5 Integration

5.4 The Fundamental Theorem of Calculus

In this section we will find connections between differential calculus (derivatives and antiderivatives) and integral calculus (definite integrals). These connections between the major ideas of calculus are important enough to be called the Fundamental Theorem of Calculus. These connections will also explain why we use the term indefinite integral for the set of all antiderivatives, and why we use such similar notations for antiderivatives and definite integrals.

Let f(t) be a continuous function defined on [a,b]. The definite integral abf(x)dx is the “area under f” on [a,b]. We can turn this concept into a function by letting the upper (or lower) bound vary.

Let F(x)=axf(t)dt. It computes the area under f on [a,x] as illustrated in Figure 5.4.1. We can study this function using our knowledge of the definite integral. For instance, F(a)=0 since aaf(t)dt=0.

margin: axbty Figure 5.4.1: The area of the shaded region is F(x)=axf(t)dt. Λ

The first part of the Fundamental Theorem of Calculus tells us how to find derivatives of these kinds of functions.

Theorem 5.4.1 The Fundamental Theorem of Calculus, Part 1

Let f be continuous on [a,b] and let F(x)=axf(t)dt. Then F is a continuous function on [a,b], differentiable on (a,b), and

  • Proof

    In order to see why this is true, we must compute limh0F(x+h)-F(x)h. Suppose x and x+h are in [a,b]. Theorem 5.2.1 implies that


    which we can rewrite as


    This allows us to simplify the numerator of the difference quotient in our limit as follows:

    F(x+h)-F(x) =ax+hf(t)dt-axf(t)dt(by the definition of F)

    so we see that


    Assume for the moment that h>0. Since x and x+h are both in [a,b] and f is continuous on [a,b], f is also continuous on [x,x+h]. Applying the Extreme Value Theorem (Theorem 3.1.1), we know that f must have an absolute minimum value f(u)=m and an absolute maximum value f(v)=M on this interval. In other words, mf(t)M whenever xtx+h. Using Theorem 5.3.3, we can now say that


    Computing the outer integrals, this becomes

    m(x+h-x)xx+hf(t) dtM(x+h-x),or
    mhxx+hf(t) dtMh.

    Since h>0, we may divide by h to obtain


    Now suppose that h<0. Preceding as before, we know that f has an absolute minimum value f(u)=m and an absolute maximum value f(v)=M on the interval [x+h,x]. We know that mf(t)M whenever x+htx, so we have


    Once again we compute to obtain


    Since -h>0, we can divide by -h to obtain:

    m-1hx+hxf(t) dtM
    f(u)=m1hxx+hf(t) dtM=f(v)(using Theorem 5.2.1(2))

    We are now ready to compute the desired limit,


    Whether h>0 or h<0, we know that


    where u and v are both between x and x+h. Note that

    limh0(x+h)=x and limh0x=x,

    so the Squeeze Theorem (Theorem 1.3.5) says that

    limh0u=x and limh0v=x.

    Since f is continuous at x, we know that

    limh0f(u)=f(x) and limh0f(v)=f(x).

    Finally, we know that


    so applying the Squeeze Theorem again tells us that


    Therefore F(x)=f(x) as desired. Because the limit exists, Exercise 35. in Section 2.1 implies that F is continuous on (a,b) as well. All that remains is to show that F is continuous at a and b. But repeating the preceding argument shows that |F(a+δ)-F(a)|(|M|+|m|)δ, and similarly for F near b.∎

Initially this seems simple, as demonstrated in the following example.

Example 5.4.1 Using the Fundamental Theorem of Calculus, Part 1

Let F(x)=-5x(t2+sint)dt. What is F(x)?

SolutionUsing the Fundamental Theorem of Calculus, we have

This simple example reveals something incredible: F(x) is an antiderivative of x2+sinx. Therefore, F(x)=13x3-cosx+C for some value of C. (We can find C, but generally we do not care. We know that F(-5)=0, which allows us to compute C. In this case, C=cos(-5)+1253.)

We have done more than found a complicated way of computing an antiderivative. Consider a function f defined on an open interval containing a, b and c. Suppose we want to compute abf(t)dt. First, let F(x)=cxf(t)dt. Using the properties of the definite integral found in Theorem 5.2.1, we know

abf(t)dt =acf(t)dt+cbf(t)dt

We now see how indefinite integrals and definite integrals are related: we can evaluate a definite integral using antiderivatives. Furthermore, Theorem 5.1.1 told us that any other antiderivative G differs from F by a constant: G(x)=F(x)+C. This means that G(b)-G(a)=(F(b)+C)-(F(a)+C)=F(b)-F(a), and the formula we’ve just found holds for any antiderivative. Consequently, it does not matter what value of C we use, and we might as well let C=0. This proves the second part of the Fundamental Theorem of Calculus.

Theorem 5.4.2 The Fundamental Theorem of Calculus, Part 2

Let f be continuous on [a,b] and let F be any antiderivative of f. Then

Example 5.4.2 Using the Fundamental Theorem of Calculus, Part 2

We spent a great deal of time in the previous section studying 04(4x-x2)dx. Using the Fundamental Theorem of Calculus, evaluate this definite integral.

SolutionWe need an antiderivative of f(x)=4x-x2. All antiderivatives of f have the form F(x)=2x2-13x3+C; for simplicity, choose C=0.

The Fundamental Theorem of Calculus states


This is the same answer we obtained using limits in the previous section, just with much less work.


A special notation is often used in the process of evaluating definite integrals using the Fundamental Theorem of Calculus. Instead of explicitly writing F(b)-F(a), the notation F(x)|ab is used. Thus the solution to Example 5.4.2 would be written as:

Example 5.4.3 Using the Fundamental Theorem of Calculus, Part 2

Evaluate the following definite integrals.

1.-22x3dx  2.0πsinxdx  3.05etdt  4.49udu  5.152dx


  1. (a)


  2. (b)

    (This is interesting; it says that the area under one “hump” of a sine curve is 2.)

  3. (c)


  4. (d)


  5. (e)


    This integral is interesting; the integrand is a constant function, hence we are finding the area of a rectangle with width (5-1)=4 and height 2. Notice how the evaluation of the definite integral led to 2(4)=8.

    In general, if c is a constant, then abcdx=c(b-a).

The Fundamental Theorem of Calculus and the Chain Rule

Part 1 of the Fundamental Theorem of Calculus (FTC) states that given F(x)=axf(t)dt, F(x)=f(x). Using other notation, ddx(F(x))=f(x). While we have just practiced evaluating definite integrals, sometimes finding antiderivatives is impossible and we need to rely on other techniques to approximate the value of a definite integral. Functions written as F(x)=axf(t)dt are useful in such situations.

It may be of further use to compose such a function with another. As an example, we may compose F(x) with g(x) to get


What is the derivative of such a function? The Chain Rule can be employed to state


An example will help us understand this.

Example 5.4.4 The FTC, Part 1, and the Chain Rule

Find the derivative of F(x)=2x2lntdt.

SolutionWe can view F(x) as being the function G(x)=2xlntdt composed with g(x)=x2; that is, F(x)=G(g(x)). The Fundamental Theorem of Calculus states that G(x)=lnx. The Chain Rule gives us

F(x) =G(g(x))g(x)

Normally, the steps defining G(x) and g(x) are skipped.

Practice this once more.

Example 5.4.5 The FTC, Part 1, and the Chain Rule

Find the derivative of F(x)=cosx5t3dt.

SolutionNote that F(x)=-5cosxt3dt. Viewed this way, the derivative of F is straightforward:


Understanding Motion with the Fundamental Theorem of Calculus

We established, starting with Key Idea 2.2.1, that the derivative of a position function is a velocity function, and the derivative of a velocity function is an acceleration function. Now consider definite integrals of velocity and acceleration functions. Specifically, if v(t) is a velocity function, what does abv(t)dt mean?

The Fundamental Theorem of Calculus states that


where V(t) is any antiderivative of v(t). Since v(t) is a velocity function, V(t) must be a position function, and V(b)-V(a) measures a change in position, or displacement.

How would we measure total distance traveled? We have to consider the intervals when v(t)0 and when v(t)0. Therefore,

total distance traveled=ab|v(t)|dt.
Example 5.4.6 Finding displacement and total distance traveled

A ball is thrown straight up with velocity given by v(t)=-32t+20ft/s, where t is measured in seconds. Find, and interpret,

1.01v(t)dt  and  2.01|v(t)|dt.


  1. (a)

    Using the Fundamental Theorem of Calculus, we have

    01v(t)dt =01(-32t+20)dt
    =4 ft.

    Thus if a ball is thrown straight up into the air with velocity v(t)=-32t+20, the height of the ball, 1 second later, will be 4 feet above the initial height. We will see in part 2. that the distance traveled is much farther. It has gone up to its peak and is falling down, but the difference between its height at t=0 and t=1 is 4 ft.

  2. (b)

    Here we are trying to find the total distance traveled by the ball. We must first consider where v(t)>0 and v(t)<0.

    v(t)=-32t+20 =0
    -32t =-20
    t =58

    This means v(t)>0 for t<58 and v(t)<0 for t>58 so we have

    01|v(t)|dt =05/8v(t)dt+5/81-v(t)dt
    =344=8.5 ft.

Integrating a rate of change function gives total change. Velocity is the rate of position change; integrating velocity gives the total change of position, i.e., displacement.

Integrating a speed function gives a similar, though different, result. Speed is also the rate of position change, but does not account for direction. So integrating a speed function gives total change of position, without the possibility of “negative position change.” Hence the integral of a speed function gives distance traveled.

As acceleration is the rate of velocity change, integrating an acceleration function gives total change in velocity. We do not have a simple term for this analogous to displacement. If a(t)=5miles/h2 and t is measured in hours, then


means the velocity has increased by 15m/h from t=0 to t=3.

margin: 1234xy (a)1234xy (b)1234xy (c)1234xy (d) Figure 5.4.2: A graph of a function f to introduce the Mean Value Theorem and differently sized rectangles giving upper and lower bounds on 14f(x)dx; the last rectangle matches the area exactly. Λ

The Mean Value Theorem and Average Value

Consider the graph of a function f in Figure 5.4.2(a) and the area defined by 14f(x)dx. Three rectangles are then drawn; in (b), the height of the rectangle is greater than f on [1,4], hence the area of this rectangle is is greater than 14f(x)dx.

In (c), the height of the rectangle is smaller than f on [1,4], hence the area of this rectangle is less than 14f(x)dx.

Finally, in (d) the height of the rectangle is such that the area of the rectangle is exactly that of 14f(x)dx. Since rectangles that are “too big”, as in (b), and rectangles that are “too little,” as in (c), give areas greater/lesser than 14f(x)dx, it makes sense that there is a rectangle, whose top intersects f(x) somewhere on [1,4], whose area is exactly that of the definite integral.

We state this idea formally in a theorem.

Theorem 5.4.3 The Mean Value Theorem of Integration

Let f be continuous on [a,b]. There exists a value c in (a,b) such that


This is an existential statement; c exists, but we do not provide a method of finding it. Theorem 5.4.3 is directly connected to the Mean Value Theorem of Differentiation, given as Theorem 3.2.1.

  • Proof

    If a=b, then aaf(x)dx=0=f(a)(a-a). Otherwise, we define the following for x in [a,b]:


    Applying Theorem 5.4.1, we know F is differentiable on (a,b) and that F(x)=f(x) for any x in (a,b). We may now apply the Mean Value Theorem for Differentiation (Theorem 3.2.1) to see that there is a value c in (a,b) such that


    Note that F(c)=f(c) and that F(b)-F(a)=abf(x)dx by Theorem 5.4.2. Therefore we can rewrite our equation as:

    f(c) =abf(x)dxb-a, or
    f(c)(b-a) =abf(x)dx.

We demonstrate the principles involved in this version of the Mean Value Theorem in the following example.

margin: 121cπsin0.69xy Figure 5.4.3: A graph of y=sinx on [0,π] and the rectangle guaranteed by the Mean Value Theorem. Λ
Example 5.4.7 Using the Mean Value Theorem

Consider 0πsinxdx. Find a value c guaranteed by the Mean Value Theorem.

SolutionWe first need to evaluate 0πsinxdx. (This was previously done in Example 5.4.3.)


Thus we seek a value c in [0,π] such that πsinc=2.


In Figure 5.4.3 sinx is sketched along with a rectangle with height sin(0.69). The area of the rectangle is the same as the area under sinx on [0,π].

Let f be a function on [a,b] with c such that f(c)(b-a)=abf(x)dx. Consider ab(f(x)-f(c))dx:

ab(f(x)-f(c))dx =abf(x)-abf(c)dx
margin: y=f(x)abcf(c)xyy=f(x)-f(c)abcf(c)xy Figure 5.4.4: On top, a graph of y=f(x) and the rectangle guaranteed by the Mean Value Theorem. Below, y=f(x) is shifted down by f(c); the resulting “area under the curve” is 0. Λ

When f(x) is shifted by -f(c), the amount of area under f above the x-axis on [a,b] is the same as the amount of area below the x-axis above f; see Figure 5.4.4 for an illustration of this. In this sense, we can say that f(c) is the average value of f on [a,b].

The value f(c) is the average value in another sense. First, recognize that the Mean Value Theorem can be rewritten as


for some value of c in [a,b]. Next, partition the interval [a,b] into n equally spaced subintervals, a=x1<x2<<xn+1=b and choose any ci in [xi,xi+1]. The average of the numbers f(c1), f(c2), …, f(cn) is:


Multiply this last expression by 1 in the form of (b-a)(b-a):

1ni=1nf(ci) =i=1nf(ci)1n
=1b-ai=1nf(ci)Δx(where Δx=(b-a)/n)

Now take the limit as n:


This tells us this: when we evaluate f at n (somewhat) equally spaced points in [a,b], the average value of these samples is f(c) as n.

This leads us to a definition.

Definition 5.4.1 The Average Value of f on [a,b]

Let f be continuous on [a,b]. The average value of  𝐟  on [𝐚,𝐛] is f(c), where c is a value in [a,b] guaranteed by the Mean Value Theorem. I.e.,

Average Value of f on [a,b]=1b-aabf(x)dx.

An application of this definition is given in the following example.

Example 5.4.8 Finding the average value of a function

An object moves back and forth along a straight line with a velocity given by v(t)=(t-1)2 on [0,3], where t is measured in seconds and v(t) is measured in ft/s.

What is the average velocity of the object?

SolutionBy our definition, the average velocity is:

13-003(t-1)2dt=1303(t2-2t+1)dt=13(13t3-t2+t)|03=1 ft/s.

We can understand the above example through a simpler situation. Suppose you drove 100 miles in 2 hours. What was your average speed? The answer is simple: displacement/time = 100 miles/2 hours = 50 mph.

What was the displacement of the object in Example 5.4.8? We calculate this by integrating its velocity function: 03(t-1)2dt=3 ft. Its final position was 3 feet from its initial position after 3 seconds: its average velocity was 1 ft/s.

This section has laid the groundwork for a lot of great mathematics to follow. The most important lesson is this: definite integrals can be evaluated using antiderivatives. Since the previous section established that definite integrals are the limit of Riemann sums, we can later create Riemann sums to approximate values other than “area under the curve,” convert the sums to definite integrals, then evaluate these using the Fundamental Theorem of Calculus. This will allow us to compute the work done by a variable force, the volume of certain solids, the arc length of curves, and more.

The downside is this: generally speaking, computing antiderivatives is much more difficult than computing derivatives. Much of our time in Chapter 8 will be devoted to techniques of finding antiderivatives so that a wide variety of definite integrals can be evaluated.

Exercises 5.4


Terms and Concepts

  1. 1.

    How are definite and indefinite integrals related?

  2. 2.

    What constant of integration is most commonly used when evaluating definite integrals?

  3. 3.

    T/F: If f is a continuous function, then F(x)=axf(t)dt is also a continuous function.

  4. 4.

    The definite integral can be used to find “the area under a curve.” Give two other uses for definite integrals.


In Exercises 5–34., use the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral.

  1. 5.


  2. 6.


  3. 7.


  4. 8.


  5. 9.


  6. 10.


  7. 11.


  8. 12.


  9. 13.


  10. 14.


  11. 15.


  12. 16.


  13. 17.


  14. 18.


  15. 19.


  16. 20.


  17. 21.


  18. 22.


  19. 23.


  20. 24.


  21. 25.


  22. 26.


  23. 27.


  24. 28.


  25. 29.


  26. 30.


  27. 31.


  28. 32.


  29. 33.


  30. 34.


  1. 35.
    Explain why: (a) -11xndx=0, when n is a positive, odd integer, and (b) -11xndx=201xndx when n is a positive, even integer.
  2. 36.

    Explain why aa+2πsintdt=0 for all values of a.

In Exercises 37–40., find a value c guaranteed by the Mean Value Theorem.

  1. 37.


  2. 38.


  3. 39.


  4. 40.


In Exercises 41–46., find the average value of the function on the given interval.

  1. 41.

    f(x)=sinx on [0,π/2]

  2. 42.

    y=sinx on [0,π]

  3. 43.

    y=x on [0,4]

  4. 44.

    y=x2 on [0,4]

  5. 45.

    y=x3 on [0,4]

  6. 46.

    g(t)=1/t on [1,e]

In Exercises 47–50., a velocity function of an object moving along a straight line is given. Find (a) the displacement of the object over the given time interval and (b) the total distance traveled by the object over the given time interval.

  1. 47.

    v(t)=-32t+20ft/s on [0,5]

  2. 48.

    v(t)=-32t+200ft/s on [0,10]

  3. 49.

    v(t)=cost ft/s on [0,3π/2]

  4. 50.

    v(t)=t4 ft/s on [0,16]

In Exercises 51–54., an acceleration function of an object moving along a straight line is given. Find the change of the object’s velocity over the given time interval.

  1. 51.

    a(t)=-32ft/s2 on [0,2]

  2. 52.

    a(t)=10ft/s2 on [0,5]

  3. 53.

    a(t)=t ft/s2 on [0,2]

  4. 54.

    a(t)=cost ft/s2 on [0,π]

In Exercises 55–62., use the Fundamental Theorem of Calculus Part 1 to find F(x).

  1. 55.


  2. 56.


  3. 57.


  4. 58.


  5. 59.


  6. 60.


  7. 61.


  8. 62.


  1. 63.
    Let g(x)=0xf(t)dt where f is the function whose graph is shown below. (a) Evaluate g(x) for x=0,1,2,3,4,5,6. (b) Estimate g(7). (c) Where does g have a minimum value? a maximum value? (d) Sketch the graph of g.
  2. 64.
    For any x>0, define g(x)=1x1tdt. (a) Show that g(x) is continuous and differentiable on (0,) with g(x)=1x. (b) Show that for any positive x and y we have g(xy)=g(x)+g(y). [Hint: Treat y as a constant, consider the derivative with respect to x of each side of the proposed equation, and apply Theorem 5.1.1.] (c) Show that for any positive x and any r we have g(xr)=rg(x). [Hint: Consider the derivative with respect to x of each side of the proposed equation and apply Theorem 5.1.1.]
  3. 65.
    Let f(x)={-1-πx<010x<π and extend this function so that it is periodic with period 2π. This function is known as a square wave and looks like -55-22xy For a positive integer n, define bn=1π-ππf(x)sin(nx)dx. (a) Find bn. (b) Graph n=1Nbnsin(nx) for various values of N. What do you observe?
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