J 2x2 dx 2 fx2 dx 2 y C J yr3 c

Example 10 Find /3x2 dx. In this case, factoring out the multiplicative constant yields j3x2 dx = 3 jx2 dx = 3( y + c'i) = +

Note that, in contrast to the preceding example, the term x3 in the final answer does not have any fractional expression attached to it. This neat result is due to the fact that 3 (the multiplicative constant of the integrand) happens to be precisely equal to 2 (the power of the function) plus 1. Referring to the power rule (Rule I), we see that the multiplicative constant (n + 1) will in such a case cancel out the fraction 1 /(« + 1), thereby yielding (x" + 1 + c) as the answer.

In general, whenever we have an expression (n + l)x" as the integrand, there is really no need to factor out the constant (rt + 1) and then integrate x"; instead, we may write x"+1 + c as the answer right away.

Example 11 Find j"|5eJ — x 2 + jcix. (x =/= 0). This example illustrates both Rules IV and V; actually, it illustrates the first three rules as well:

The correctness of the result can again be verified by differentiation.

Rules Involving Substitution

Now we shall introduce two more rules of integration which seek to simplify the process of integration, when the circumstances are appropriate, by a substitution of the original variable of integration. Whenever the newly introduced variable of integration makes the integration process easier than under the old, these rules will become of service.

Rule VI (the substitution rule) The integral of f(u)(du/dx) with respect to the variable x is the integral of f(u) with respect to the variable u:

jf(u)^dx = jf{u) du = F(u) + c where the operation / du has been substituted for the operation / dx.

This rule, the integral-calculus counterpart of the chain rule, may be proved by means of the chain rule itself. Given a function F(u), where u = u(x), the chain rule states that d V! \ d T7l \du T?'t \du ft \ du

Since f(u)(du/dx) is the derivative of F(u), it follows from (13.3) that the integral (antiderivative) of the former must be

You may note that this result, in fact, follows also from the canceling of the two dx expressions on the left.

Example 12 Find /2x(x2 + 1) dx. The answer to this can be obtained by first multiplying out the integrand:

jlx(x2 + 1) dx = j(2x3 + 2x) dx = y + x2 + c but let us now do it by the substitution rule. Let m = x2 + 1; then du/dx = 2x, or dx = du/2x. Substitution of du/2x for dx will yield j2x(x2 + 1) dx = J2xu~ = Judu = y + c,

= ~ (x4 + 2x2 + 1) + c, = j-x4 + x2 + c where c = 4 + c,. The same answer can also be obtained by substituting du/dx for 2x (instead of du/2x for dx).

Example 13 Find J6x2(x3 + 2)" dx. The integrand of this example is not easily multiplied out, and thus the substitution rule now has a better opportunity to display its effectiveness. Let u — x3 + 2; then du/dx = 3x2, so that j 6x2(x3 + 2 fdx = f^ — ^u^dx = f2u"du

Example 14 Find f%e2x+3 dx. Let u = 2x + 3; then du/dx = 2, or dx = du/2. Hence, j%e2x^3dx = j 8e"y = 4 feudu = 4 e" + c = 4e2x+3 + c

As these examples show, this rule is of help whenever we can—by the judicious choice of a function u = u(x)—express the integrand (a function of x) as the product of f(u) (a function of u) and du/dx (the derivative of the u function which we have chosen). However, as illustrated by the last two examples, this rule can be used also when the original integrand is transformable into a constant multiple of f(u)(du/dx). This would not affect the applicability because the constant multiplier can be factored out of the integral sign, which would then leave an integrand of the form f(u)(du/dx), as required in the substitution rule. When the substitution of variables results in a variable multiple of f(u)(du/dx), say, x times the latter, however, factoring is not permissible, and this rule will be of no help. In fact, there exists no general formula giving the integral of a product of two functions in terms of the separate integrals of those functions; nor do we have a general formula giving the integral of a quotient of two functions in terms of their separate integrals. Herein lies the reason why integration, on the whole, is more difficult than differentiation and why, with complicated integrands, it is more convenient to look up the answer in prepared tables of integration formulas rather than to undertake the integration by oneself.

Rule VII (integration by parts) The integral of v with respect to u is equal to uv less the integral of u with respect to v:

The essence of this rule is to replace the operation f du by the operation / dv.

The rationale behind this result is relatively simple. First, the product rule of differentials gives us d(uv) = v du + udv

If we integrate both sides of the equation (i.e., integrate each differential), we get a new equation or uv = / v du + /udv [no constant is needed on the left (why?)]

Then, by subtracting fu dv from both sides, the result stated above emerges.

Example 15 Find fx(x + 1 )'/2 dx. Unlike Examples 12 and 13, the present example is not amenable to the type of substitution used in Rule VI. (Why?) However, we may consider the given integral to be in the form of jv du, and apply Rule VII. To this end, we shall let v = x, implying dv = dx, and also let u = f(x + 1)3/2, so that du = (x + 1)1/2 dx. Then we can find the integral to be j x (x + 1)'dx = jv du = uv — Iu dv

0 0

Post a comment