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Fundamental Theorem of Calculus ~ p. 1
FUNDAMENTAL THEOREM OF CALCULUS
Objective: Investigate the link between differentiation and integration as given by the Fundamental Theorem of Calculus
Two properties of integrals
b a
∫ f(x) dx
a b
− ∫ f(x) dx
- If a = b, then )x = 0, so = 0
a a
∫ f(x) dx
Comparison properties of integrals
- If f(x) ≥ 0 for a ≤ x ≤ b, then
b a
∫ f(x) dx^ ≥^0
This can be interpreted as the area under the graph of a function where the area is entirely above the x-axis.
- If f(x) ≥ g(x) for a ≤ x ≤ b, then
b a
∫ f(x) dx^ ≥
b a
∫ g(x) dx
This means that a “bigger” function has a “bigger” integral.
- If m (^) ≤ f(x) (^) ≤ M for a (^) ≤ x (^) ≤ b, then m(b (^) − a) (^) ≤ M(b a)
b a
∫ f(x) dx ≤^ −
This means that the area under the graph of f is greater than the area of the rectangle with height m and less than the area of the rectangle with height M
Motivation for the fundamental theorem: let f be a continuous function on [a, b] and define
a function g by g(x) = where a x b.
x a
∫ f(t) dt ≤^ ≤
- g depends only on x
- If x is a fixed number, then is a definite number.
x a
∫ f(t) dt
- If x is a variable, then also varies and defines a function of x, denoted by
x a
∫ f(t) dt
g(x)
- Example: Suppose f(t) = t^2 and a = 1
x 3 2 1
x 1 t dt 3
- g (x)′ = x ,^2 that is g′ ( )x =f x( )
- g is defined as the integral of f, then g turns out to be an antiderivative of f.
Fundamental Theorem of Calculus ~ p. 2
In general, consider any continuous function f with f(x) ≥ 0.
- g(x) = can be interpreted as the area under the graph of f from a to x,
x a
∫ f(t) dt
where x can vary from a to b.
- For h > 0, g(x + h) – g(x) is the area under the graph of f from x to (x + h).
- If h is small, this area is approximately equal to the area of the rectangle with height f(x) and width h, that is g(x + h) – g(x) ≈ h[f(x)].
- Then g(x h) g(x) f(x) h
- It appears that (^) h 0 g (x) lim g(x^ h)^ g(x) f(x)! → h
′ = +^ − =
The fundamental theorem of calculus : Suppose f is continuous on [a, b]
Part 1. If g(x) = then
x
∫a f(t) dt,
g (x)′^ = f(x)
If f is integrated and the result is then differentiated, we arrive back at the original function f.
Part 2. = F(b) F(a), where F is any antiderivative of f, that is,
b a
∫ f(x) dx −^ F^ ′ = f.
If we differentiate a function F, and then integrate the result, we arrive back at the original function F in the form of F(b) – F(a).
OPTIONAL : Proof of FTCI: Let g(x) =
x a
∫ f(t) dt
A. If x and (x + h) are in the open interval (a, b), then g(x + h) – g(x) = = =
x h a f(t) dt
x a
∫ f(t) dt
x x h a x f(t) dt f(t) dt
⎜ +^ ⎟−
⎝ ∫^ ∫^ ⎠
x a
∫ f(t) dt
x h a f(t) dt
B. Then, for h ≠ 0, g(x h) g(x) 1 h h
x h a f(t) dt
C. Assume that h > 0.
- Since f is continuous on [x, x + h], there are numbers u and v in [x, x + h] such that f(u) = m and f(v) = M, where m and M are the absolute minimum and maximum
Fundamental Theorem of Calculus ~ p. 4
Examples using the Fundamental Theorem
x (^2) 0
g(x) = ∫ (1 +t ) dt
x (^2) 0
g(x) = ∫ (1 +t ) dt t t^ x x^ g x x
x
3
0
3 2 3 3
- Using FTCI: Let f(t) = 1 + t 2 ; then
x (^2) 0
g(x) = ∫ (1 +t ) dt ⇒ g′ ( )x = f x( ) = 1 +x^2
- Find g (x)′^ if g(x) = cos t dt
x
- Using FTXII: g(x) = (^) cos t dt = x
∫ π [^ sin^ t]^ sin^ x sin
x π =^ −^ π⇒^ g′^ ( )x^ =cosx
• Using FTCI: Let f(t) =cos t; then g(x) = ∫ π
x cos t dt ⇒ g′ ( )x = f x( ) =cosx
Find the derivative of each of the following functions.
x
g(x) 1 ln t dt = ∫ θ θ
10 F(x) (^) x tan d
Solutions
Let = = ∫ ⇒ = =
x 1 f(t) ln t; then g(x) ln t dt g'(x) f(x) ln x
Let θ = θ = ∫ θ θ = − ∫ θ θ⇒ = − = −
10 x f( ) tan ; then F(x) (^) x tan d 10 tan d F '(x) f(x) tan x
