The fundamental theorem of calculus

The example below calculates the accumulation function of $\displaystyle {f(t)}={at+b}$ from 0, where the line is above the $\displaystyle x$-axis: the area between 0 and $\displaystyle x$ is a trapezium whose parallel sides are $\displaystyle {f(0)}={b}$ and $\displaystyle {f(x)}={ax+b}$ and whose height is $\displaystyle x$:

• $\displaystyle {F_0'(x)}={ax+b}={f(x)}$: the accumulation function is an antiderivative of $\displaystyle f$.
• Does this happen for every function? The theorem below says it does.

The fundamental theorem of calculus: if $\displaystyle f$ is continuous, the derivative of its accumulation function is $\displaystyle f$:

That is: the rate of change of the accumulated quantity is the given rate. For example, the derivative of the distance is the speed.

The proof below assumes that $\displaystyle f$ is positive and rising, and shows that the difference quotient of $\displaystyle F_a$ tends to $\displaystyle f(x)$ (when the function falls, or changes sign, the proof is similar).

• $\displaystyle F_a(x+h)-F_a(x)$, for $\displaystyle h>0$, is the area under the graph over the small piece $\displaystyle [x,x+h]$.
• $\displaystyle f$ rises, so on that piece $\displaystyle f(x)\le f\le f(x+h)$: the area is more than the rectangle of height $\displaystyle f(x)$ and less than the rectangle of height $\displaystyle f(x+h)$:

• Divide by $\displaystyle h$:

• As $\displaystyle h\to0$, $\displaystyle f(x+h)\to f(x)$, because $\displaystyle f$ is continuous. The quotient in the middle is trapped between two expressions that tend to $\displaystyle f(x)$, so it tends to $\displaystyle f(x)$ too.
• The limit of the difference quotient is the derivative, so $\displaystyle {F_a'(x)}={f(x)}$.

The accumulation function is an antiderivative of $\displaystyle f$, and the antiderivatives of $\displaystyle f$ are all $\displaystyle F(x)+C$. Of them, $\displaystyle F_a$ is the one whose value at $\displaystyle a$ is 0:

• $\displaystyle {F_a(x)}={F(x)+C}$, and $\displaystyle {F_a(a)}={0}$, so $\displaystyle {F(a)+C}={0}$ and $\displaystyle {C}={-F(a)}$.
• So, for any antiderivative $\displaystyle F$ of $\displaystyle f$:

• In short, write $\displaystyle {\left[F(x)\right]_a^b}={F(b)-F(a)}$.
• Any antiderivative will do: $\displaystyle C$ cancels in the difference.

In the last lesson the graph of $\displaystyle {F_0(x)}={\int_0^x\left(4t-t^2\right)dt}$ was given with no formula. The example below finds the formula, and the distance the car covered in the 4 seconds:

• An antiderivative of $\displaystyle 4t-t^2$: $\displaystyle 2t^2-\frac{t^3}{3}$.
• $\displaystyle {F_0(x)}={\left[2t^2-\frac{t^3}{3}\right]_0^x}={2x^2-\frac{x^3}{3}-0}={2x^2-\frac{x^3}{3}}$.
• $\displaystyle {F_0(4)}={2\cdot16-\frac{64}{3}}={\frac{32}{3}}$: the car covered $\displaystyle 10\frac{2}{3}$ metres.

Calculate the accumulation functions.

Calculate the definite integrals.

The board below shows the graph of $\displaystyle {f(x)}={2x}$.
$\displaystyle {F(x)}={\int_1^x f(t)\,dt}$.
Write $\displaystyle F(x)$ on the board, and make sure its graph passes through the point $\displaystyle (1,0)$.

$\displaystyle {G(x)}={\int_2^x\sqrt{t^2+5}\,dt}$.
$\displaystyle G(x)$ cannot be written with the functions you have learnt, but its derivative can be found by the fundamental theorem.

Find $\displaystyle G'(x)$.

Calculate $\displaystyle G'(2)$ and $\displaystyle G(2)$.