DEFINITION Antiderivatives

A function \(F(x)\) is an antiderivative of \(f(x)\) on \((a, b)\) if \(F'(x) = f(x)\) for all \(x \in (a, b)\).

THEOREM 1 The General Antiderivative

Let \(F(x)\) be an antiderivative of \(f(x)\) on \((a, b)\). Then every other antiderivative on \((a, b)\) is of the form \(F(x) + C\) for some constant \(C\).

Proof

If \(G(x)\) is a second antiderivative of \(f(x)\), set \(H(x) = G(x) - F(x)\). Then \[ H'(x) = G'(x) - F'(x) = f(x) - f(x) = 0. \] Therefore \(H(x)\) must be a constant—say, \(H(x) = C\) —and therefore \(G(x) = F(x) + C\).

GRAPHICAL INSIGHT

The graph of \(F(x) + C\) is obtained by shifting the graph of \(F(x)\) vertically by \(C\) units. Since vertical shifting moves the tangent lines without changing their slopes, it makes sense that all of the functions \(F(x) + C\) have the same derivative (Figure 5.28). Theorem 1 tells us that conversely, if two graphs have parallel tangent lines, then one graph is obtained from the other by a vertical shift.

Figure 5.28: The tangent lines to the graphs of \(y = F(x)\) and \(y = F(x) + C\) are parallel.

Example 1

Find two antiderivatives of \(f(x) = \cos x\). Then determine the general antiderivative.

Solution The functions \(F(x) = \sin x\) and \(G(x) = \sin x + 2\) are both antiderivatives of \(f(x)\). The general antiderivative is \(F(x) = \sin x + C\), where \(C\) is any constant.

THEOREM 2 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 \[ \boxed{\bbox[#FAF8ED,5pt]{\int\limits_a^b f(x) \ dx = F(b) - F(a)} }\tag{1} \]

\(F(x)\) is called an antiderivative of \(f(x)\) if \(F'(x) = f(x)\).

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.

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 \([a,b]\) into \(N\) equal parts: \[ P\colon x_0=a < x_1 < x_2 < \cdots < x_N = b \] We can break up \(F(b) - F(a)\) as a sum of changes over the intervals \([x_{i-1},x_i]\): \[ F(b) - F(a) = \left( F(x_1) - F(a)\right) + \left( F(x_2) - F(x_1)\right) + \cdots + \left( F(b) - F(x_{N-1})\right) \] On the right-hand side, \(F(x_1)\) is canceled by \(- F(x_1)\) in the second term, \(F(x_2)\) is canceled by \(-F(x_2)\), etc. (Figure 5.29). In summation notation, \[ F(b) - F(a) = \sum\limits_{i=1}^N \left( F(x_i) - F(x_{i-1})\right)\tag{2} \]

Figure 5.29: Note the cancelation when we write \(F(b) - F(a)\) as a sum of small changes \(F(x_i) - F(x_{i-1})\).

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

The Mean Value Theorem tells us that there is a point \(x_i^*\) in \([x_{i-1},x_i]\) such that \[ F(x_i) - F(x_{i-1}) = F'(x_i^*)(x_i - x_{i-1}) = f(x_i^*)(x_i - x_{i-1}) = f(x_i^*)\Delta x_i \] Therefore, Eq. (2) can be written \[ F(b) - F(a) = \sum\limits_{i=1}^N f(x_i^*)\Delta x_i \] This sum is the Riemann sum \(\mbox{Rie}_N(f,X^*)\) with sample points \(X^* = \left\{x_i^*\right\}\).

Now, \(f(x)\) is integrable (Theorem 1, Section 5.2), so \(\mbox{Rie}_N(f,X^*)\) approaches \(\int_a^bf(x) dx\) as \(N\) tends to infinity. On the other hand, \(\mbox{Rie}_N(f,X^*)\) is equal to \(F(b) - F(a)\) with our particular choice \(X^*\) of sample points. This proves the desired result: \[ F(b) - F(a) = \lim\limits_{N \rightarrow \infty} \mbox{Rie}_N(f,X^*) = \int\limits_a^b f(x) dx \]

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 \(F'(x) = f(x)\): \[ F(x_i) - F(x_{i-1}) = f(c_i^*)\Delta x_i \] But \(f(c_i^*)\Delta x_i\) is the area of a thin rectangle that approximates a sliver of area under the graph of \(f(x)\) (Figure 5.30). This is the essence of the Fundamental Theorem: the total change \(F(b) -F(a)\) is equal to the sum of small changes \(F(x_{i}) -F(x_{i-1})\), 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 5.30

Notation

\(F(b) -F(a)\) is denoted \(F(x)\Big|_a^b\). In this notation, the FTC reads \[ \int\limits_a^b f(x)dx = F(x)\bigg|_a^b \]

REMINDER The Power Rule for Integrals (valid for \(n \neq -1\)) states: \[ \int x^ndx = \frac{x^{n+1}}{n+1}+C \]

EXAMPLE 2

Calculate the area under the graph of \(f(x) = x^{3}\) over \([2, 4]\).

Solution Since \(F(x)=\frac14x^4\) is an antiderivative of \(f(x) = x^{3}\), FTC I gives us \[ \int\limits_2^4 x^3dx = F(4) - F(2) = \frac14x^4\bigg|_2^4 = \frac144^4 - \frac142^4 = 60 \]

EXAMPLE 3

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

Solution The function \(G(x) = 4x^{1/4} + \tfrac98x^{8/3}\) is an antiderivative of g(x). The area (Figure 5.31) is equal to \[ \begin{aligned} \int\limits_1^3 \left(x^{-3/4} + 3x^{5/3}\right)dx & = G(x)\bigg|_1^3 = \left(4x^{1/4} + \tfrac98x^{8/3}\right)\bigg|_1^3\\ & = \left(4\cdot3^{1/4} + \tfrac98\cdot3^{8/3}\right) - \left(4\cdot1^{1/4} + \tfrac98\cdot1^{8/3}\right)\\ & \approx 26.325 - 5.125 =21.1 \end{aligned} \]

Figure 5.31: Region under the graph of \(g(x) = x^{-3/4} + 3x^{5/3}\) over \([1, 3]\).

EXAMPLE 4

Calculate \(\int_{-\pi/4}^{\pi/4}\sec^2x dx\) and sketch the corresponding region.

Solution Figure 5.32 shows the region. Recall that \((\tan x)' = \sec^{2}x\). Therefore,

Figure 5.32: Graph of \(y = \sec^{2} x\).

EXAMPLE 5

Evaluate

• (a) \(\int_0^{\pi} \sin x dx \)

  and

• (b) \(\int_0^{2\pi} \sin x dx \).

Solution

• (a) Since \((-\cos x)' = \sin x\), the area of one “hump” (see Figure 5.33 below) is \[ \int\limits_0^{\pi} \sin x dx = -\cos x \bigg|_0^{\pi} = -\cos\pi - (-\cos 0) = - (-1) - (-1) = 2 \]
• (b) We expect the integral over \([0, 2\pi]\) to be zero since the second hump lies below the \(x\)-axis, and, indeed, \[ \int\limits_0^{2\pi} \sin x dx = -\cos x \bigg|_0^{2\pi} = -\cos2\pi - (-\cos 0) = - 1 - (-1) = 0 \]

Figure 5.33: Use the slider to see how the value of the integral changes. The area of one “hump” is 2 and therefore, because of symmetry, the integral over \([0,2\pi]\) has value zero.

NOTATION Indefinite Integral

The notation

We say that \(F(x) + C\) is the general antiderivative or indefinite integral of \(f(x)\).

There are no Product, Quotient, or Chain Rules for integrals. However, we will see that the Product Rule for derivatives leads to an important technique called Integration by Parts (Section 7.1) and the Chain Rule leads to the Substitution Method (Section 5.6).

THEOREM 3 Power Rule for Integrals

Proof

We just need to verify that \(F(x)=\dfrac{x^{n+1}}{n+1}\) is an antiderivative of \(f(x) = x^{n}\):

Notice that in integral notation, we treat dx as a movable variable, and thus we write \(\int\frac{1}{x}dx\) as \(\int\frac{dx}{x}\).

THEOREM 4 Antiderivative of \(y=\frac{1}{x}\)

The function \(F(x) = \ln |x|\) is an antiderivative of \(y=\frac{1}{x}\) in the domain \(\{x : x \neq 0\}\); that is,

THEOREM 5 Linearity of the Indefinite Integral

• Sum Rule: \(\displaystyle\int(f(x)+g(x))dx = \int f(x)dx + \int g(x)dx\)
• Scalar Multiples Rule: \(\displaystyle\int cf(x) dx = c\int f(x)dx\)

Example 6

Evaluate \(\displaystyle\int(3x^4-5x^{\frac{2}{3}}+x^{-3})dx\).

Solution We integrate term by term and use the Power Rule:

To check the answer, we verify that the derivative is equal to the integrand:

Example 7

Evaluate \(\displaystyle\int\left(\frac{5}{x}-3x^{-10}\right) \ dx\).

Solution Apply Eq. (3) and the Power Rule:

Basic Trigonometric Integrals

Example 8

Evaluate \(\displaystyle\int \big(\sin(8t-3)+20\cos9t \big)dt\).

Solution

CONCEPTUAL INSIGHT Which Antiderivative?

Antiderivatives are unique only to within an additive constant. 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 \[ F(b) - F(a) = \underbrace{(G(b) + C) - (G(a) + C)}_{\text{The constant cancels}} = G(b) - G(a) \] The two antiderivatives yield the same value for the definite integral: \[ \int\limits_a^b f(x) dx = F(b) - F(a) = G(b) - G(a) \]