Question

the Fundamental Theorem of Calculus (Part I) states that: If a function f is continuous on [a, b], then the function g defined by g ( x ) = ∫ a x f ( t ) d t for a ≤ x ≤ b is an antiderivative of f, and for any a ≤ x ≤ b:

Answer 1

Answer:

g(x) is continuous on [a,b] and can be differentiated on (a,b), and that is the same as g'(x)= f(x)

Step-by-step explanation:

1) Let's take a continuous function [a,b] e x ranging between [a,b], g depends only on x. If

2) The Fundamental Theorem of Calculus in its first part says, in other words, that the derivative of a definite integral with respect to its upper bound is equal to its integrand calculated with respect its upper bound.

Suppose in a function, we need to calculate an area below the curve of any given function within an interval (a,b). According to Riemann, we can do it by estimating the area of several (n )rectangles with

$f(x)\approx g(x+h)-g(x)\\f(x)\ \approx \frac{g(x+h)-g(x)}{h} \\g'(x)=lim _h{\rightarrow0} \frac{g(x+h)-g(x)}{h}$

3) Then this is the same as to say that g(x) is continuous on the closed interval [a,b] and it is differentiable on open interval (a,b),id est g'(x)= f(x)

$g(x)=\int_{a}^{b}f(t)dt \:\:\:a\leqx \leq x \leq b$