University:
California State University, Northridge
Course:
MATH 370 | Foundations of Geometry
Academic year:

2004
Views:

125

Pages:

38
Author:

reescamerianwj

Fundamental Theorem of Calculus. Part I: Connection between integration and differentiation Motivation: Problem of finding antiderivatives 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. 1 Example 2. Find an antiderivative of . 2 1+x An answer: arctan x. !!..??!..!? HOW DO YOU KNOW? Some antiderivatives can be found by reading differentiation formulas backwards: α 0 α−1 [x ] = αx [cos x]0 = − sin x [tan x]0 = [arctan x]0 = 0 [arcsin x] = 1 cos2 x 1 1+x2 √ 1 1−x2 Limitation: there aren’t formulas for 2 (x2) sin(x ), e , q 2 x + sin (1 − x3) ?!? Do these functions have antiderivatives? 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) 2 will give the antiderivative of sin(t ). 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, t3 t0 4 hc v(t0 ) h4 ∆t v(t0 ) ∆t t1 t 4 h2 4 (t1 = t0 + ∆t) 4 h After two moments of time d(t2) ≈ d(t0) + v(t0)∆t + v(t1)∆t, (t2 = t1 + ∆t) t3 t0 h 4 h4 ∆t v(t0 ) ∆t t1 t2 h 4 4 ∆t hc v(t1 ) 4 v(t1 ) ∆t h After n moments of time d(tn) ≈ d(t0) + v(t0)∆t + . . . + v(tn−1)∆t = d(t0) + n−1 X v(ti)∆t i=0 (tn = t0 + n∆t) t3 t0 h 4 v(tn-1 )∆t h4 ∆t v(t0 ) ∆t t1 4 4 h ∆t h t2 4 v(t1 ) ∆t hc v(tn ) As ∆t → 0, (∆t = (t − t0)/n). d(t) − d(t0) = lim n−1 X ∆t→0 v(ti)∆t i=0 Compare this to Z t n−1 X v(ti)∆t v(τ )dτ = lim t0 ∆t→0 i=0 Combining the two formulas Z t d(t) − d(t0) = v(τ )dτ t0 while d0(t) = v(t) Assuming t0 = 0, d(t0) = 0, distance can be calculated from velocity by Z t d(t) = v(τ )dτ 0 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 Z x F (x) = f (τ )dτ. a is an antiderivative of f (x), F 0(x) = f (x). Z x d f (τ )dτ = f (x) dx a EXAMPLES d x Z f (τ )dτ = f (x) dx a Example 3. Find an antiderivative of 2 f (x) = sin(x ). Z An answer: F (x) = x 2 sin(τ )dτ . 0 d x Z f (τ )dτ = f (x) dx a Example 4. Find an antiderivative of (x2) f (x) = e Z An answer: G(x) = x e 0 (τ 2) . dτ . d dx Z x f (τ )dτ = f (x) a Example 5. Find an antiderivative of q 2 f (x) = x + sin (1 − x3). An answer: Z xq 2 H(x) = τ + sin (1 − τ 3)dτ . 0 d dx Z x f (τ )dτ = f (x) a Example 6. Calculate F 0(x) if Z x F (x) = sin(τ 2)dτ 0 Answer: sin(x2). d dx Z x f (τ )dτ = f (x) a Example 7. Calculate G0(x) if Z x (τ 2) G(x) = e dτ 0 Answer: e (x2) . Z d dx x f (τ )dτ = f (x) a Example 8. Calculate H 0(x) if Z xq 2 H(x) = τ + sin (1 − τ 3)dτ 0 Answer: q 2 x + sin (1 − x3). d dx Z x f (τ )dτ = f (x) a 0 Example 9. Calculate F (x) if Z 0 F (x) = sin(τ 2)dτ x Solution: d F (x) = d "Z # 0 sin(τ 2)dτ dx dx x Z x d − sin(τ 2)dτ = − sin(x2) dx = d Z x f (τ )dτ = f (x) dx a 0 Example 10. Calculate G (x) if Z 5x G(x) = sin(τ 2)dτ 1 Trick. Introduce F (x) = G(x) = F (5x). 0 Rx 1 2 sin(τ )dτ , so, 2 Use FTC to calculate F (x) = sin(x ). Use the Chain Rule to find the derivative d dx 0 G(x) = 0 d dx F (5x) 2 = F (5x)[5x] = sin((5x) )5. d dx Z x f (τ )dτ = f (x) a 0 Example 11. Calculate H (x) if Z x3 2 H(x) = sin(τ )dτ π Trick. Introduce F (x) = H(x) = F (x3). Rx π sin(τ 2)dτ , so, Use FTC to calculate F 0(x) = sin(x2). Use the Chain Rule to find the derivative d dx 0 H(x) = 3 3 0 d dx 3 F (x ) 3 2 2 = F (x )[x ] = sin((x ) )3x . Example 12. Calculate G0(x) if Z x3 2 G(x) = √ sin(τ )dτ x Solution: First, notice that Z 0 Z G(x) = √ sin(τ 2)dτ + x x3 sin(τ 2)dτ 0 Rx Trick. Introduce F (x) = 0 sin(τ 2)dτ , so, √ 3 G(x) = −F ( x) + F (x ). Use FTC to calculate F 0(x) = sin(x2). Use the Chain Rule to find the derivative d G(x) = − d √ F ( x) + d 3 F (x ) dx dx √ √ 0 0 = −F ( x)[ x] + F 0(x3)[x3]0 √ 2 1 3 2 2 = − sin(( x) ) √ + sin((x ) )3x . 2 x dx Control exercises Example 13. Calculate F 0(x) if Z 1+x2 F (x) = sin(τ 2)dτ 1−x2 Example 14. Calculate G0(x) if Z x G(x) = x2 sin(τ 2)dτ 1 Fundamental Theorem of Calculus Proof of Part I We will be using two facts Interval Additive Property: Z b Z c Z f (τ )dτ = f (τ )dτ + a a b f (τ )dτ c Comparison Property: If m ≤ f (x) ≤ M on [a, b] Z b m(b − a) ≤ f (τ )dτ ≤ M (b − a) a x Z Let F (x) = f (τ )dτ . a We will now show that F 0(x) = f (x). By definition 0 F (x) = lim F (x + h) − F (x) h→0 R x+h = lim h→0 a f (τ )dτ − h h Rx a f (τ )dτ Notice, that by the Interval Additive Property Z x+h Z x Z x+h f (τ )dτ − f (τ )dτ = f (τ )dτ a a x R x+h Rx Therefore, 0 F (x) = lim a f (τ )dτ − h h→0 = lim h→0 a 1 h Z x+h f (τ )dτ x f (τ )dτ Also, by the Comparison Property, for m= min τ ∈[x,x+h] f (τ ), M = max f (τ ) τ ∈[x,x+h] one has Z x+h mh ≤ f (τ )dτ ≤ M h x or, dividing by h, Z x+h 1 m≤ f (τ )dτ ≤ M h x Finally, by the Squeeze Theorem, expression Z x+h 1 m≤ f (τ )dτ ≤ M h x converges to f (x) ≤ lim h→0 1 h Z x+h f (τ )dτ ≤ f (x) 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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