The FTC was first stated clearly by Isaac Newton in 1666, although other mathematicians, including Newton’s teacher Isaac Barrow, had discovered versions of it earlier.

REMINDER

F(x) is called an antiderivative of f(x) if F′(x) = f(x). We say also that F(x) is an indefinite integral of f(x), and we use the notation

THEOREM 1 The Fundamental Theorem of Calculus, Part I

Assume that f(x) is continuous on [a, b]. If f(x) is an antiderivative of f(x) on [a, b], then

Proof

The quantity F(b) −F(a) is the total change in F (also called the “net change”) over the interval [a, b]. Our task is to relate it to the integral of F′(x) = f(x). There are two main steps.

Step 1. Write total change as a sum of small changes.

Given any partition P of [a, b]:

we can break up F(b) −F(a) as a sum of changes over the intervals [x_{[em]i[/em]−1}, x_i]:

On the right-hand side, F(x₁) is canceled by −F(x₁) in the second term, F(x₂) is canceled by −F(x₂), etc. (Figure 1). In summation notation,

Figure 5.28: Note the cancelation when we write F(b) − F(a) as a sum of small changes F(x_i) − F(x_{[em]i[/em]−1}).

Step 2. Interpret Eq. (2) as a Riemann sum.

The Mean Value Theorem tells us that there is a point in [x_{[em]i[/em]−1}, x_i] such that

Therefore, Eq. (2) can be written

This sum is the Riemann sum R(f, P, C*) with sample points .

Now, f(x) is integrable (Theorem 1, Section 5.2), so R(f, P, C*) approaches as the norm ‖P‖ tends to zero. On the other hand, R(f, P, C*) is equal to F(b) − F(a) with our particular choice C* of sample points. This proves the desired result:

CONCEPTUAL INSIGHT A Tale of Two Graphs

In the proof of FTC I, we used the MVT to write a small change in F(x) in terms of the derivative F′(x) = f(x):

But is the area of a thin rectangle that approximates a sliver of area under the graph of f(x) (Figure 2). This is the essence of the Fundamental Theorem: the total change F(b) −F(a) is equal to the sum of small changes F(x_i) −F(x_{[em]i[/em]−1}), which in turn is equal to the sum of the areas of rectangles in a Riemann sum approximation for f(x). We derive the Fundamental Theorem itself by taking the limit as the width of the rectangles tends to zero.

Figure 5.29

Notation

F(b) −F(a) is denoted . In this notation, the FTC reads

REMINDER The Power Rule for Integrals (valid for n ≠ −1) states:

EXAMPLE 1

Calculate the area under the graph of f(x) = x³ over [2, 4].

Solution Since is an antiderivative of f(x) = x³, FTC I gives us

EXAMPLE 2

Find the area under g(x) = x^−3/4 + 3x^5/3 over [1, 3]

Solution The function is an antiderivative of g(x). The area (Figure 3) is equal to

Figure 5.30: Region under the graph of g(x) = x^−3/4 + 3x^5/3 over [1, 3].

EXAMPLE 3

Calculate and sketch the corresponding region.

Solution Figure 4 shows the region. Recall that (tan x)′ = sec²x. Therefore,

Figure 5.31: Graph of y = sec² x.

EXAMPLE 4

Evaluate

• (a)

  and

• (b) .

Solution

• (a) Since (−cos x)′ = sin x, the area of one “hump” (Figure 5) is
  

  Figure 5.32: The area of one hump is 2. The signed area over [0, 2π] is zero.
  
• (b) We expect the signed area over [0, 2π] to be zero since the second hump lies below the x-axis, and, indeed,
  

EXAMPLE 5

Evaluate .

Solution The function is an antiderivative of f(x) = e^{3[em]x[/em]−1}, so the definite integral (the shaded area in Figure 6) is

Figure 5.33

EXAMPLE 6 The Logarithm as an Antiderivative

Evaluate

• (a)
  and
• (b) .

Solution By Eq. (3),

• (a)
• (b)

The areas represented by these integrals is shown in Figures 7(B) and (C).

CONCEPTUAL INSIGHT Which Antiderivative?

Antiderivatives are unique only to within an additive constant (Section 4.9). Does it matter which antiderivative is used in the FTC? The answer is no. If F(x) and G(x) are both antiderivatives of f(x), then F(x) = G(x) + C for some constant C, and

The two antiderivatives yield the same value for the definite integral: