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The Fundamental Theorem of Calculus, Both Parts Explained

Most of a first calculus course is spent on two problems that look completely unrelated: finding the slope of a curve, and finding the area under one. The Fundamental Theorem of Calculus is the result that says those are the same problem viewed from opposite ends. It is the most important theorem in the subject, and it is the reason you can evaluate a definite integral without ever computing a limit of Riemann sums by hand.

The theorem comes in two parts that students routinely mix up. This guide separates them clearly, states the hypotheses that actually matter, and works through the version that appears most often on exams: differentiating an integral whose limits are functions of x.

Two Statements, One Relationship

Start with a function f that is continuous on a closed interval, and build a new function by accumulating signed area from a fixed left endpoint a out to a moving right endpoint x. Part 1 says that this accumulation function is differentiable and that its derivative is f itself. Part 2 says that if you already know any antiderivative of f, the total accumulation across the interval is simply the difference of that antiderivative's values at the two endpoints.

Read side by side, the two parts say that differentiation and integration undo each other. Part 1 starts with an integral and hands back a derivative; Part 2 starts with a derivative and hands back an integral's value. Part 2 is in fact a consequence of Part 1, which is why some textbooks call Part 1 the theorem and Part 2 the evaluation theorem. Whatever the naming, know which statement you are invoking.

\frac{d}{dx}\left[\int_a^x f(t)\, dt\right] = f(x)

Part 1: differentiating an accumulation returns the original function.

\int_a^b f(x)\, dx = F(b) - F(a), \quad F'(x) = f(x)

Part 2: evaluate a definite integral with any antiderivative F.

Part 1: Differentiating an Accumulation

Formally, let f be continuous on the closed interval from a to b, and define F at each x in that interval as the integral of f from a to x. Then F is continuous on the whole interval, differentiable on its interior, and its derivative at every such x equals f evaluated at x. The continuity of f is not decoration; it is what makes the derivative exist and equal f everywhere, and dropping it breaks the conclusion.

Notice the change of letter inside the integral. The variable of integration is a dummy, so it must be named something other than the limit, which is why the integrand is written in terms of t while the upper limit is x. Writing x in both places is a notation error that leads to real confusion about what is being held fixed.

The geometric reading is worth carrying around. F measures accumulated signed area up to the moving right edge, and the rate at which that area grows as the edge slides right is exactly the height of the graph at the edge. A tall graph means area piles up quickly; a graph below the axis means the accumulated total decreases. There is also a quiet but powerful consequence: every continuous function has an antiderivative, namely its own accumulation function, even when no formula in terms of familiar functions exists for it.

Part 2: Evaluating a Definite Integral

Part 2 is the computational workhorse. If f is continuous on the interval from a to b and F is any antiderivative of f, then the definite integral of f from a to b equals F at b minus F at a. That is the entire procedure behind every definite integral you have evaluated: find an antiderivative, plug in the top limit, plug in the bottom limit, subtract.

The word any deserves emphasis. Two antiderivatives of the same function differ by a constant, and that constant appears in both F at b and F at a, so it cancels in the subtraction. This is exactly why you drop the constant of integration when evaluating a definite integral, and why nobody has to agree on which antiderivative to use.

A concrete pass: to integrate three x squared plus two x from zero to two, take the antiderivative x cubed plus x squared. At the upper limit that is eight plus four, or twelve; at the lower limit it is zero. The definite integral equals twelve. Similarly, the integral of one over the square root of x from one to four uses the antiderivative two times the square root of x, giving four minus two, which is two. Neither computation required a Riemann sum, and that is the whole point.

Variable Limits and the Chain Rule

Exams rarely ask you to differentiate an accumulation with a bare x on top. Far more often the upper limit is itself a function of x, and then Part 1 combines with the chain rule. If the upper limit is g of x, the derivative of the integral is f evaluated at g of x, multiplied by the derivative of g. Structurally you are differentiating the composition of the accumulation function with g, so the outer derivative is f and the inner derivative is g prime.

For example, the derivative with respect to x of the integral of sine of t from zero to x squared is sine of x squared times two x. Students who forget the chain-rule factor lose the two x and get a wrong answer that looks almost right, which is the most expensive kind of mistake.

When both limits vary, split the integral at any convenient constant c inside the domain. The integral from u of x to v of x equals the integral from c to v of x minus the integral from c to u of x, and differentiating each piece gives f at v times v prime minus f at u times u prime. So the derivative of the integral of the natural logarithm of t from x to x squared, for x greater than one, is the logarithm of x squared times two x minus the logarithm of x, which simplifies to four x times the logarithm of x minus the logarithm of x.

\frac{d}{dx}\int_{u(x)}^{v(x)} f(t)\, dt = f(v(x))\, v'(x) - f(u(x))\, u'(x)

The Leibniz rule: chain-rule factors for both moving limits.

Where the Hypotheses Bite

The continuity requirement is the part most often skipped, and it produces spectacular errors. Consider the integral of one over x squared from negative one to one. Applying Part 2 mechanically with the antiderivative negative one over x gives negative one minus one, or negative two. That is impossible: the integrand is positive everywhere it is defined, so no honest area can be negative. The catch is that the integrand blows up at zero, which sits inside the interval, so Part 2 simply does not apply. Treated properly as an improper integral, it diverges.

A second habit worth breaking is equating a definite integral with area. A definite integral computes net signed area, counting regions below the axis as negative. The integral of sine from zero to two pi is zero, because the hump above the axis exactly cancels the trough below it, even though the total area enclosed is four. When a problem asks for total area or total distance travelled, integrate the absolute value, which usually means splitting at the zeros of the integrand.

Finally, Part 1 promises that an antiderivative exists, not that you can write it down. The accumulation function of e to the negative t squared is a perfectly well-defined, differentiable function, and it is genuinely useful in statistics, but no combination of elementary functions expresses it. Existence and expressibility are different questions.

Putting It to Work

Rewriting Part 2 with f prime in place of f gives the net change theorem: integrating a rate of change over an interval returns the net change in the quantity itself. That single sentence is why integrals show up throughout physics, biology, and economics. Integrate velocity and you get displacement; integrate the absolute value of velocity and you get distance travelled; integrate a marginal cost and you get the added total cost.

When you meet a definite integral, run a short mental checklist before computing anything.

Is the integrand continuous on the whole closed interval? If not, it is an improper integral, not a Part 2 problem.
Do you need an antiderivative, or is the answer a known area or a symmetry argument?
Does the problem want net signed area or total area? Total area needs absolute values and a split at the zeros.
If you are differentiating an integral, are the limits constant or variable? Variable limits need chain-rule factors.
After substitution, did you convert the limits, or convert the antiderivative back before evaluating?

Frequently asked questions

What is the difference between Part 1 and Part 2?

Part 1 differentiates an integral with a variable upper limit and returns the integrand. Part 2 evaluates a definite integral as the difference of any antiderivative's values at the endpoints. Part 1 is about building a function; Part 2 is about computing a number.

Does the function really have to be continuous?

Yes for the standard statements. If the integrand has an infinite discontinuity inside the interval, applying Part 2 produces nonsense, such as a negative value for a positive integrand. Handle those as improper integrals instead.

How do I differentiate an integral whose upper limit is a function of x?

Use Part 1 together with the chain rule: the derivative is the integrand evaluated at the upper limit, times the derivative of that limit. If both limits vary, subtract the same expression built from the lower limit.

Why can I ignore the constant of integration in a definite integral?

Any two antiderivatives differ by a constant, and that constant appears in both endpoint evaluations, so it cancels when you subtract. Every antiderivative gives the same definite integral.

Does a definite integral always equal the area under the curve?

Only when the function is non-negative. In general it gives net signed area, with regions below the axis counted negatively. For total area, integrate the absolute value and split the interval at the zeros of the function.