1.5 Methods Of Integration

Since there are no foolproof formulae that can be used for all integrals (as there are for derivatives), the following integration methods all attempt to change a difficult integral into a simpler one. All examples use definite integrals.

1.5.1 Substitution

This method substitutes each part of a definite integral with a counterpart so that the result is kept the same. With the substitution $u = h(x)$, the integral

$$\int^b_a f(x)dx$$

becomes transformed into

$$\int^{h(b)}_{h(a)}g(u)du$$ where $g(u)du = f(x)dx$. Thus the integrand, $dx$, and the limits have been transformed into equivalent counterparts. We obtain $g(u)$ by finding the inverse function $h^{-1} (u)$ so that

$$\begin{align*} x &= h^{-1}(u) \\ dx &= \frac{d[h^{-1}(u)]}{du} du \\ g(u)du &= [f(x)][dx] \\ g(u) &= [f(h^{-1}(u))][\frac{dh^{-1}(u)}{du}] \\ \end{align*}$$

Often, $h(x)$ appears explicitly in the integrand so that $u$ is substituted directly.

Example 3

The problem presented above on the oxygen depletion due to sewage dumping can now be solved. The integral in eqn. (1.5) is first separated.

$$\begin{equation} \int^{\infty}_0(te^{-t^2} +e^{-t}) dt = \int^{\infty}_0 te^{-t^2}dt + \int^{\infty}_0|e^{-t}dt \tag{1.6} \end{equation}$$

The antiderivative of $te^{-t^2}$ is not obvious, so we substitute a new function into the integral. Let $u = t^2$. Then the differential $du$ is

$$du = (du/dt)dt = 2tdt$$ so that $$\frac{du}{2} = tdt$$ The limits remain, since $u = 0$ when $t = 0$ and $u = \infty$ when $t = \infty$. Direct substitution then gives

$$\begin{align*} \int^{\infty}_0 te^{-t^2}dt &= \int^{\infty}_0e^{-t^2}(tdt) \\ &= \int^{\infty}_0 e^{-u} (\frac{du}{2}) = \frac{1}{2} \int^{\infty}_0 e^{-u}du \\ &= \frac{1}{2} [-e^{-u}]^{\infty}_0 = \frac{1}{2}[0-(-1)] \\ &= \frac{1}{2} \\ \end{align*}$$

Since the second term in eqn. (1.6) is now obvious $\int^{\infty}_0 e^{-u}du = \int^{\infty}_0 e^{-t}dt=1$, we obtain the solution to eqn. (1.5) of $$C(\infty) = 1/2 + 1 = 3/2$$

1.5.2 Integration by Parts

This method utilizes a relation from differential calculus concerning the total differential. Recall that the differential of a product of two functions $u(x)$, $v(x)$ can be written $$d(uv) = udv + vdu$$ If we evaluate antiderivatives, we obtain

$$\begin{equation} uv = \int udv + \int vdu \tag{1.7} \end{equation}$$

Rearrangement of eqn. (1.7) gives the integration by parts formula $$\int udv = uv - \int vdu$$ With definite integrals, we usually write this formula as

$$\begin{equation} \int^b_a [u(x)\frac{dv}{dx}]dx = [uv]^b_a - \int^b_a [v(x)\frac{du}{dx}]dx \tag{1.8} \end{equation}$$

Again, the goal is to change a difficult integral, the left side of eqn. (1.8), into a simpler one, the right side of eqn. (1.8). An example is presented later.

1.5.3 Partial Fractions

When the integrand is a ratio of two polynomials, it often can be decomposed into a sum of simpler terms. Only the case of non-repeated linear terms in the denominator is treated here. For more complicated cases in an ecological setting, see Clow and Urquhart (1974) p. 559.

The integrand is assumed to be of the form $F(x)/G(x)$ where $F(x)$ and $G(x)$ are polynomial functions and where $G(x)$ is the product of linear factors. For example, $$G(x) = (1 + 2x)(2 + 2x)(1+x)$$ is a polynomial composed of factors linear in $x$. The partial fraction technique replaces the single rational expression by a sum of terms where each denomination is one of the linear factors of $G(x)$. If we have two linear factors, $$G(x) = A(x)B(x)$$ then a partial fractions decomposition gives $$\frac{F(x)}{G(x)} = \frac{F(x)}{A(x)B(x)} = \frac{C_1}{A(x)} + \frac{C_2}{B(x)}$$ where $C_1$ and $C_2$ are constants. Multiplication by $A(x)B(x)$ gives

$$\begin{equation} F(x) = C_1B(x) + C_2A(x) \tag{1.9} \end{equation}$$

Let $r_1, r_2$ be zeros of $A(x), B(x)$, respectively, i.e. $A(r_1 ) = B(r_2 ) = 0$. Then, with $x = r_1$, eqn. (1.9) is $$F(r_1) = C_1 B(r_1)$$

and with $x = r_2$, eqn. (1.9) becomes $$F(r_2) = C_2 A(r_2)$$ so that $C_1,C_2$ can be easily determined.

Example 4

The logistic growth model for animal populations is represented as a differential equation $$\frac{dN}{dt} = rN(1 - N/K), \;\;\;\text{where }\; r,K = \mbox{constant}$$

Separating variables (see the section, Differential Equations) gives, using the differentials $dN$ and $dt$, $$\frac{dN}{N(1 - N/K)} = rdt$$ which is integrated to yield the equation

$$\begin{equation} \int \frac{dN}{N(1 - N/K)} = \int rdt \tag{1.10} \end{equation}$$

The right side of eqn. (1.10) is easily integrated. The left side, however, must be reworked. First rewrite the integral as $$\frac{1}{N(1 - N/K)} = \frac{K}{N(K-N)}$$

Now expand in partial fractions as $$\frac{K}{N(K-N)} = \frac{a_1}{N} + \frac{a_2}{K-N}$$ Multiplying both sides by $N(K - N)$ gives

$$\begin{equation} K = a_1(K - N) + a_2N \tag{1.11} \end{equation}$$

Since (1.11) must hold for all values of $N$ (Clow and Urquhart 1974, p. 562), then setting $N = 0$ gives $$a_1 = 1$$ and $N = K$ gives

$$a_2 = 1$$ so that
$$\frac{K}{N(K-N)} = \frac{1}{N} + \frac{1}{K-N}$$ Then eqn. (1.10) is $$\int \frac{dN}{N} + \int \frac{dN}{K-N} = \int rdt$$ which integrates to

$$\ln N - \ln (K-N) = rt + C$$

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