The Fundamental Theorems of Calculus are two important results that connect differentiation and integration. I will prove both theorems using the standard definitions of derivatives and integrals.
First, let's recall the definitions:
Derivative: If f(x) is a function, then the derivative of f(x) with respect to x, denoted as f'(x), is the limit:
f'(x) = lim (h -> 0) [(f(x + h) - f(x)) / h]
Integral: If f(x) is a function and [a, b] is a closed interval, then the definite integral of f(x) over [a, b], denoted as ∫(a to b) f(x) dx, is the limit:
∫(a to b) f(x) dx = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
where x_0 = a, x_n = b, and x_i are points in the interval [a, b] such that x_(i-1) < x_i.
Now, let's prove the two Fundamental Theorems of Calculus:
Fundamental Theorem of Calculus, Part 1 (Differentiation):
If f(x) is continuous on [a, b], and g(x) is its antiderivative, then g'(x) = f(x).
Proof:
By definition, we need to show that the limit:
g'(x) = lim (h -> 0) [g(x + h) - g(x)] / h
exists and equals f(x) for all x in [a, b].
Since g(x) is an antiderivative of f(x), we have:
g(x + h) - g(x) = ∫(x to x+h) f(t) dt
Now, we can use the definition of the integral to rewrite this as:
The Fundamental Theorems of Calculus are two important results that connect differentiation and integration. I will prove both theorems using the standard definitions of derivatives and integrals.
First, let's recall the definitions:
f'(x) = lim (h -> 0) [(f(x + h) - f(x)) / h]
∫(a to b) f(x) dx = lim (n -> ∞) [Σ (from i = 1 to n) f(x_i) * (x_i - x_(i-1))]
where x_0 = a, x_n = b, and x_i are points in the interval [a, b] such that x_(i-1) < x_i.
Now, let's prove the two Fundamental Theorems of Calculus:
Fundamental Theorem of Calculus, Part 1 (Differentiation):
If f(x) is continuous on [a, b], and g(x) is its antiderivative, then g'(x) = f(x).
Proof:
By definition, we need to show that the limit:
g'(x) = lim (h -> 0) [g(x + h) - g(x)] / h
exists and equals f(x) for all x in [a, b].
Since g(x) is an antiderivative of f(x), we have:
g(x + h) - g(x) = ∫(x to x+h) f(t) dt
Now, we can use the definition of the integral to rewrite this as:
g(x + h) - g(x) = lim (n ->