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Fundamental Theorem of Calculus

The fundamental theorem of calculus links derivatives and integrals. Part 1 says the derivative of an accumulated area is the original function. Part 2 says you can evaluate a definite integral by finding an antiderivative F and computing F(b) − F(a), with no Riemann sums.

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Chapter 5 · Lesson 4 of 6Calculus 1 · Integrals
  1. 1.Antiderivatives and Indefinite Integrals
  2. 2.Riemann Sums
  3. 3.The Definite Integral
  4. 4.Fundamental Theorem of Calculus
  5. 5.U-Substitution
  6. 6.Average Value of a Function
See all calculus chapters →

The fundamental theorem of calculus (FTC) says that differentiation and integration are inverse processes. It comes in two parts:

Part 1. If is continuous on , then the function has derivative :

Part 2. If is continuous on and is any antiderivative of , then

Part 2 is the one you’ll use most. It turns the limit of Riemann sums from the definite integral lesson into a two-step calculation: find an antiderivative, then subtract its values at the endpoints.

Part 1: the derivative with respect to x of the integral from a to x of f of t d t equals f of x. Part 2: the integral from a to b of f of x d x equals F of b minus F of a
The two parts of the fundamental theorem of calculus

The Accumulation Function

Part 1 is about a function defined by an integral. Fix a starting point and let the right endpoint vary:

is the net area under from to , so it’s called an accumulation function (or area function). The letter is a dummy variable; it’s used so that can be the endpoint.

Example. For and , the region from 0 to is a trapezoid (for ), and

Left: the line f of t equals 2 minus t over 2, with the area from 0 to x equals 3 shaded blue. Right: the graph of the accumulation function A of x equals 2 x minus x squared over 4, a downward parabola. The point at x equals 3 has height 3.75, the shaded area. A rises while f is positive, peaks at x equals 4 where f equals 0, and falls after that, where f is negative
A(x) collects the area under f; its slope at each x is f(x)

Look at how the two graphs match up. While is positive, area is being added, so increases. At , and has its maximum. After that is negative, area is subtracted, and decreases. That’s the behaviour of a function whose derivative is , and indeed .

FTC Part 1 (the First Fundamental Theorem)

Fundamental theorem of calculus, Part 1. If is continuous on , then is differentiable on , and .

In words: the rate at which area accumulates is the height of the curve. Integrating and then differentiating gets you back to . It also shows that every continuous function has an antiderivative, namely .

Why it works. Increase by a small amount . The extra area is a thin strip under the curve between and . Its width is and its height is close to :

A curve y equals f of t with the area from a to x shaded light blue. A thin amber strip from x to x plus h sits on top of it. The strip is almost a rectangle of width h and height f of x, so its area is about f of x times h
The new area from x to x + h is almost a rectangle of height f(x)

Divide by and let . Since is continuous, the strip’s height closes in on , and the difference quotient becomes the derivative:

A full proof uses the mean value theorem for integrals to pin the strip’s height between the minimum and maximum of on .

Derivative of an Integral

Part 1 lets you differentiate an integral without evaluating it. That matters when the integrand has no simple antiderivative.

Example 1. Find .

The lower limit is a constant and the upper limit is , so substitute for in the integrand:

The lower limit doesn’t matter: changing it only changes the integral by a constant.

When the upper limit is a function of , combine Part 1 with the chain rule:

Example 2. Find .

Here and :

When is in the lower limit, reverse the limits first. That changes the sign.

Example 3. Find .

When both limits depend on , split the integral at a constant and handle each piece:

Example 4. Find .

Neither of the last two integrals can be evaluated with elementary functions, but their derivatives take one line.

FTC Part 2 (the Second Fundamental Theorem)

Fundamental theorem of calculus, Part 2. If is continuous on and is any antiderivative of , then .

Part 2 is also called the evaluation theorem. The difference is often written with brackets or a bar:

Why it works. By Part 1, is an antiderivative of . Any other antiderivative differs from it by a constant, . Then

and . The constant cancels, which is also why you never need in a definite integral.

Naming. Most textbooks call the evaluation formula the second fundamental theorem of calculus, as here. Some courses number the two parts the other way round, so check which statement your teacher means.

How to Evaluate a Definite Integral

  1. Find an antiderivative of the integrand, using the antiderivative rules. Leave off .
  2. Evaluate at the upper limit and at the lower limit .
  3. Subtract: , upper minus lower.
  4. Check that is continuous on . If it isn’t, the theorem doesn’t apply.

Example 5. Evaluate .

Example 6. Evaluate .

One arch of the sine curve has area exactly 2.

Example 7. Evaluate .

Rewrite as powers: . An antiderivative is :

Example 8. Evaluate .

Example 9 (checking an earlier answer). In the Riemann sums lesson, sums for on approached . With Part 2:

The limit that took a table of sums is now one subtraction.

The Net Change Theorem

Part 2 has a useful restatement. If is the rate of change of a quantity , then integrating the rate gives the total change in the quantity:

This is the net change theorem: the integral of a rate of change is the net change. It turns rate information into amounts:

Rate Integral gives
velocity displacement, the net change in position
flow rate into a tanknet change in the amount of liquid
population growth ratenet change in population
marginal cost change in cost from to

To find the amount at the end, add the net change to the starting amount: .

Example 10. Water flows into a tank at liters per minute for . The tank holds 50 L at . How much does it hold at ?

The amount rises by 100 L, so the tank holds L.

Displacement vs distance. The integral of velocity counts motion in the negative direction as negative, so it gives displacement. For the total distance traveled, integrate speed, : split the interval where changes sign and add the absolute values.

Example 11. A particle moves along a line with velocity m/s for . Find its displacement and the total distance it travels.

The velocity curve v of t equals t squared minus 4 t plus 3 from t equals 0 to 4. It is above the axis from 0 to 1 and from 3 to 4, with blue regions of area four thirds each, and below the axis from 1 to 3, with a rose region of area four thirds. Displacement is four thirds; total distance is 4
Displacement adds signed areas; distance adds their sizes

An antiderivative is . The displacement is

For distance, changes sign at and . The three pieces are

so the total distance is m. The particle goes forward, comes back, then goes forward again, ending only m from where it started.

Common Mistakes

  • Subtracting in the wrong order. It’s : upper limit minus lower limit.
  • Dropping parentheses at the lower limit. . Without the parentheses, the signs of the lower-limit terms go wrong.
  • Forgetting the chain rule factor. , not .
  • Forgetting the sign when is the lower limit. Reverse the limits first.
  • Ignoring discontinuities. is not . The integrand is positive, so a negative answer is impossible. isn’t continuous at 0, so Part 2 doesn’t apply.
  • Confusing displacement with distance. Integrating gives displacement; integrate for distance.

For proofs of both parts and more examples, see Sections 5.3 and 5.4 of the free textbook OpenStax Calculus Volume 1.

Practice Problems

Try each one before opening the answer. The practice set has 12 more with instant checking.

Problem 1. Evaluate .

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.

Answer: 4

Problem 2. Evaluate .

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.

Answer: 1

Problem 3. Evaluate .

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. Then .

Answer:

Problem 4. Find .

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By Part 1, replace with .

Answer:

Problem 5. Find .

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The upper limit is , so multiply by .

Answer:

Problem 6. A car’s velocity is m/s for . How far does it travel?

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on , so distance equals displacement: .

Answer: 36 m

With the fundamental theorem, a definite integral is only as hard as finding an antiderivative. Next, u-substitution, the chain rule in reverse, extends the antiderivative rules to composite functions like .

Frequently Asked Questions

What is the fundamental theorem of calculus?+

It's the theorem that connects differentiation and integration. Part 1: if is continuous, then , so integrating and then differentiating gives back . Part 2: if is any antiderivative of , then . Part 2 is how definite integrals are computed in practice.

What is the second fundamental theorem of calculus?+

Most textbooks call the evaluation formula the second fundamental theorem (Part 2), where is any antiderivative of a continuous . Some books, and some AP materials, number the parts the other way and call the second theorem. Check which statement your course means; the mathematics is the same.

How do you find the derivative of an integral?+

If the lower limit is a constant and the upper limit is , the derivative is the integrand with replaced by : . If the upper limit is a function , use the chain rule: . If is in the lower limit, flip the limits first, which changes the sign.

Why don't you add + C when evaluating a definite integral?+

Any constant cancels: . So you can use the simplest antiderivative, with . The answer to a definite integral is a number, not a family of functions.

What is the net change theorem?+

The net change theorem says the integral of a rate of change gives the total change: . For example, integrating velocity over a time interval gives displacement (net change in position), and integrating a flow rate gives the net change in the amount of liquid.

Does the fundamental theorem of calculus always work?+

It needs to be continuous on . If has a vertical asymptote inside the interval, can give a wrong answer. For example, is not , even though is an antiderivative: the integrand is positive, and the integral doesn't exist as a definite integral.

Related lessons

The Definite IntegralAntiderivatives and Indefinite IntegralsRiemann SumsChain Rule