second fundamental theorem of calculus
Second fundamental theorem of calculus
The second fundamental theorem of calculus, also known as the Newton-Leibniz theorem, gives a practical method to evaluate a definite integral, as it replaces the potentially difficult limit-of-sums definition of the integral with a simpler subtraction of antiderivative values.
Let \[f\] be a real-valued function on a closed interval \[\left[ a,b \right]\] and \[F\] a continuous function on \[\left[ a,b \right]\] which is an antiderivative of \[f\] in \[\left( a,b \right)\]: \[F^{\prime}(x)=f(x)\]. If \[f\] is Riemann integrable on \[\left[ a,b \right]\], then \[\int_{a}^{b}f(x)\,dx=F(b)-F(a)\].
Intuitively, the second fundamental theorem of calculus tells us that "the total change is the sum of all the little changes". When \[dx\] is a tiny (infinitesimal) increment in \[x\], the product \[F^{\prime}(x)\cdot dx\] tells us how much \[F\] would change if it changed exactly at that constant slope, and when we add all these tiny changes we get the total change \[F(b)-F(a)\].
Proof
Partition \[\left[ a,b \right]\] into endpoints, \[a=x_{0}<x_{1}<\cdots<x_{n}=b\]. Note that the total change in the value of \[F\] across the interval \[\left[ a,b \right]\] is the sum of the changes in the value of \[F\] across all the tiny subintervals \[\left[ x_{k},x_{k+1} \right]\]: \[F(b)-F(a)=\sum_{k=0}^{n-1}\left( F(x_{k+1})-F(x_{k}) \right)\].
Now, as we chop \[\left[ a,b \right]\] into increasingly tiny slices, \[F(x_{k+1})-F(x_{k})\approx F^{\prime}(x_{k})(x_{k+1}-x_{k})\]. We can show this with the definition of derivatives:
Which implies that \[f(x+\Delta x)-f(x)\approx f^{\prime}(x)\Delta x\] when \[\Delta x\] is very small.
A more rigorous proof for this approximation is by using the mean value theorem. By mean value theorem, there must exist a \[c_{k}\in\left( x_{k},x_{k+1} \right)\] such that \[F^{\prime}(c_{k})=\frac{F(x_{k+1})-F(x_{k})}{x_{k+1}-x_{k}}\implies F^{\prime}(c_{k})(x_{k+1}-x_{k})=F(x_{k+1})-F(x_{k})\].
Substituting it back, \[F(b)-F(a)=\sum_{k=0}^{n-1}F^{\prime}(c_{k})(x_{k+1}-x_{k})=\sum_{k=0}^{n-1}F^{\prime}(c_{k})\Delta x_{k}\]. Then, we apply \[F^{\prime}(x)=f(x)\] and take the limit of both sides. This gives us \[\lim_{\Vert \Delta x_{k}\to0 \Vert}F(b)-F(a)=\lim_{\Vert \Delta x_{k}\to0 \Vert}\sum_{k=0}^{n-1}f(c_{k})\Delta x_{k}\]. However, since \[F(b)\] and \[F(a)\] isn't dependent on \[\Vert \Delta x_{k} \Vert\], we can simplify it to \[F(b)-F(a)=\lim_{\Vert \Delta x_{k}\to0 \Vert}\sum_{k=0}^{n-1}f(c_{k})\Delta x_{k}\].
The expression on the right side of the equation matches the definition of the integral over \[f\] from \[a\] to \[b\]. Therefore, we have proven that \[\int_{a}^{b}f(x)\,dx\].
Do note that it is perfectly fine to end up with \[c_{k}\] as the definition of Riemann integral only requires it (\[x_{k}^{\ast}\]) to be within the subinterval.