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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.

Theorem 7.1.5.

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.

Theorem 7.2.4.

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

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

$\Box$

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 $b$.

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

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

$\Box$

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