Fundamental Theorem Of Calculus

The fundamental theorem of calculus is the statement that the two central operations of Calculus , Differentiation and Integration , are inverses of each other. It is of such central importance in calculus that it is called the Fundamental Theorem for the entire field of study. This means that if a Continuous Function is first integrated and then differentiated, the original function is retrieved. An important consequence of this, sometimes called the second fundamental theorem of calculus, allows one to compute integrals by using an Antiderivative of the function to be integrated. In his 2003 book (page 340), James Stewart credits the idea that led to the fundamental theorem to the English mathematician Isaac Barrow although the first known proof of the fundamental theorem was due to the Scottish mathematician James Gregory . INTUITION Intuitively, the theorem simply states that the sum of Infinitesimal changes in a quantity over time (or some other quantity) add up to the net change in the quantity. To comprehend this statement, we will start with an example. Suppose a particle travels in a straight line with its position given by ''x''(''t'') where ''t'' is time. The derivative of this function is equal to the infinitesimal change in ''x'' per infinitesimal change in time (of course, the derivative itself is dependent on time). Let us define this change in distance per time as the speed ''v'' of the particle. In Leibniz's Notation : : $rac{dx}{dt} = v(t)$ Rearranging that equation, it is clear that: : $dx = v(t)\,dt$ By the logic above, a change in ''x'', call it $\Delta x$ , is the sum of the infinitesimal changes ''dx''. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; hence, the integration operation allows the recovery of the original function from its derivative. As one can reasonably infer, this operation works in reverse as we can differentiate the result of our integral to recover the original function. FORMAL STATEMENTS '' in popular culture. To impress the judges in a contest Donald Duck Figure Skate s the theorem on ice. The theorem is written differently and is slightly wrong: one of the " $=$ " should have been a " $-$ ". It should have been : $\int_b^af'(x)dx = f(a)-f(b)$ ]] Stated formally, the theorem says the following. Let ''f'' be a continuous real-valued function defined on a closed interval ''b'' . Let ''F'' be the function defined for ''x'' in ''b'' by : $F(x) = \int_a^x f(t)\, dt$ then : $F'(x) = f(x)\,$ for every ''x'' in ''b'' . Let ''f'' be a continuous real-valued function defined on a closed interval ''b'' . Let ''F'' be a function such that : $f(x) = F'(x)\,$ for all ''x'' in ''b'' then : $\int_a^b f(x)\,dx = F(b) - F(a)$ . Corollary Let ''f'' be a real-valued function defined on a closed interval ''b'' . Let ''F'' be a function such that : $f(x) = F'(x)\,$ for all ''x'' in ''b'' then : $F(x) = \int_a^x f(t)\,dt + F(a)$ and : $f(x) = rac{d}{dx} \int_a^x f(t)\,dt$ . PROOF Part I It is given that : $F(x) = \int_{a}^{x} f(t) dt$ Let there be two numbers ''x''₁ and ''x''₁ + Δ''x'' in ''b'' . So we have : $F(x_1) = \int_{a}^{x_1} f(t) dt$ and : $F(x_1 + \Delta x) = \int_{a}^{x_1 + \Delta x} f(t) dt$ . Subtracting the two equations gives : $F(x_1 + \Delta x) - F(x_1) = \int_{a}^{x_1 + \Delta x} f(t) dt - \int_{a}^{x_1} f(t) dt \qquad (1)$ . It can be shown that : $\int_{a}^{x_1} f(t) dt + \int_{x_1}^{x_1 + \Delta x} f(t) dt = \int_{a}^{x_1 + \Delta x} f(t) dt$ . :(The sum of the areas of two adjacent regions is equal to the area of both regions combined.) Manipulating this equation gives : $\int_{a}^{x_1 + \Delta x} f(t) dt - \int_{a}^{x_1} f(t) dt = \int_{x_1}^{x_1 + \Delta x} f(t) dt$ . Substituting the above into (1) results in : $F(x_1 + \Delta x) - F(x_1) = \int_{x_1}^{x_1 + \Delta x} f(t) dt \qquad (2)$ . According to the Mean Value Theorem for integration, there exists a ''c'' in ''x''₁ + Δ''x'' such that : $\int_{x_1}^{x_1 + \Delta x} f(t) dt = f(c) \Delta x$ . Substituting the above into (2) we get : $F(x_1 + \Delta x) - F(x_1) = f(c) \Delta x \,$ . Dividing both sides by Δ''x'' gives : $rac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = f(c)$ . :Notice that the expression on the left side of the equation is Newton's Difference Quotient for ''F'' at ''x''₁. Take the limit as Δ''x'' → 0 on both sides of the equation. : $\lim_{\Delta x o 0} rac{F(x_1 + \Delta x) - F(x_1)}{\Delta x} = \lim_{\Delta x o 0} f(c)$ The expression on the left side of the equation is the definition of the derivative of ''F'' at ''x''₁. : $F'(x_1) = \lim_{\Delta x o 0} f(c) \qquad (3)$ . To find the other limit, we will use the Squeeze Theorem . The number ''c'' is in the interval ''x''₁ + Δ''x'' , so ''x''₁ ≤ ''c'' ≤ ''x''₁ + Δ''x''. Also, $\lim_{\Delta x o 0} x_1 = x_1$ and $\lim_{\Delta x o 0} x_1 + \Delta x = x_1$ . Therefore, according to the squeeze theorem, : $\lim_{\Delta x o 0} c = x_1$ . Substituting into (3), we get : $F'(x_1) = \lim_{c o x_1} f(c)$ . The function ''f'' is continuous at ''c'', so the limit can be taken inside the function. Therefore, we get : $F'(x_1) = f(x_1) \,$ . which completes the proof. (Leithold et al, 1996) Part II This is a limit proof by Riemann Sums . Let ''f'' be continuous on the interval ''b'' , and let ''F'' be an antiderivative of ''f''. Begin with the quantity : $F(b) - F(a)\,$ . Let there be numbers x such that : $a = x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b$ . It follows that : $F(b) - F(a) = F(x_n) - F(x_0) \,$ . Now, we add each ''F''(''x''_''i'') along with its additive inverse, so that the resulting quantity is equal: : $\begin{matrix} F(b) - F(a) & = & F(x_n)\,+\, + F(x_1) \,-\,F(x_0) \, \ & = &$ {Link without Title} \,+\, {Link without Title} \,+\, {Link without Title} \, \end{matrix} The above quantity can be written as the following sum: : $\qquad (1)$ Next we will employ the Mean Value Theorem . Stated briefly, Let ''F'' be continuous on the closed interval ''b'' and differentiable on the open interval (''a'', ''b''). Then there exists some ''c'' in (''a'', ''b'') such that : $F'(c) = rac{F(b) - F(a)}{b - a}.$ It follows that : $F'(c)(b - a) = F(b) - F(a). \,$ The function ''F'' is differentiable on the interval ''b'' ; therefore, it is also differentiable and continuous on each interval ''x''_''i''-1. Therefore, according to the mean value theorem (above), : $F(x_i) - F(x_{i-1}) = F'(c_i)(x_i - x_{i-1}). \,$ Substituting the above into (1), we get : $F(b) - F(a) = \sum_{i=1}^n - x_{i-1}) .$ The assumption implies $F'(c_i) = f(c_i).$ Also, $x_i - x_{i-1}$ can be expressed as $\Delta x$ of partition $i$ . : $\qquad (2)$ Notice that we are describing the area of a rectangle, with the width times the height, and we are adding the areas together. Each rectangle, by virtue of the Mean Value Theorem , describes an approximation of the curve section it is drawn over. Also notice that $\Delta x_i$ does not need to be the same for any value of $i$ , or in other words that the width of the rectangles can differ. What we have to do is approximate the curve with $n$ rectangles. Now, as the size of the partitions get smaller and n increases, resulting in more partitions to cover the space, we will get closer and closer to the actual area of the curve. By taking the limit of the expression as the norm of the partitions approaches zero, we arrive at the Riemann Integral . That is, we take the limit as the largest of the partitions approaches zero in size, so that all other partitions are smaller and the number of partitions approaches infinity. So, we take the limit on both sides of (3). This gives us
	:<math>F(b) - F(a)	\lim_{\ \Delta \ o 0} \sum_{i=1}^n x_i) </math>

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Fundamental Theorem Of Calculus