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Differentiating an accumulation at a point of continuity, evaluating by an antiderivative, and why the two halves are different theorems.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to prove both halves of the fundamental theorem — the first from additivity, the size estimate and continuity at a point, the second from the mean value theorem and a telescoping sum — and to say which hypotheses each needs. You will also be able to spot the standard misuse, an antiderivative that fails at an interior point, and to say what the accumulation does where the integrand jumps.
The integral with its additivity and size estimate, the mean value theorem, and the corollary that two functions with the same derivative on an interval differ by a constant. The two halves of this theorem use those three and nothing else.
The accumulation of a function is the function sending a point to the integral from a fixed start up to it. An antiderivative on an interval is a function differentiable there whose derivative is the given function at every point of it, the ends included.
The differentiation half. Let $f$ be integrable on $[a,b]$ and set $F(x) = \int_a^x f$. Then $F$ is continuous on $[a,b]$; and at every point $c$ where $f$ is continuous, $F$ is differentiable with $F'(c) = f(c)$.
Proof. Additivity gives $F(c+h) - F(c) = \int_c^{c+h} f$, so $$F(c+h) - F(c) - f(c)h = \int_c^{c+h}(f(t) - f(c))\,dt.$$ Given $\varepsilon$, continuity at $c$ makes the integrand smaller than $\varepsilon$ on the interval, and the size estimate bounds the integral by $\varepsilon|h|$. Divide by $h$.
The evaluation half. Let $f$ be integrable on $[a,b]$ and let $G$ be differentiable on $[a,b]$ with $G' = f$. Then $\int_a^b f = G(b) - G(a)$.
Proof. Partition, and write $G(b) - G(a)$ as a telescoping sum of changes across the pieces. The mean value theorem turns each change into $f(t_i)\Delta x_i$ for some $t_i$ in the piece, so the total is a Riemann sum, caught between the lower and upper sums. Refining squeezes those onto $\int_a^b f$ while the total never changes.
They are different theorems. The first accumulates and needs continuity at a point; the second evaluates and needs an antiderivative on the whole interval. Neither is a restatement of the other, and each has hypotheses the other does not.
Another way: picture
For the first half: nudge the right-hand end of a shaded region by $h$ and the area gained is a thin strip, whose area is about its height $f(c)$ times its width $h$. Dividing by $h$ leaves the height. That is the whole theorem, and the proof is that picture made into an inequality.
Another way: steps
To evaluate $\int_a^b f$:
Step 2 is where misuse happens. An antiderivative that fails at one interior point, or an integrand that is unbounded there, makes step 3 produce a number with no meaning.
Continuous integrands are the easy case. If $f$ is continuous on $[a,b]$, the first half says $F(x) = \int_a^x f$ is an antiderivative of $f$ on the whole interval. So every continuous function has an antiderivative, and the second half then applies with $G = F$ or with any other antiderivative, since they differ by a constant.
Integrable is not enough for an antiderivative. A function with a jump is integrable and is no function's derivative, because derivatives have the intermediate value property (Darboux's theorem, met in the intermediate value lesson). So the second half has a genuine hypothesis that the first half's conclusion does not supply.
Having an antiderivative is not enough for integrability either. The function $x^2\sin(1/x^2)$, extended by $0$, is differentiable everywhere, and its derivative is unbounded near $0$ — so that derivative has an antiderivative and is not Riemann integrable.
The two hypotheses are therefore independent, and the theorem's two halves overlap exactly on the continuous functions. That overlap is why elementary calculus can treat integration and antidifferentiation as the same operation without anyone noticing the gap.
| Result | Proof |
|---|---|
| substitution | apply the evaluation half to $G \circ u$ and use the chain rule |
| integration by parts | apply it to $uv$ and use the product rule |
| differentiating a variable-limit integral | the differentiation half plus the chain rule |
| the mean value theorem for integrals | the differentiation half plus the mean value theorem |
| an antiderivative is unique up to a constant | the constancy corollary from unit 4 |
Every technique of a first calculus course is in that table, and each is a one-line consequence once the theorem is available. That is the sense in which this lesson is the summit of the course's first five units: the rules were learned long ago, and this is where they are earned.
Using an antiderivative that fails inside the interval. $-1/x$ is an antiderivative of $1/x^2$ away from the origin, and applying the evaluation half across the origin gives a negative number for a positive integrand.
Assuming the accumulation is differentiable everywhere. It is differentiable where the integrand is continuous. At a jump the accumulation has a corner.
Treating the two halves as one theorem. They have different hypotheses, and there are integrable functions with no antiderivative and differentiable functions whose derivative is not integrable.
Forgetting that the integrand must be integrable. An unbounded integrand is not Riemann integrable at all, and the question then belongs to improper integrals, which are defined by a separate limit.
Let $H(x) = \int_0^{x^2} \cos t\,dt$. The integrand is continuous, so the first half applies to $F(u) = \int_0^u \cos t\,dt$.
Identify the accumulation inside.
So $F'(u) = \cos u$, and $H(x) = F(x^2)$.
The theorem, then the composition.
By the chain rule $H'(x) = \cos(x^2) \cdot 2x$. The chain rule is doing the outer work; the theorem supplies $F'$.
Two theorems, each in its place.
Evaluate $\int_{-1}^{1} \frac{dx}{x^2}$ by the antiderivative $-1/x$: the difference of end values is $-2$.
A computation that looks routine.
The integrand is positive wherever defined, so a negative answer is impossible; and it is unbounded near $0$.
The answer contradicts the order property.
The integral does not exist as a Riemann integral. The antiderivative was not differentiable at $0$, and neither hypothesis of the evaluation half held.
The hypotheses, not the arithmetic, failed.
The integrand is bounded on any bounded interval and has finitely many jumps there, so it is integrable and $F$ is defined and continuous.
Integrability first, then continuity of $F$.
The first half gives $F'(x) = \lfloor x \rfloor$ at every non-integer $x$, since the integrand is continuous there.
The theorem, at its points of continuity.
At each integer the one-sided derivatives are the two values of the floor function there, so $F$ has a corner and no derivative — one corner per integer, and $F$ is differentiable exactly on the non-integers. The accumulation is one degree smoother than its integrand and no more.
Build the proof that the accumulation $F(x) = \int_a^x f$ of a continuous $f$ satisfies $F'(c) = f(c)$.
This task has no paper form; do it on a device.
Compute $\displaystyle\int_0^{3} 3x\,dx$ using the evaluation half of the theorem.
Answer:
Put the steps of the proof that the integral of $f$ equals the difference of an antiderivative's end values into order.
Number the steps in order (write the number in the box):
Match each statement to the half of the theorem it belongs to, or to neither.
| the differentiation half | the evaluation half | a corollary of the differentiation half | false | |
|---|---|---|---|---|
| the accumulation of $f$ has derivative $f$ at each point of continuity | ||||
| the integral of $f$ is the difference of an antiderivative's end values | ||||
| every continuous function on an interval has an antiderivative | ||||
| every integrable function has an antiderivative |
Here is a computation of $\displaystyle\int_{-1}^{1} \dfrac{1}{x^2}\,dx$. Mark the one step that is not permitted.
This task has no paper form; do it on a device.
Let $f$ be $0$ below $1$ and $6$ from $1$ on, and let $F(x) = \int_0^x f$. What happens to $F$ at $x = 1$?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Build the proof that the accumulation $F(x) = \int_a^x f$ of a continuous $f$ satisfies $F'(c) = f(c)$.
This task has no paper form; do it on a device.
You can prove both halves, evaluate an integral by an antiderivative, and say why an integrable function need not have one. Say in your own words what goes wrong when an antiderivative fails at one interior point. Next: what happens when a whole sequence of functions is allowed to move.
10. Your turn: where is $F(x) = \int_0^x \lfloor t \rfloor\,dt$ differentiable?, step 3