Lecture 13 Notes: Differentials and Indefinite Integrals

18.01 Calculus Jason Starr Fall 2005 Lecture 13. October 13, 2005 Homework. Problem Set 4 Part I: (a) and (b); Part II: Problem 3. Practice Problems. Course Reader: 3A­1, 3A­2, 3A­3. 1. Differentials. An alternative notation for derivatives is differential notation. The differential notation, dF (x) = f (x)dx, is shorthand for the sentence “The derivative of F (x) with respect to x equals f (x).” Formally, this is related to the Leibniz notation for the derivative, dF (x) = f (x), dx which means the same thing as the differential notation. It may look like the first and second equation are obtained by dividing and multiplying by the quantity dx. It is crucial to remember that dF/dx is not a fraction, although the notation suggests otherwise. In differential notation, some derivative rules have a very simple form, and are thus easier to remember. Here are a few derivative rules in differential notation. dF (x) d(F (x) + G(x)) d(cF (x)) d(F (x)G(x)) d(F (x)/G(x)) = F � (x)dx = dF (x) + dG(x) = cdF (x) = G(x)dF (x) + F (x)dG(x) = 1/(G(x))2 (G(x)dF (x) − F (x)dG(x)) The chain rule has a particularly simple form, d(F (u)) = dF dF du du = dx. du du dx 18.01 Calculus Jason Starr Fall 2005 √ Example. Using differential notation, the derivative of sin( x2 + 1) is, d sin((x2 + 1)1/2 ) = cos((x2 + 1)1/2 )d(x2 + 1)1/2 = cos((x2 + 1)1/2 )( 21 (x2 + 1)−1/2 )d(x2 + 1) = cos((x2 + 1)1/2 ) 21 (x2 + 1)−1/2 (2xdx) = x(x2 + 1)−1/2 cos((x2 + 1)1/2 )dx . 2. Antidifferentiation. Recall, the basic problem of differentiation is the following. Problem (Differentiation). Given a function F (x), find the function f (x) satisfying dF dx = f (x). The bais problem of antidifferentiation is the inverse problem. Problem (Antidifferentiation). Given a function f (x), find a function F (x) satisfying f (x). dF dx = A function F (x) solving the problem is called an antiderivative of f (x), or sometimes an indefinite integral of f (x). The notation for this is, � F (x) = f (x)dx. The expression f (x) is called the integrand. It is important to note, if F (x) is one antiderivative of f (x), then for each constant C, F (x) + C is also an antiderivative of f (x). The constant C is called a constant of integration. In a sense that can be made precise, the problem of differentiation has a complete solution whenever F (x) is a “simple expression”, i.e., a function built from the differentiable functions we have seen so far. Unfortunately, for very many simple functions f (x), no antiderivative of f (x) has a simple expression. In large part, this is what makes antidifferentiation difficult. Luckily, many of the most important simple functions f (x) do have an antiderivative with a simple expression. One goal of this unit is to learn how to recognize when a simple antiderivative exists, and some tools to compute the antiderivative. 3. Antidifferentiation. Guess­and­check. The main technique for antidifferentiation is edu­ cated guessing. Example. Find an antiderivative of f (x) = x2 + 2x + 1. Since the derivative of xn is nxn−1 , it is reasonable to guess there is an antiderivative of the form F (x) = Ax3 + Bx2 + Cx. Differentiation gives, dF = 3Ax2 + 2Bx + C. dx Thus, F (x) is an antiderivative of f (x) if and only if, 3A = 1, 2B = 2, and C = 1. This gives an antiderivative, � (x2 + 2x + 1)dx = 1 3 x 3 + x2 + x + E , 18.01 Calculus Jason Starr Fall 2005 where E is any constant. Guess­and­check is a game we can lose, as well as win. However, there are a few rules that better the odds in this guessing game. In fact, they are basically the same rules for derivatives in differential notation, simply written backwards. � � � (f �(x) + g(x))dx = f (x)dx + g(x)dx � cf (x)dx = c f (x)dx � � f (u(x))u� (x)dx = f (u)du 4. Antidifferentiation. Integration by substitution. The last rule above is very important, and called integration by substitution. Example. Find an antiderivative of x sin(x2 ). This time guess­and­check is much less effective. By roughly the same logic in the last example, we might guess an antiderivative has the form Ax3 sin(x2 ). The derivative is 3Ax2 sin(x2 ) + 2Ax4 cos(x2 ). The first term is good, but the second term is bad. We can try to correct our guess by adding a term, Ax3 sin(x2 ) − 2/5Ax5 cos(x2 ), whose derivative is now 3Ax2 sin(x2 ) + 4/5Ax6 sin(x2 ). This still doesn’t work, and is leading in the wrong direction. A better solution is to use integration by substitution. Observe part of f (x) can be written as a function of u(x) = x2 . Also, the derivative u� (x) = 2x occurs in f (x) through x = 1/2(2x) = u� (x)/2. Thus, x sin(x2 ) = sin(u(x))u� (x)/2, u(x) = x2 . Applying integration by substitution, � � � x sin(x2 )dx = sin(u(x)) 21 u� (x)dx = 12 sin(u)du = −1 cos(u) + C = −1 cos(x2 ) + C. 2 2 Here is a checklist for applying integration by substition to find the antiderivative of f (x). (i) Find an expression u(x) so that most of the integrand f (x) can be expressed as a simpler function of u(x). (ii) Compute the differential du(x) = u� (x)dx. (iii) Inside the differential f (x)dx, try to find du = u� (x)dx as a factor. (iv) Try to write f (x)dx as g(u)du. If you cannot do this, the method does not apply with the given choice of u. � (v) Find an antiderivative G(u) = g(u)du for the simpler integrand g(u) (if this is possible). (vi) Back­substitute u = u(x) to get an antiderivative F (x) = G(u(x)) for f (x). 18.01 Calculus Example. Compute the antiderivative, � Jason Starr Fall 2005 sin(x)3 cos(x)dx. Most of the integrand is a function of sin(x). So substitute u(x) = sin(x). The differential of u is du = cos(x)dx. The differential sin(x)3 cos(x)dx contains du = cos(x)dx as a factor. The remainder of the integrand is sin(x)3 = u3 . So, according to integration by substitution, � � 1 3 sin(x) cos(x)dx = u3 du = u4 + C. 4 Finally, back­substitute u = sin(x) to get, � sin(x)3 cos(x)dx = (sin(x))4 /4 + C.