Fundamental Theorems Of Integral Calculus
Fundamental Theorems Of Integral Calculus :
First Fundamental Theorem Of Integral Calculus :
Theorem 1 :
If a function f is bounded and integrable on [ a,b ] , then the function F defined as x
F(x) = ∫ f(t) dt , a≤x≤b
0
is continuous on [ a,b ] and further more , if f is continuous at a point of [ a,b ] , then F is derivable at c and F'(c) = f(c) .
Proof :
It is given that the function f is bounded . Then by definition ∃ a number k such that |f(x)|≤ k for x ∈ [ a,b ] ......(1)
Let x₁ , x₂ ∈[ a,b ] such that a≤x₁≤x₂≤b .
x₂ x₁
Then |F(x₂) - F(x₁) | = |∫ f(t) dt - ∫ f(t) dt |
a a
x₂ a
= |∫ f(t) dt + ∫ f(t) dt |
a x₁
a x₂
= |∫ f(t) dt + ∫ f(t) dt |
x₁ a
x₂ x₂
= |∫ f(t) dt | = |f(t) ||∫ dt|
x₁ x₁
≤ k(x₂ - x₁ ) using equation (1)
So for a given ε > 0 , we find
|F(x₂) - F(x₁)|< ε if |x₂ - x₁|< ε/k
Here the function F is continuous on [ a,b ] .
To Prove F'(c) = f(c) :
Let f be continuous at a point c ∈ [a,b].
So for any ε > 0 ∃ a δ > 0 such that |f(x) - f(c) |< ε for |x- c| < δ
Let c-δ < s ≤ c ≤ t < c+δ
Now |F(t) - F(s) / t-s - f(c) |
t s
= |∫ f(x) dx - ∫ f(x) dx / t-s - f(c) |
a a
t a
= |[∫ f(x) dx + ∫ f(x) dx - (t-s) f(c)] / t-s |
a s
a t
= |[ ∫ f(x) dx + ∫ f(x) dx - f(c) (t-s) ] / t- s|
s a
t
≤ |∫ {f(x) - f(c) } dx| / t-s
s
t
≤ [∫ |f(x) - f(c) |dx] / t-s
s
t
< ε ∫ dt / t- s
s
= ε(t - s ) / (t - s ) = ε
∴ | F(t) - F(s) / t - s - f(c) |< ε
⇒ F'(c) = f(c)
i.e continuity of f at any point of [ a,b ]
⇒ F is derivable at that point (Proved)
Second Fundamental Theorem Of Integral Calculus :
Theorem 2 :
A function f is bounded and integrable on [ a,b] , then
b
∫ f dx = F(b) - F(a)
a
Proof :
Given F' = f is bounded and integrable .
Then for every given ε > 0 , ∃ a δ > 0 such that for every partition
P = { a = x₀,x₁,x₂,......xₙ = b} with norm μ(P) <δ
n b
|Σ f(tᵢ) Δxᵢ - ∫ f dx |< ε
i=1 a
n b
or , lim Σ f(tᵢ) Δxᵢ = ∫ f(x) dx ......(1)
μ(P)-->0 i= 1 a
for any tᵢ ∈ Δxᵢ
Let us choose tᵢ ∈ Δxᵢ , such that by Lagrange's Mean Value Theorem , we have
F(xᵢ) - F(xᵢ₋₁) / xᵢ - xᵢ₋₁ = F'(tᵢ) i= 1,2,3.....,n
⇒ F(xᵢ) - F(xᵢ₋₁) = f(tᵢ) (xᵢ - xᵢ₋₁) = f(tᵢ)Δxᵢ
n n
⇒ Σ f(tᵢ) Δxᵢ = Σ [ F(xᵢ) - F(xᵢ₋₁)]
i=1 i = 1
= F(b) - F(a) ..........(2)
From equations (1) and (2) , we get
b
∫ f(x) dx = F(b) - F(a) (Proved)
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