The Fundamental Theorem of Calculus (FTC)

There are four somewhat different but equivalent versions of the Fundamental Theorem of Calculus.

• FTCI:
  Let $f(t)$ be continuous on $[a,b]$ and for $x$ in the interval $[a,b]$, define a function $F(x)$ by the definite integral:
  $$F(x) = \int_a^x f(t) dt.$$
  Then $F(x)$ is differentiable on $[a,b]$ and $F'(x) = f(x)$, for any $x$ in $[a,b]$.
• FTCII:
  Let $f(t)$ be continuous on $[a,b]$.
  Let $G(t)$ be any function such that $G'(t) = f(t)$, for any $t$ in $[a,b]$.
  Then we have:
  $$\int_a^b f(t) dt = G(b) - G(a).$$

• FTCIII:
  Let $g(t)$ have a continuous derivative on the interval $[a,b]$.
  Then we have the integral:
  $$\int_a^b g'(t) dt = g(b) - g(a).$$

• FTCIV:
  For $g$ a differentiable function of $t$ introduce its differential $dg = g'(t) dt$.
  Then we have the integral, for $g$ a differentiable function with continuous derivative:
  $$\int_a^b dg = g(b) - g(a).$$

Note that FTCIII and FTCIV are just rewrites of each other.
FTCIII has the intuitive phrasing: the integral of the instantaneous rate of change of a quantity is the total change.

Partial proof:

We defer for a moment the proof of FTCI.
We assume FTCI and prove the other parts of FTC from it.

• Proof of FTCII.
  For FTCII,assume given a function $G(t)$, such that $G'(t) = f(t)$, for any $t$ in $[a,b]$.
  Then we want to show that:
  $$\int_a^b f(t) dt = G(b) - G(a).$$
  As in FTCI, define $F(x)$ for $x$ in $[a,b]$ by the formula:
  $$F(x) = \int_a^x f(t) dt.$$
  Then by FTCI, we have $F'(x) = f(x)$ for any $x$ in $[a,b]$, so $F'(t) = f(t) = G'(t)$, for any $t$ in $[a.b]$.
  So $(G - F)'(t) = G'(t) - F'(t) = f(t) - f(t) = 0$, for any $t$ in $[a,b]$.
  So the function $G(t) - F(t)$ is constant.
  So $G(b) - F(b) = G(a) - F(a)$.
  So we get:
  $$G(b) - G(a) = F(b) - F(a) = \int_a^b f(t) dt - \int_a^a f(t) dt = \int_a^b f(t)dt.$$

• Proof of FTCIII and FTCIV.
  Since FTCIV is just a rewrite of FTCIII, it suffices to prove FTCIII.
  By FTCII, we have:
  $$\int_a^b g'(t) dt = G(b) - G(a),$$
  where $G(t)$ is any function such that $G'(t) = g'(t)$.
  So $G(t) = g(t)$ will do.
  With this choice of $G(t)$, we have:
  $$\int_a^b g'(t) dt = G(b) - G(a) = g(b) - g(a),$$
  giving us the required result and we are done.

Sample uses of FTC.

• Find a formula for the function $I(x)$ defined for any real $x$ by the integral:
  $$I(x) = \int_0^x t^2 dt.$$
  By FTCI, we have: $I'(x) = x^2$.
  But a function whose derivative is $x^2$ is the function $\frac{x^3}{3}$.
  Put $J(x) = I(x) - \frac{x^3}{3}$.
  Then $J'(x) = I'(x) - x^2 = x^2 - x^2 = 0$.
  So $J(x)$ is constant.
  But $J(0) = I(0) - 0 = 0$.
  So $J(x) = 0$ for all $x$.
  So $I(x) = \frac{x^3}{3}$, for any real $x$.
  We have shown that, for any real $x$, we have:
  $$I(x) = \int_0^x t^2 dt = \frac{x^3}{3}.$$
  Then for example we have:
  $$\int_0^3 t^2 dt = \frac{3^3}{3} = 9.$$

• Evaluate the definite integral:
  $$\int_0^\pi \sin(t) dt.$$
  By FTCII, we have:
  $$\int_0^\pi \sin(t) dt = G(\pi) - G(0),$$
  where $G(t)$ is any function such that $G'(t) = \sin(t)$.
  But one such function is $G(t) = - \cos(t)$, since $\frac{d}{dt}\cos(t) = - \sin(t)$.
  So we get:
  $$\int_0^\pi \sin(t) = (- \cos(\pi)) - (-\cos(0)) = (-(-1)) - (- 1) = 1 + 1 = 2.$$

• The velocity of a particle moving along the $x$-axis is $v(t) = t^2 - 5t + 6$.
  
  • If the particle is at the origin at time $t = 0$, where is the particle at time $t = 3$?
    
    By FTCIII, the change in position from $t = 0$ to $t = 3$, is given by:
    $$x(3) - x(0) = \int_0^3 x'(t) dt,$$
    where $x(t)$ is the position at time $t$.
    But also we have $x'(t) = v(t)$, so we get:
    $$x(3) - x(0) = \int_0^3 x'(t) dt = \int_0^3 v(t) dt = \int_0^3 (t^2 - 5t + 6)dt,$$
    $$x(3) = \int_0^3 (t^2 - 5t + 6)dt.$$
    Here we used the given fact that $x(0) = 0$.
    To do the remaining integral, we observe that $G(t) = \frac{t^3}{3} - \frac{5t^2}{2} + 6t$ obeys the equation: $G'(t) = t^2 - 5t + 6$, so FTCII applies to give:
    $$x(3) = \int_0^3 (t^2 - 5t + 6)dt = G(3) - G(0) = G(3)$$
    $$= \frac{3^3}{3} - \frac{5(3)^2}{2} + 6(3) = 9 - \frac{45}{2} + 18 = \frac{9}{2}.$$
    So the particle is at position $x = \frac{9}{2}$ at time $t = 3$.
  • How much distance does the particle cover in the time interval $[0, 3]$?
    The total distance covered $D$ is given by the integral of the absolute value of the velocity:
    $$D = \int_0^3 \vert v(t)\vert dt.$$
    Here $v(t) = t^2 - 5t + 6 = (t - 2)( t - 3)$.
    We see that in the interval $[0, 2]$, we have $v(t)\ge 0$ and $\vert v(t)\vert = v(t)$, whereas in the interval $[2, 3]$, we have $v(t)\le 0$ and $\vert v(t)\vert = -v(t)$, so we get:
    $$D = \int_0^3 \vert v(t)\vert dt = \int_0^2 v(t) dt + \int_2^3 - v(t) dt = \int_0^2 v(t) dt - \int_2^3 v(t) dt$$
    $$= (G(2) - G(0)) - (G(3) - G(2)) = 2G(2) - G(0) - G(3) = 2G(2) - \frac{9}{2}$$
    $$= 2( \frac{2^3}{3} - \frac{5(2)^2}{2} + 6(2)) - \frac{9}{2} = 2(\frac{8}{3} - 10 + 12) - \frac{9}{2} = \frac{28}{3} - \frac{9}{2} = \frac{29}{6}.$$