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Both parts, why each is true, and why it is a surprising claim rather than a definition restated.
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 state both parts of the fundamental theorem with the continuity hypothesis each needs, differentiate an accumulation function including the chain rule factor when the upper limit is not $x$ and the sign change when the variable is in the lower limit, evaluate a definite integral by finding any antiderivative and taking top minus bottom, say why any antiderivative will do, refuse to apply the second part across a discontinuity inside the interval, and say why the theorem is a claim that could have been false rather than a definition being restated.
The derivative was defined in unit 3 as a limit of difference quotients — a slope. The integral was defined in this unit as a limit of Darboux sums — an area. Nothing so far connects them, and there is no reason from the definitions to expect a connection: one is about how steeply a function rises, the other about how much room is underneath it.
The theorem below says they are inverse operations. That is not a definition being unpacked; it is a genuinely surprising fact, and it is what makes calculus a subject rather than two subjects.
Part 1. If $f$ is continuous on $[a,b]$, define the accumulation function $$F(x) = \int_a^x f(t)\,dt.$$ Then $F$ is differentiable on $(a,b)$ and $$F'(x) = f(x).$$
Part 2. If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then $$\int_a^b f(x)\,dx = F(b) - F(a).$$
The two point in opposite directions and both are needed.
Part 1 builds an antiderivative out of an integral. It is therefore an existence theorem: every continuous function has an antiderivative — its own accumulation function — even when no formula for it can be written. $\int_0^x e^{-t^2}dt$ is a perfectly good antiderivative of $e^{-t^2}$, and no elementary formula for it exists.
Part 2 evaluates an integral out of an antiderivative. It is what makes integration computable: a limit of sums, replaced by two substitutions.
"Any" antiderivative. The constant cancels in $F(b) - F(a)$, which is why the second part does not care which member of the family you pick.
With a variable upper limit that is not $x$: $\dfrac{d}{dx}\displaystyle\int_a^{u(x)} f(t)\,dt = f(u(x))\,u'(x)$ — the chain rule, applied to Part 1.
Another way: picture
The accumulation function $F(x) = \int_a^x f$ as a region whose right edge slides. Push the edge right by $dx$ and the region gains a sliver of height $f(x)$ and width $dx$ — so the area grows at rate $f(x)$. That sliver is Part 1, and the proof is that argument made precise with the mean value theorem for integrals.
Another way: steps
To evaluate $\int_a^b f$: check $f$ is continuous on $[a,b]$; find any antiderivative $F$; compute $F(b) - F(a)$.
To differentiate $\int_a^{u(x)} f$: substitute $u(x)$ into $f$, then multiply by $u'(x)$. Do not integrate.
If the variable is in the lower limit: flip it with the orientation rule first, picking up a minus sign.
The argument is three lines and worth following, because it is where the mean value theorem pays off for the last time.
$$\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\,dt$$ by additivity. The right side is the average value of $f$ on $[x, x+h]$.
Since $f$ is continuous, the mean value theorem for integrals gives a $c_h$ in $[x, x+h]$ with that average equal to $f(c_h)$. As $h \to 0$, $c_h \to x$, and continuity gives $f(c_h) \to f(x)$.
So $F'(x) = f(x)$. Continuity is used twice — once to get $c_h$, once to pass to the limit — and both uses are essential.
Part 2 follows from Part 1 plus the constant theorem. If $G$ is any antiderivative of $f$, then $G$ and the accumulation $F$ differ by a constant (lesson 26, resting on lesson 22). So $$G(b) - G(a) = F(b) - F(a) = \int_a^b f - 0 = \int_a^b f.$$
The whole chain — completeness, extreme value theorem, Fermat, Rolle, mean value theorem, the constant theorem, Part 1, Part 2 — is the argument this course has been building since lesson 1.
Ignoring the continuity hypothesis. $\int_{-1}^{1} x^{-2}dx$ evaluated as $[-x^{-1}]$ gives $-2$: negative, for a positive integrand. The integrand is undefined at $0$, inside the interval, so the theorem never applied. Check for poles inside the limits before evaluating.
Forgetting the chain rule on a variable limit. $\dfrac{d}{dx}\int_0^{x^2} f = f(x^2)\cdot 2x$, not $f(x^2)$.
Leaving the variable in the lower limit. $\dfrac{d}{dx}\int_x^b f = -f(x)$. Flip first.
Integrating when Part 1 would answer it. $\dfrac{d}{dx}\int_1^x \sqrt{1+t^3}\,dt$ is $\sqrt{1+x^3}$ immediately. Trying to find the antiderivative first is work that cannot be completed.
Reading $+C$ into a definite integral. The constant cancels; a definite integral is a number.
Confusing the variable of integration with the limit. In $\int_a^x f(t)dt$ the $t$ is bound and internal; the $x$ is the variable the result depends on. Writing $\int_a^x f(x)dx$ uses one letter for two things.
Because integration is taught as "anti-differentiation", the fundamental theorem can read as a restatement of a definition. It is not. Two entirely separate constructions were made — a limit of difference quotients and a limit of Darboux sums, with no reference to each other — and the theorem asserts that one undoes the other.
Nothing in the definitions made that inevitable. The Darboux construction would have defined a perfectly sensible number for a function with no antiderivative at all, and it does: $\int_0^1 e^{-x^2}dx$ exists and is about $0.7468$. What the theorem adds is that when an antiderivative can be found, the area is a difference of its values — turning an infinite process into two substitutions.
That is why it earns the name, and why the sliver picture is worth keeping: the area grows at the rate the curve is high, and everything else is bookkeeping on top of that one observation.
$H(x) = \displaystyle\int_{x}^{x^2}\sin(t^2)\,dt$. There is no antiderivative to find, so Part 1 is the only route.
Split at a constant first.
$H(x) = \int_{x}^{0} + \int_0^{x^2} = -\int_0^{x}\sin(t^2)dt + \int_0^{x^2}\sin(t^2)dt$, using additivity and orientation.
Now each has a constant lower limit.
Differentiating each with Part 1 and the chain rule: $H'(x) = -\sin(x^2) + \sin(x^4)\cdot 2x$.
One factor per variable limit.
$\displaystyle\int_{-2}^{2}\frac{1}{x}dx$. An antiderivative is $\ln|x|$, and $\ln|2| - \ln|-2| = 0$, which looks tidy and symmetric.
The tempting calculation.
But $1/x$ is undefined at $0$, which is inside $[-2,2]$, so the theorem does not apply and the number means nothing.
Check the interval before evaluating.
The integral is divergent: the two halves are $+\infty$ and $-\infty$ and do not cancel in any sense the definition allows. Calculus II makes this precise; here it is enough to know the answer $0$ is not one.
Part 1 with a variable upper limit $u = 3x$: no antiderivative is needed, and none exists.
Recognise the shape.
Substitute the limit into the integrand: $\cos((3x)^2) = \cos(9x^2)$.
Multiply by $u' = 3$: the answer is $3\cos(9x^2)$.
Match each FTC situation to the necessary method or hypothesis.
| $A'(x)=f(x)$ | Derivative is $f(g(x))g'(x)$ | Use $F(b)-F(a)$ | They differ by a constant, which cancels in a definite integral | |
|---|---|---|---|---|
| $A(x)=\int_0^x f(t)dt$ with continuous $f$ | ||||
| $\int_0^{g(x)}f(t)dt$ | ||||
| $\int_a^b f(x)dx$ with antiderivative $F$ | ||||
| Two antiderivatives of the same function |
Match each situation to the Fundamental Theorem step that resolves it.
| It is $\sqrt{1+x^4}$ | It is $2x\cos(x^2)$ | Use $[x^2+x]_0^3$ | $A$ is decreasing there | |
|---|---|---|---|---|
| $\frac{d}{dx}\int_0^x \sqrt{1+t^4}\,dt$ | ||||
| $\frac{d}{dx}\int_1^{x^2} \cos t\,dt$ | ||||
| $\int_0^3 (2x+1)\,dx$ | ||||
| $A(x)=\int_0^x f(t)dt$ and $f(x)<0$ |
Evaluate $\displaystyle\int_0^{1} 6x^2 \, dx$.
Answer:
Let $G(x) = \displaystyle\int_0^{x^{4}} 5t \, dt$. Find $G'(2)$.
Answer:
To find $\dfrac{d}{dx}\displaystyle\int_1^{x} \sqrt{1 + t^{8}}\,dt$, which part of the theorem is needed?
| Use FTC Part I | Use FTC Part II: $F(b)-F(a)$ | Use Part I and multiply by $g'(x)$ | Its accumulation function still has derivative equal to the integrand | |
|---|---|---|---|---|
| Differentiate $\int_a^x f(t)dt$ | ||||
| Evaluate $\int_a^b f(x)dx$ with antiderivative $F$ | ||||
| Upper bound is $g(x)$ | ||||
| Continuous integrand has no elementary antiderivative formula |
$A(x) = \displaystyle\int_0^x (4 - t)\,dt$. At which $x > 0$ is $A$ largest?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A student writes $\displaystyle\int_{-1}^{1}\frac{1}{x^2}dx = \left[-\frac1x\right]_{-1}^{1} = -1 - 1 = -2$. What is wrong?
| FTC endpoint evaluation is permitted | Use of ordinary FTC is invalid; investigate improper behavior | A finite integral cannot be negative | Definite value is $F(b)-F(a)$ when hypotheses hold | |
|---|---|---|---|---|
| Integrand continuous on $[a,b]$ | ||||
| Integrand has a pole inside $[a,b]$ | ||||
| Integrand is positive throughout | ||||
| An antiderivative $F$ exists on the interval |
You can use both parts, and say which one a question needs. Say in your own words why the first part is an existence theorem. Next: substitution, which is the chain rule read backwards.
9. Your turn: $\dfrac{d}{dx}\displaystyle\int_2^{3x}\cos(t^2)\,dt$, step 3
One factor for the chain rule.