ d(t₀) + v(t₀)∆t, (t₁ = t₀ + ∆t) 4 4 4 4 4 h h h c t₀ t₁ t2 t₃ v(t )₀ ∆t v(t )₀ ∆t h

After two moments of time d(t₂) ≈ d(t₀) + v(t₀)∆t + v(t₁)∆t, (t₂ = t₁ + ∆t) 4 4 4 4 4 h h _c t₀ t₁ t2 v(t )₁ ∆t ∆t v(t )₀ ∆t v(t )₁ ∆t 4 4 4 h h t₃ h

After n moments of time d(t_n) ≈ d(t₀) + v(t₀)∆t + . . . + v(t_n−1)∆t = d(t₀) + n−1 X i=0 v(t_i)∆t (t_n = t₀ + n∆t) 4 4 4 4 4 h h h h h _c t₀ t₁ t2 t₃ v(t )_n ∆t ∆t v(t )₀ ∆t v(t )₁ ∆t v(t )_n-1 ∆t

As ∆t → 0, (∆t = (t − t₀)/n). d(t) − d(t₀) = lim ∆t→0 n−1 X i=0 v(t_i)∆t Compare this to Z t t₀ v(τ )dτ = lim ∆t→0 n−1 X i=0 v(t_i)∆t

Combining the two formulas d(t) − d(t₀) = Z t t₀ v(τ )dτ while d0(t) = v(t)

Assuming t₀ = 0, d(t₀) = 0, distance can be calculated from velocity by

d(t) =

Z t

0

v(τ )dτ

Is this a function in our regular sense? Yes, for each value of t it defines a unique number.

If d0(t) = v(t)? Yes. This constitutes the assertion of Fundamental Theorem of Calculus.

THEOREM. Let f (x) be a continuous function (so, the definite integral of f (x) exists). Then the function

F (x) = Z x a f (τ )dτ. is an antiderivative of f (x), F 0(x) = f (x). d dx Z x a f (τ )dτ  = f (x)

d dx Z x a f (τ )dτ  = f (x)

Example 3. Find an antiderivative of

f (x) = sin(x2).

An answer: F (x) =

Z x

0

d dx Z x a f (τ )dτ  = f (x)

Example 4. Find an antiderivative of

f (x) = e(x2).

An answer: G(x) =

Z x

0

d dx Z x a f (τ )dτ  = f (x)

Example 5. Find an antiderivative of

f (x) = q x + sin2(1 − x3). An answer: H(x) = Z x 0 q τ + sin2(1 − τ 3)dτ .

d dx Z x a f (τ )dτ  = f (x) Example 6. Calculate F 0(x) if F (x) = Z x 0 sin(τ 2)dτ Answer: sin(x2).

d dx Z x a f (τ )dτ  = f (x) Example 7. Calculate G0(x) if G(x) = Z x 0 e(τ2)dτ Answer: e(x2).

d dx Z x a f (τ )dτ  = f (x) Example 8. Calculate H0(x) if H(x) = Z x 0 q τ + sin2(1 − τ 3)dτ Answer: q x + sin2(1 − x3).

d dx Z x a f (τ )dτ  = f (x) Example 9. Calculate F 0(x) if F (x) = Z 0 x sin(τ 2)dτ Solution: d dxF (x) = d dx " Z 0 x sin(τ 2)dτ # = d  − Z x sin(τ 2)dτ  = − sin(x2)

d dx Z x a f (τ )dτ  = f (x) Example 10. Calculate G0(x) if G(x) = Z 5x 1 sin(τ 2)dτ

Trick. Introduce F (x) = R₁x sin(τ 2)dτ , so, G(x) = F (5x).

Use the Chain Rule to find the derivative d dxG(x) = d dxF (5x) = F 0(5x)[5x]0 = sin((5x)2)5.

d dx Z x a f (τ )dτ  = f (x) Example 11. Calculate H0(x) if H(x) = Z x3 π sin(τ 2)dτ

Trick. Introduce F (x) = R_πx sin(τ 2)dτ , so, H(x) = F (x3).

Use the Chain Rule to find the derivative d dxH(x) = d dxF (x 3 ) = F 0(x3)[x3]0 = sin((x3)2)3x2.

Example 12. Calculate G0(x) if G(x) = Z x3 √ x sin(τ 2)dτ

Solution: First, notice that

G(x) = Z 0 √ x sin(τ 2)dτ + Z x3 0 sin(τ 2)dτ

Trick. Introduce F (x) = R₀x sin(τ 2)dτ , so,

Use FTC to calculate F 0(x) = sin(x2). Use the Chain Rule to find the derivative

d dxG(x) = − d dxF ( √ x) + d dxF (x 3 ) = −F 0(√x)[√x]0 + F 0(x3)[x3]0 = − sin((√x)2) 1 2√x + sin((x 3 )2)3x2.

Control exercises Example 13. Calculate F 0(x) if F (x) = Z 1+x2 1−x2 sin(τ 2)dτ Example 14. Calculate G0(x) if G(x) = Z x 1 x2 sin(τ 2)dτ

Fundamental Theorem of Calculus

Proof of Part I

We will be using two facts

Interval Additive Property:

Z b a f (τ )dτ = Z c a f (τ )dτ + Z b c f (τ )dτ Comparison Property: If m ≤ f (x) ≤ M on [a, b] m(b − a) ≤ Z b a f (τ )dτ ≤ M (b − a)

Let F (x) =

Z x

a

f (τ )dτ .

We will now show that F 0(x) = f (x).

By definition F 0(x) = lim h→0 F (x + h) − F (x) h = lim h→0 R x+h a f (τ )dτ − R x a f (τ )dτ h

Notice, that by the Interval Additive Property Z x+h a f (τ )dτ − Z x a f (τ )dτ = Z x+h x f (τ )dτ Therefore, F 0(x) = lim h→0 R x+h a f (τ )dτ − R x a f (τ )dτ h = lim h→0 1 h Z x+h x f (τ )dτ

Also, by the Comparison Property, for m = min τ ∈[x,x+h] f (τ ), M = max τ ∈[x,x+h] f (τ ) one has mh ≤ Z x+h x f (τ )dτ ≤ M h or, dividing by h, m ≤ 1 h Z x+h x f (τ )dτ ≤ M

Finally, by the Squeeze Theorem, expression m ≤ 1 h Z x+h x f (τ )dτ ≤ M converges to f (x) ≤ lim h→0 1 h Z 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