The Fundamental Theorem of Calculus: Part 2

The Fundamental Theorem of Calculus has two parts. Part 1 begins with an integral and differentiates it: the accumulation function $A(x) = \displaystyle\int_a^x f(t)\,dt$ satisfies $A' = f$, which is what guarantees that every continuous $f$ has an antiderivative.

Part 2 begins with an antiderivative and evaluates an integral: given any $F$ with $F' = f$, it returns the value of $\displaystyle\int_a^b f(x)\,dx$ as a number, with no Riemann sums to compute.

What does this mean?

Imagine climbing a staircase whose steps vary in height. Adding up every step's rise gives the same number as reading the height at the top and subtracting the height at the bottom. The individual steps never need to be counted, provided both heights are known.

Here the steps are $f$, the rate at which height is gained, and the running height is $F$. Summing the rises across $[a, b]$ is $\displaystyle\int_a^b f(x)\,dx$, and the net change in height is $F(b) - F(a)$. Part 2 says the two are equal.

The staircase also shows why a subtraction appears. What the integral gives is a change in $F$, never a single value of $F$: reading $F(b)$ on its own would depend on the height $F$ started from, and subtracting $F(a)$ removes it. On the staircase that difference is the climb rather than the height reached.

Suppose $f$ is continuous on $[0, 5]$ and $F$ is an antiderivative of $f$. Given that $\displaystyle\int_0^5 f(x)\,dx = 14$ and $F(0) = 3$, find $F(5)$.

Solution

No integrand is supplied, so there is nothing here to integrate. What we are given is the equation FTC Part 2 asserts between the integral and two values of $F$, with one of those values unknown.

Since $f$ is continuous on $[0, 5]$ and $F' = f$, Part 2 applies with lower limit $0$ and upper limit $5$:

We substitute the two numbers we were given, $14$ for the integral and $3$ for $F(0)$:

We add $3$ to both sides, which gives $F(5) = 17$.

Read as net change, this says that $F$ has value $3$ at $x = 0$, and the integral records a net change of $14$ across $[0, 5]$, so the value at $x = 5$ is $17$.

Part 2 is one equation relating three quantities: the integral, the value of $F$ at the upper limit, and its value at the lower limit. Any one of them follows from the other two.

FTC Part 2 turns a definite integral into two steps: find an antiderivative of the integrand, then take the difference of its values at the two limits of integration.

A definite integral is defined as a limit of Riemann sums, so evaluating one from the definition means summing rectangles and taking a limit. Part 2 replaces that limit with two evaluations of a single function and one subtraction.

It also completes a connection. A definite integral is a number, fixed by net signed area; an antiderivative is a function, found by reversing differentiation. Part 1 joined them in one direction, building an antiderivative out of accumulated area. Part 2 uses the join in the other direction: an antiderivative is what computes the number.

Finding the antiderivative needs only the facts already established: the power rule, and the antiderivatives of $e^x$, $\cos x$ and $\sin x$ etc...

Evaluate $\displaystyle\int_1^4 3x^2\,dx$.

Solution

The integrand is $f(x) = 3x^2$, a polynomial, so $f$ is continuous on $[1, 4]$ and FTC Part 2 applies.

We first need an antiderivative. Taking the constant $3$ outside and applying the power rule for antiderivatives,

so we may take $F(x) = x^3$. Differentiating confirms it: $F'(x) = 3x^2 = f(x)$.

Next we evaluate $F$ at each limit of integration. The upper limit is $4$ and the lower limit is $1$:

Finally we subtract, upper value first: