#readwise # Fundamental Theorem of Calculus  ## Metadata - Author: [[brilliant.org]] - Full Title: Fundamental Theorem of Calculus - URL: https://brilliant.org/wiki/fundamental-theorem-of-calculus/ ## Summary The Fundamental Theorem of Calculus connects differentiation and integration. It states that if a function is continuous, the area function $S(x)$ defined by the integral of that function is continuous and differentiable. The first part of the theorem shows that the derivative of $S(x)$ equals the original function $f(x)$. The second part states that the definite integral of $f$ from $a$ to $b$ equals the difference of its antiderivatives at those points. ## Highlights We have learned about indefinite integrals, which was the process of finding the antiderivative of a function. In contrast to the indefinite integral, the result of a definite integral will be a number, instead of a function. The definite integral of a function is the signed area under the graph of the function, and is expressed in the form of $\int_a^b f(x) \, dx$:  Now, suppose that we formed an area function $S(x)$ in such a way that it is dependent on the function $f(x)$ as $S(x)=\int _{ a }^{ x }{ f(t)\, dt }$ where $f$ is continuous on the interval $[a,b]$. Now, suppose we wanted to find the the rate of change of the area with respect to x:  We can see from the figure above that the area of the shaded region is equal to the area under the curve $f(t)$ from $a$ to $x+\Delta x$ minus the area under $f(t)$ from $a$ to $x$. Thus, $\begin{aligned} \Delta S&=A(x+\Delta x)-A(x)\\\\ \frac{\Delta S}{\Delta x}&=\frac{A(x+\Delta x)-A(x)}{\Delta x}. \end{aligned}$​ So, the rate of change of area becomes $S'(x)=\frac{dS}{dx}=\lim_ {\Delta x\rightarrow 0 } \frac { S(x+\Delta x)-S(x) }{ \Delta x }$ We know that there is an $\overline{x}$ found between $x$ and $x+\Delta x$ such that the area of the shaded region is equal to $f(\overline{x})\Delta x$: $ \begin{aligned} S'(x) &=\lim_{\Delta x\rightarrow 0 }\frac { S(x+\Delta x)-S(x) }{ \Delta x } \\ &=\lim_{\Delta x\rightarrow 0 }\frac { f(\overline { x } )\Delta x }{ \Delta x } \\ &=\lim_{\Delta x\rightarrow 0 } f(\overline { x } )\\ &=f(x). \end{aligned} $ The last step is true because, as $\Delta x\rightarrow 0$, anything found between $x$ and $x+\Delta x$ approaches $x$. So, now we are ready to state the first fundamental theorem of calculus: If $f$ is continuous on $[a,b]$, then the function defined by $S(x)=\int _{ a }^{ x }{ f(t)\, dt }$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and $S'(x)=f(x)$. ([View Highlight](https://read.readwise.io/read/01jfmra6b4v0ez1a95y5q32ehe)) ---