Fundamental theorem of calculus

The fundamental theorem(s) of calculus relate the concepts of differentiation (or calculating instantaneous rates of change) to integration (or accumulating the product of an integrand and differential).

Differentiation of the Integral

First fundamental theorem of calculus

For a continuous function $f$ , an antiderivative or indefinite integral $F$ can be obtained as the integral of $f$ over an interval with a variable upper bound.

Let’s break it down and try and digest it bit by bit.

Take a Function $f$ to be a continuous function that is defined by a real value on the closed interval $[t_i,t_n]$. Then there exists some indefinite integral or antiderivative function $F$ which is the integral of $f(t)$ with a variable upper bound $x$. The lower bound is fixed to some point within $[t_i,t_n]$ and is usually denoted as a constant $a$.

$$F(x)={\int }_{t_i}^{x}f(t)dt={\int }_a^{x}f(t)dt$$

We take the function $F$ to be defined for all of $x$ in the $[t_i,t_n]$ interval; we call this an indefinite integral. This can also be seen as an Area function or Accumulation function.

Note that we’ve defined $f$ within the interval $[t_i,t_n]$ and our independent variable $x$ is ‘constrained’ within this interval. i.e. $x\in [t_i,t_n]$.

Thus we can rewrite our integrand and differential in ^6d3128 in terms of $x$. Notation wise, we can also remove the bounds as we are implicitly saying the indefinite integral is a set of all the functions (within our arbitrary interval) that satisfies first theorem:

$$\int f(x)dx$$ $$\int f(x)dx=F(x)+C$$

The antiderivative $F$, when differentiated is equal to the original function $f$.

$$\begin{array}{rl}{F}^{\prime }(x)& =\frac{d}{dx}F(x)\\ & =\frac{d}{dx}{\int }_a^{x}f(x)dx\end{array}$$ $$F^{\prime }(x)=f(x)\text{for x∈[ti,tn]}$$

Evaluation of the Definite Integral

Second fundamental theorem of calculus

The integral of a function $f$ over a fixed interval is equal to the change of any antiderivative $F$ between the ends of the interval.

If $f$ is integrable on $[t_i,t_n]$ and $F$ is any antiderivative of $f$ on $[t_i,t_n]$ (meaning ^fafe59), then:

$$\text{require}physics{\int }_a^{b}f(x)dx=F(b)-F(a)=\text{eval}{F(x)}_a^b$$

So any definite integral can be evaluated by calculating the antiderivative/Area function, evaluating it at the two bounds, and taking the difference.

Recall the antiderivative function represents the area under the curve of the integrand (the function we are integrating over). If we plot both, we can see how they correspond - the value of $F$ plots out the area of $f$ across the domain.

Using the sine function

\begin{align} f(t) &= \sin(t) \ F’(t) &= f(t) \ F(t) &= -\cos(t) \ \frac{d}{dt} (-cos(t)) &= sin(t) \end{align}

\begin{document}
\begin{tikzpicture}[domain=-5:5]
    % Draw grid
    \draw[very thin, color=gray] (-5,-2.1) grid (5,2.1);
 
    
    % Draw axes
    \draw[->] (-5,0) -- (5,0) node[right] {$t$};
    \draw[->] (0,-2) -- (0,2) node[above] {$f(t)$};
 
    % Shade the area under the curve between 0 and pi
    \fill[blue!20, opacity=.5] 
         plot[smooth, domain=0:pi] (\x, {sin(deg(\x))}) 
         -- (3.14159,0) 
         -- (0,0) 
         -- cycle;
 
    % Label t0 and t1 on the x-axis
    \draw (0,0) node[below] {$t_0$};
    \draw (pi,0) node[below] {$t_1$};
 
    % Draw the graph of f(t) = sin(t)
    \draw[color=black, thick, smooth] plot (\x, {sin(deg(\x))}) node[right] {$f(t)=\sin(t)$};
 
    % Optionally, draw an antiderivative if desired
    \draw[color=black!50, thick, smooth] plot (\x, {-cos(deg(\x))}) node[above, yshift=20] {$F(t)=-\cos(t)$};
\end{tikzpicture}
\end{document}

Let’s try finding the area of $sin(t)$ between $0$ and $\pi$ by evaluating the definite integral:

$${\int }_0^{\pi }\sin (t)dt=F(\pi )-F(0)$$ $$\begin{array}{rl}F(\pi )& =1\\ F(0)& =-1\end{array}$$ $$F(\pi )-F(0)=1-(-1)=2$$

alternative interpretation

We can also visualise this integral in a geometric way with a unit circle (using radians as the unit of $\theta$):

$${\int }_0^{\pi }\sin (\theta )d\theta$$

As we go from $0$ to $\pi$ - on a unit Circle, a small angle $d\theta$ sweeps an arc with length $d\theta$ (since radius = 1 for unit circle, arclength is equal to the angle swept).

\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}%[scale=2, transform shape]

\draw [thick] (6,0) arc [start angle=0, end angle=180, radius=3]; \draw [thick] (0,0) — (6,0); % Define key points \coordinate (O) at (3,0); \coordinate (A) at (0,0); \coordinate (END) at (6,0); \coordinate (B) at ({3+3cos(40)}, {3sin(40)}); \coordinate (C) at ({3+3cos(40)}, 0); \coordinate (D) at ({3+3cos(48)}, {3sin(48)}); \coordinate (E) at ({3+3cos(48)}, 0); % Draw angle theta \draw [thick] (O) — (B); \draw [dashed] (B) — (C); \draw [dashed] (D) — (E); % Mark theta \draw [blue!20, thick] (3.4, 0.01) arc [start angle=0, end angle=48, radius=0.3] node [right] {$\theta$}; % Draw dTheta arc \draw [red, thick] (B) arc [start angle=40, end angle=48, radius=3]; % Add labels \node [below] at (O) {$.$}; \node [below] at (END) {$0$}; \node [below] at (A) {$\pi$}; \node [above right] at (B) {$d\theta$}; \node [below, scale=0.6, xshift=-10, opacity=0.8] at (C) {$\sin \theta \cdot d\theta$}; % Right-angle marker \draw (C) ++(-90:-0.1) — ++(-0.1,0) — ++(0,-0.1);

\end{tikzpicture} \end{document}

If we 'zoom in' non-rigorously and look at a right angle triangle where our arc length represents the hypotenuse:
```tikz

\usetikzlibrary{angles,quotes} \begin{document} \begin{tikzpicture} \coordinate (orig) at (0,0); \coordinate (B) at (4,0); \coordinate (C) at (0,3); \draw[thick,gray] (orig) — ++(4,0) node[below left, xshift=-40] {$sin\theta \cdot d\theta$}; \draw[thick,gray] (orig) — ++(0,3); \draw[red, thick] (B) — (C) node[above right, yshift=-40, xshift=60, scale=1.2] (z) {$d\theta$}; \pic [draw, →, "$\theta$", angle eccentricity=1.5] {angle = orig—C—B}; \draw (orig) ++(90:0.2) — ++(0.2,0) — ++(0,-0.2); \end{tikzpicture} \end{document}

$$

\begin{align} \sin(\theta) &= \frac{opp}{hyp} \ opp &= \sin(\theta)\cdot hyp \ &= \sin(\theta) \cdot d\theta \end{align}

The integrand and differential product is giving us the change in the horizontal length component, or if look at $d\theta$ as a vector, then the vector projection onto a straight line parallel to the x-axis. Integrating $sin(\theta)\,d\theta$ from $0$ to $\pi$ sums all these vertical bits over the upper half‐circle.

the total comes out to 2 – which is exactly the diameter of the circle.