First Fundamental Theorem of Calculus: If f is continuous on [a, b], then F(x) = integral a to x f(t) dt has a derivative at every point of [a, b] and dF/dx = d/dx integral a to x f(t) dt = f(x), a is less than and equal to x is less than equal to b. Let us understand Proof of Fundamental Theorem of Calculus: We prove the fundamental theorem of calculus by applying the definition of derivative directly to the function F(x).
This means writing out the difference quotient F(x + h) – F(x) /h and showing that its limits as h approaches to 0 is the number f(x). When we replace F(x + h) and F(x) by their definite integrals, the numerator in above equation becomes F(x + h) – F(x) = integral a to x+h f(t) dt – integral a to x f(t) dt. The additive rule for integrals simplifies the right hand side to Integral x to x+h f(t) dt So that the above equation becomes F(x + h) – F(x) /h = 1/h [F(x + h) – F(x)] = 1/h integral x to x+h f(t) dt. According to the mean value theorem for definite integrals, the value of the last expression in the above equation is one of the values taken on by f in the interval joining x and x + h. That is for some number c in this interval, 1/h integral x to x + h f(t) dt = f(c).
We can therefore find out what happens to (1/h) times the integral as h approaches to 0 by watching what happens to f(c) as h approaches to 0. As h approaches to 0, the endpoint x + h approaches x, pushing c ahead of it like a bead on a wire. So c approaches x, and since f is continuous at x, f(c) approaches f(x) .Lim h approaches 0 f(c) = f(x).Going back to the beginning, then we have dF/dx = lim h approaches to 0 [F(x+h) – F(x)] /h= lim h approaches 0 1/h integral x to x+h f(t) dt = lim h approaches 0 f(c)= f(x). This concludes the proof.
Let us more understand this through Fundamental Theorem of Calculus problems: let us take few fundamental theorem of calculus examples .suppose we have Find dy/dx if y = integral 1 to x^2 cos t dt.
Now to understand this solution suppose let us notice that the upper limit of integration is not x but x^2. To find dy/dx we must therefore treat y as the composite of y = integral 1 to u cos t dt and u = x^2 and apply the chain rule: dy/dx = (dy/du).(du/dx) = d/du integral 1 to u cos t dt . du/dx = cos u . du/dx= cos x^2 . 2x = 2x cos x^2.
Second Fundamental Theorem of Calculus: If f is continuous at every point of [a, b] and F is any anti-derivative of f on [a, b], then Integral a to b f(X) dx = F(b) – F(a). Let us understand this by second Fundamental Theorem of Calculus Examples suppose we have Evaluate integral 0 to pi cos x dx. Now let us solve this integral 0 to pi cos x dx = sin x 0 to pi = sin pi – sin 0 = 0 – 0 = 0.
This means writing out the difference quotient F(x + h) – F(x) /h and showing that its limits as h approaches to 0 is the number f(x). When we replace F(x + h) and F(x) by their definite integrals, the numerator in above equation becomes F(x + h) – F(x) = integral a to x+h f(t) dt – integral a to x f(t) dt. The additive rule for integrals simplifies the right hand side to Integral x to x+h f(t) dt So that the above equation becomes F(x + h) – F(x) /h = 1/h [F(x + h) – F(x)] = 1/h integral x to x+h f(t) dt. According to the mean value theorem for definite integrals, the value of the last expression in the above equation is one of the values taken on by f in the interval joining x and x + h. That is for some number c in this interval, 1/h integral x to x + h f(t) dt = f(c).
We can therefore find out what happens to (1/h) times the integral as h approaches to 0 by watching what happens to f(c) as h approaches to 0. As h approaches to 0, the endpoint x + h approaches x, pushing c ahead of it like a bead on a wire. So c approaches x, and since f is continuous at x, f(c) approaches f(x) .Lim h approaches 0 f(c) = f(x).Going back to the beginning, then we have dF/dx = lim h approaches to 0 [F(x+h) – F(x)] /h= lim h approaches 0 1/h integral x to x+h f(t) dt = lim h approaches 0 f(c)= f(x). This concludes the proof.
Let us more understand this through Fundamental Theorem of Calculus problems: let us take few fundamental theorem of calculus examples .suppose we have Find dy/dx if y = integral 1 to x^2 cos t dt.
Now to understand this solution suppose let us notice that the upper limit of integration is not x but x^2. To find dy/dx we must therefore treat y as the composite of y = integral 1 to u cos t dt and u = x^2 and apply the chain rule: dy/dx = (dy/du).(du/dx) = d/du integral 1 to u cos t dt . du/dx = cos u . du/dx= cos x^2 . 2x = 2x cos x^2.
Second Fundamental Theorem of Calculus: If f is continuous at every point of [a, b] and F is any anti-derivative of f on [a, b], then Integral a to b f(X) dx = F(b) – F(a). Let us understand this by second Fundamental Theorem of Calculus Examples suppose we have Evaluate integral 0 to pi cos x dx. Now let us solve this integral 0 to pi cos x dx = sin x 0 to pi = sin pi – sin 0 = 0 – 0 = 0.
