[Calculus] VII. Integrals

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Indefinite Integrals

Definition 7.1.1.  A function \( F \) is called an antiderivative of \( f \) on an interval \( I \) if \( F'(x) = f(x) \) for all \( x \) in \( I \).

Theorem 7.1.2.  If \( F \) is an antiderivative of \( f \) on an interval \( I \), then the most general antiderivative of \( f \) on \( I \) is

\( \displaystyle F(x) + C \)

where \( C \) is an arbitrary constant.

Definition 7.1.3.  The notation \( \int f(x)\operatorname{d}\!x \) is used for an antiderivative of \( f \) and is called an indefinite integral. Thus

\( \displaystyle \int f(x)\operatorname{d}\!x = F(x) \qquad \text{means} \qquad F'(x) = f(x) \)

Theorem 7.1.4.

\( \displaystyle (1) \hspace{2.5em} \int cf(x)\operatorname{d}\!x = c \int f(x)\operatorname{d}\!x \\[14pt] \displaystyle (2) \hspace{2.5em} \int \left[f(x) + g(x)\right]\operatorname{d}\!x = \int f(x)\operatorname{d}\!x + \int g(x)\operatorname{d}\!x \\[14pt] \displaystyle (3) \hspace{2.5em} \int k\operatorname{d}\!x = kx + C \\[14pt] \displaystyle (4) \hspace{2.5em} \int x^{n}\operatorname{d}\!x = \frac{x^{n + 1}}{n + 1} + C \quad (n \neq -1) \\[14pt] \)

Theorem 7.1.5.

\( \displaystyle (1) \hspace{2.5em} \int \sin x \operatorname{d}\!x = -\cos x + C \\[14pt] \displaystyle (2) \hspace{2.5em} \int \cos x \operatorname{d}\!x = \sin x + C \\[14pt] \displaystyle (3) \hspace{2.5em} \int {\sec}^{2} x \operatorname{d}\!x = \tan x + C \\[14pt] \displaystyle (4) \hspace{2.5em} \int {\csc}^{2} x \operatorname{d}\!x = -\cot x + C \\[14pt] \displaystyle (5) \hspace{2.5em} \int \sec x \tan x \operatorname{d}\!x = \sec x + C \\[14pt] \displaystyle (6) \hspace{2.5em} \int \csc x \cot x \operatorname{d}\!x = -\csc x + C \\[14pt] \)

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The Definite Integral

Definition 7.2.1.  If \( f \) is a function defined for \( a \ge x \ge b \), we divide the interval \( \left[a,\ b \right] \) into \( n \) subintervals of equal width \( \Delta x = (b - a)/n \).

We let \( x_{0}\ (=a),\ x_{1},\ x_{2},\ \cdots \ ,\ x_{n}\ (=b) \) be the endpoints of these subintervals and we let \( x_{1}^{*},\ x_{2}^{*},\ \cdots \ ,\ x_{n}^{*} \) be any sample points in these subintervals, so \( x_{i}^{*} \) lies in the \( i \)th subinterval \( \left[ x_{i - 1},\ x_{i} \right] \). Then the definite integral of \( f \) from \( a \) to \( b \) is

\( \displaystyle \int_{a}^{b} f(x) \operatorname{d}\!x = \lim_{n \to \infty} \sum_{i = 1}^{n} f(x_{i}^{*}) \Delta x \)

provided that this limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that \( f \) is integrable on \( \left[a,\ b \right] \). The precise meaning of the limit that defines the integral is as follows:

For every number \( \varepsilon > 0 \) there is an integer \( N \) such that

\( \displaystyle \left\vert \ \int_{a}^{b} f(x) \operatorname{d}\!x - \sum_{i = 1}^{n} f(x_{i}^{*}) \Delta x \ \ \right\vert < \varepsilon \)

for every integer \( n > N \) and for every choice of \( x_{i}^{*} \) in \( \left[ x_{i - 1},\ x_{i} \right] \).

Figure 1 : \( \displaystyle \sum_{i = 1}^{n}f(x_{i}^{*})\Delta x \)

Figure 2 : \( \displaystyle \int_{a}^{b}f(x) \operatorname{d}\!x \)

Remark 7.2.2.  The symbol \( \int \) was introduced by Leibniz and is called an integral sign. It is an elongated \( S \) and was chosen because an integral is a limit of sums. In the notation \( \int_{a}^{b} f(x) \operatorname{d}\!x \), \( f(x) \) is called the integrand and \( a \) and \( b \) are called the limits of integration; \( a \) is the lower limit and \( b \) is the upper limit. The procedure of calculating an integral is called integration.

The sum

\( \displaystyle \sum_{i = 1}^{n} f(x_{i}^{*}) \Delta x \)

is called a Riemann sum. So the definition says that the definite integral of an integrable function can be approximated to within any desired degree of accuracy by a Riemann sum.

Also, notice that If \( f(x) \ge 0 \), the integral \( \int_{a}^{b}f(x) \operatorname{d}\!x \) is the area under the curve \( y = f(x) \) from \( a \) to \( b \).

Theorem 7.2.3.

\( \displaystyle (1) \hspace{2.5em} \int_{a}^{b} c \operatorname{d}\!x = c(b - a) \quad \text{where} \ c \ \text{is any constant} \\[14pt] \displaystyle (2) \hspace{2.5em} \int_{b}^{a} f(x) \operatorname{d}\!x = -\int_{a}^{b} f(x) \operatorname{d}\!x \\[14pt] \displaystyle (3) \hspace{2.5em} \int_{a}^{a} f(x) \operatorname{d}\!x = 0 \\[14pt] \displaystyle (4) \hspace{2.5em} \int_{a}^{b} \left[f(x) + g(x)\right] \operatorname{d}\!x = \int_{a}^{b} f(x) \operatorname{d}\!x + \int_{a}^{b} g(x) \operatorname{d}\!x \\[14pt] \displaystyle (5) \hspace{2.5em} \int_{a}^{b} cf(x) \operatorname{d}\!x = c \int_{a}^{b} f(x) \operatorname{d}\!x \quad \text{where} \ c \ \text{is any constant} \\[14pt] \displaystyle (6) \hspace{2.5em} \int_{a}^{c} f(x) \operatorname{d}\!x + \int_{c}^{b} f(x) \operatorname{d}\!x = \int_{a}^{b} f(x) \operatorname{d}\!x \)

Theorem 7.2.4.

\( \displaystyle (1) \hspace{3em} \)If \( f(x) \ge 0 \) for \( a \le x \le b \), then \( \displaystyle \int_{a}^{b} f(x) \operatorname{d}\!x \ge 0 \)

\( \displaystyle (2) \hspace{3em} \)If \( f(x) \ge g(x) \) for \( a \le x \le b \), then \( \displaystyle \int_{a}^{b} f(x) \operatorname{d}\!x \ge \int_{a}^{b} g(x) \operatorname{d}\!x \)

\( \displaystyle (3) \hspace{3em} \)If \( m \le f(x) \le M \) for \( a \le x \le b \), then \( \displaystyle m(b - a) \le \int_{a}^{b} f(x) \operatorname{d}\!x \le M(b - a) \)

Proof.  \( (3) \) Since \( m \le f(x) \le M \), \( (2) \) gives

\( \displaystyle \int_{a}^{b} m \operatorname{d}\!x \le \int_{a}^{b} f(x) \operatorname{d}\!x \le \int_{a}^{b} M \operatorname{d}\!x \)

Using Theorem 7.2.3 - \( (1) \) to evaluate the integrals on the left and right sides, we obtain

\( \displaystyle m(b - a) \le \int_{a}^{b} f(x) \operatorname{d}\!x \le M(b - a) \)

\( \Box \)

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The Fundamental Theorem of Calculus (FTC)

Theorem 7.3.1. (FTC1)  If \( f \) is continuous on \( \left[a,\ b\right] \), then the function \( g \) defined by

\( \displaystyle g(x) = \int_{a}^{x} f(t) \operatorname{d}\!t \quad a \le x \le b \)

is continuous on \( \left[a,\ b\right] \) and differentiable on \( \left(a,\ b\right) \), and

\( \displaystyle g'(x) = f(x) \)

Proof.  If \( x \) and \( x + h \) are in \( \left(a,\ b\right) \), then

\( \displaystyle \begin{align} g(x + h) - g(x) &= \int_{a}^{x + h} f(t) \operatorname{d}\!t - \int_{a}^{x} f(t) \operatorname{d}\!t \\[7pt] &= \left( \int_{a}^{x} f(t) \operatorname{d}\!t + \int_{x}^{x + h} f(t) \operatorname{d}\!t \right) - \int_{a}^{x} f(t) \operatorname{d}\!t \\[7pt] &= \int_{x}^{x + h} f(t) \operatorname{d}\!t \end{align} \)

and so, for \( h \neq 0 \),

\( \displaystyle \frac{g(x + h) - g(x)}{h} = \frac{1}{h}\int_{x}^{x + h}f(t) \operatorname{d}\!t \qquad \cdots \qquad \text{(i)} \)

For now let’s assume that \( h > 0 \). Since \( f \) is continuous on \( \left[x,\ x + h\right] \), Theorem 6.1.5 says that there are numbers \( u \) and \( v \) in \( \left[x,\ x + h\right] \) such that \( f(u) = m \) and \( f(v) = M \), where \( m \) and \( M \) are the absolute minimum and maximum values of \( f \) on \( \left[x,\ x + h\right] \).

By Theorem 7.2.4 - \( (3) \), we have

\( \displaystyle mh \le \int_{x}^{x + h}f(t) \operatorname{d}\!t \le Mh \)

that is,

\( \displaystyle f(u)h \le \int_{x}^{x + h}f(t) \operatorname{d}\!t \le f(v)h \)

Since \( h > 0 \), we can divide this inequality by \( h \):

\( \displaystyle f(u) \le \frac{1}{h}\int_{x}^{x + h}f(t) \operatorname{d}\!t \le f(v) \)

Now we use Equation \( \text{(i)} \) to replace the middle part of this inequality:

\( \displaystyle f(u) \le \frac{g(x + h) - g(x)}{h} \le f(v) \qquad \cdots \qquad \text{(ii)} \)

Inequality \( \text{(ii)} \) can be proved in a similar manner for the case where \( h < 0 \).

Now we let \( h \to 0 \). Then \( u \to x \) and \( v \to x \), since \( u \) and \( v \) lie between \( x \) and \( x + h \). Therefore

\( \displaystyle \begin{align} &\lim_{h \to 0}f(u) = \lim_{u \to x}f(u) = f(x) \\[7pt] &\lim_{h \to 0}f(v) = \lim_{v \to x}f(v) = f(x) \end{align} \)

because \( f \) is continuous at \( x \). We conclude, from \( \text{(ii)} \) and Theorem 1.2.3, that

\( \displaystyle g'(x) = \lim_{h \to 0}\frac{g(x + h) - g(x)}{h} = f(x) \qquad \cdots \qquad \text{(iii)} \)

If \( x = a \) or \( b \), then Equation \( \text{(iii)} \) can be interpreted as a one-sided limit. Then Theorem 3.1.7 (modified for one-sided limits) shows that \( g \) is continuous on \( \left[a,\ b\right] \).

\( \Box \)

Theorem 7.3.2. (FTC2)  If \( f \) is continuous on \( \left[a,\ b\right] \), then

\( \displaystyle \int_{a}^{b} f(x)\operatorname{d}\!x = F(b) - F(a) \)

where \( F \) is any antiderivative of \( f \), that is, a function \( F \) such that \( F' = f \).

Proof.  Let \( g(x) = \int_{a}^{x} f(t)\operatorname{d}\!t \). We know from Theorem 7.3.1 that \( g'(x) = f(x) \); that is, \( g \) is an antiderivative of \( f \). If \( F \) is any other antiderivative of \( f \) on \( \left[a,\ b\right] \), then we know from Corollary 6.2.4 that \( F \) and \( g \) differ by a constant:

\( \displaystyle F(x) = g(x) + C \qquad \cdots \qquad \text{(iv)} \)

for \( a < x < b \). But both \( F \) and \( g \) are continuous on \( \left[a,\ b\right] \) and so, by taking limits of both sides of Equation \( \text{(iv)} \) (as \( x \to a^{+} \) and \( x \to b^{-} \)), we see that it also holds when \( x = a \) and \( x = b \). So \( F(x) = g(x) + C \) for all \( x \) in \( \left[a,\ b\right] \).

If we put \( x = a \) in the formula for \( g(x) \), we get

\( \displaystyle g(a) = \int_{a}^{a} f(t)\operatorname{d}\!t = 0 \)

So, using Equation \( \text{(iv)} \) with \( x = b \) and \( x = a \), we have

\( \displaystyle \begin{align} F(b) - F(a) &= \left[\ g(b) + C\ \right] - \left[\ g(a) + C\ \right] \\[7pt] &= g(b) - g(a) = g(b) = \int_{a}^{b} f(t)\operatorname{d}\!t \end{align} \)

\( \Box \)

Remark 7.3.3.  Theorem 7.3.2 states that if we know an antiderivative \( F \) of \( f \), then we can evaluate \( \int_{a}^{b} f(x)\operatorname{d}\!x \) simply by subtracting the values of \( F \) at the endpoints of the interval \( \left[a,\ b\right] \). It’s very surprising that \( \int_{a}^{b} f(x)\operatorname{d}\!x \), which was defined by a complicated procedure involving all of the values of \( f(x) \) for \( a \le x \le b \), can be found by knowing the values of \( F(x) \) at only two points, \( a \) and \( b \).

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The Substitution Rule

Theorem 7.4.1. (The Substitution Rule)  If \( u = g(x) \) is a differentiable function whose range is an interval \( I \) and \( f \) is continuous on \( I \), then

\( \displaystyle \int f(g(x))g'(x) \operatorname{d}\!x = \int f(u) \operatorname{d}\!u \)

Proof.  If \( F' = f \), then

\( \displaystyle \int F'(g(x))g'(x) \operatorname{d}\!x = F(g(x)) + C \qquad \cdots \qquad \text{(v)} \)

because, by Theorem 4.2.1,

\( \displaystyle {\operatorname{d}\over\operatorname{d}\!x}\!\left[F(g(x))\right] = F'(g(x))g'(x) \)

If we make the “change of variable” or “substitution” \( u = g(x) \), then from Equation \( \text{(v)} \) we have

\( \displaystyle \begin{align} \int F'(g(x))g'(x) \operatorname{d}\!x &= F(g(x)) + C \\[7pt] &= F(u) + C \\[7pt] &= \int F'(u) \operatorname{d}\!u \end{align} \)

or, writing \( F' = f \), we get

\( \displaystyle \int f(g(x))g'(x) \operatorname{d}\!x = \int f(u) \operatorname{d}\!u \)

Thus we have proved the following rule.

\( \Box \)

Remark 7.4.2.  Notice that Theorem 7.4.1 for integration was proved using Theorem 4.2.1 for differentiation. Notice also that if \( u = g(x) \), then \( \operatorname{d}\!u = g'(x) \operatorname{d}\!x \), so a way to remember Theorem 7.4.1 is to think of \( \operatorname{d}\!x \) and \( \operatorname{d}\!u \) as differentials.

Thus Theorem 7.4.1 says: it is permissible to operate with \( \operatorname{d}\!x \) and \( \operatorname{d}\!u \) after integral signs as if they were differentials.

Example 7.4.3.  Find \( \displaystyle \int \frac{x}{\sqrt{1 - 4x^{2}}} \, \operatorname{d}\!x \).

Solution.  Let \( u = 1 - 4x^{2} \). Then \( \operatorname{d}\!u = -8x \operatorname{d}\!x \), so \( \displaystyle x \operatorname{d}\!x = - \frac{1}{8} \operatorname{d}\!u \) and

\( \displaystyle \begin{align} \int \frac{x}{\sqrt{1 - 4x^{2}}} \, \operatorname{d}\!x &= -\frac{1}{8} \int \frac{1}{\sqrt{u}} \, \operatorname{d}\!u = -\frac{1}{8} \int u^{-1/2} \operatorname{d}\!u \\[7pt] &= -\frac{1}{8} \left(2 \sqrt{u}\right) + C = -\frac{1}{4} \sqrt{1 - 4x^{2}} + C \end{align} \)

\( \Box \)

Theorem 7.4.4.  If \( g' \) is continuous on \( \left[a,\ b\right] \) and \( f \) is continuous on the range of \( u = g(x) \), then

\( \displaystyle \int_{a}^{b} f(g(x))g'(x) \operatorname{d}\!x = \int_{g(a)}^{g(b)} f(u) \operatorname{d}\!u \)

Proof.  Let \( F \) be an antiderivative of \( f \). Then, by \( \text{(v)} \), \( F(g(x)) \) is an antiderivative of \( f(g(x))g'(x) \), so by Theorem 7.3.2, we have

\( \displaystyle \int_{a}^{b} f(g(x))g'(x) \operatorname{d}\!x = F(g(x)) \Big]_{a}^{b} = F(g(b)) - F(g(a)) \)

But, applying Theorem 7.3.2 a second time, we also have

\( \displaystyle \int_{g(a)}^{g(b)} f(u) \operatorname{d}\!u = F(u) \Big]_{g(a)}^{g(b)} = F(g(b)) - F(g(a)) \)

\( \Box \)

Theorem 7.4.5.  Suppose \( f \) is continuous on \( \left[-a,\ a\right] \).

(a)  If \( f \) is even \( \left( \ f(-x) = f(x) \ \right) \), then \( \int_{-a}^{a}f(x) \operatorname{d}\!x = 2 \int_{0}^{a}f(x) \operatorname{d}\!x \).

(b)  If \( f \) is odd \( \left( \ f(-x) = -f(x) \ \right) \), then \( \int_{-a}^{a}f(x) \operatorname{d}\!x = 0 \).

Proof.  We split the integral in two:

\( \displaystyle \int_{-a}^{a}f(x) \operatorname{d}\!x = \int_{-a}^{0}f(x) \operatorname{d}\!x + \int_{0}^{a}f(x) \operatorname{d}\!x = -\int_{0}^{-a}f(x) \operatorname{d}\!x + \int_{0}^{a}f(x) \operatorname{d}\!x \qquad \cdots \qquad \text{(vi)} \)

In the first integral on the far right side we make the substitution \( u = -x \). Then \( \operatorname{d}\!u = -\operatorname{d}\!x \) and when \( x = -a \), \( u = a \). Therefore

\( \displaystyle -\int_{0}^{-a}f(x) \operatorname{d}\!x = -\int_{0}^{a}f(-u)\,(-\operatorname{d}\!u) = \int_{0}^{a}f(-u) \operatorname{d}\!u \)

and so Equation \( \text{(vi)} \) becomes

\( \displaystyle \int_{-a}^{a}f(x) \operatorname{d}\!x = \int_{0}^{a}f(-u) \operatorname{d}\!u + \int_{0}^{a}f(x) \operatorname{d}\!x \qquad \cdots \qquad \text{(vii)} \)

(a)  If \( f \) is even, then \( f(-u) = f(u) \) so Equation \( \text{(vii)} \) gives

\( \displaystyle \int_{-a}^{a}f(x) \operatorname{d}\!x = \int_{0}^{a}f(u) \operatorname{d}\!u + \int_{0}^{a}f(x) \operatorname{d}\!x = 2\int_{0}^{a}f(x) \operatorname{d}\!x \)

(b)  If \( f \) is odd, then \( f(-u) = -f(u) \) so Equation \( \text{(vii)} \) gives

\( \displaystyle \int_{-a}^{a}f(x) \operatorname{d}\!x = -\int_{0}^{a}f(u) \operatorname{d}\!u + \int_{0}^{a}f(x) \operatorname{d}\!x = 0 \)

\( \Box \)

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The body text and the figures were excerpted from James Stewart, Calculus 8E.