# The Fundamental Theorem of Calculus ![[Integral#^ia5q62]] The fundamental theorem of calculus shows us how to derive a function that outpus the area under a curve for a particular section. $ \begin{gather} A(f,a,b) =\int \limits_a^b f(x)dx = F(b)-F(a) \\ \text {where $F$ is antiderivative of $f$} \\ \text {i. e. } F’(x) = f(x) \end{gather} $ ^xr2kow ## Proof Start by assuming that a function that determines the area between 0 and $x$ for a particular curve $f(x)$ exists and is denoted as $A(x)$. For an infinitesimal change of $x$, the shape of the additional area approximates a rectangle with the width of $dx$ and height of $f(x)$. That can be described as: $dA = f(x) \, dx$ This means that the derivative of $A(x)$, the function we're looking for, is $f(x)$! This fact by itself does not get us very var because there exist many antiderivatives of $f(x)$, each with a different constant term, and we don't know which one to use. Luckily, one more fact we know, by definition, is that when $x=0$, $A(x)$ is also 0, because the area between 0 and 0 must also be 0. So if we evaluate any antiderivative of $f$ for both $x$ and 0, and subtract the two values, any constants will cancel out. This gives us $A(x) = F(x) - F(0)$ where $F(x) = \int f(x) \, dx$. This can then be expanded to evaluate the area under any section and not just those sections with the lower bound at 0 by subtracting the smaller section from the bigger one: $A(a, b) = (F(b) - F(0)) - (F(a) - F(0)) = F(b) - F(a)$ Because concepts we used to come to these conclusions are applicable to all smooth functions we can say that the formula for computing the integral is valid for all smooth functions as well. The above reasoning is presented by [[3Blue1Brown]] in [[The Essence of Calculus]]. Here are the relevant highlights: - [[The Essence of Calculus#^6bzelb]] - [[Integration and The Fundamental Theorem of Calculus#^l17wwx]] - [[Integration and The Fundamental Theorem of Calculus#^hxrhvt]] For a more rigorous proof see [[Fundamental Theorem of Calculus]] from Brilliant.