Differentials and Antiderivatives

Lecture 15 18.01 Fall 2006 Lecture 15: Differentials and Antiderivatives Differentials New notation: dy = f � (x)dx (y = f (x)) Both dy and f � (x)dx are called differentials. You can think of dy = f � (x) dx as a quotient of differentials. One way this is used is for linear approximations. Δy dy ≈ Δx dx Example 1. Approximate 651/3 Method 1 (review of linear approximation method) x1/3 1 −2/3 f � (x) = x 3 f (x) ≈ f (a) + f � (a)(x − a) 1 x1/3 ≈ a1/3 + a−2/3 (x − a) 3 f (x) = A good base point is a = 64, because 641/3 = 4. Let x = 65. 1/3 65 1/3 = 64 1 1 + 64−2/3 (65 − 64) = 4 + 3 3 � 1 16 � (1) = 4 + 1 ≈ 4.02 48 Similarly, (64.1)1/3 ≈ 4 + 1 480 Method 2 (review) 651/3 = (64 + 1)1/3 = [64(1 + � �1/3 1 1/3 1 1 )] = 641/3 [1 + ]1/3 = 4 1 + 64 64 64 Next, use the approximation (1 + x)r ≈ 1 + rx with r = 1 1 and x = . 3 64 1 1 1 651/3 ≈ 4(1 + ( )) = 4 + 3 64 48 This is the same result that we got from Method 1. 1 Lecture 15 18.01 Fall 2006 Method 3 (with differential notation) y dy x1/3 |x=64 = 4 � � 1 −2/3 1 1 1 = x dx|x=64 = dx = dx 3 3 16 48 = 1 when dx = 1. 48 1 (65)1/3 = 4 + 48 We want dx = 1, since (x + dx) = 65. dy = What underlies all three of these methods is y dy dx = = x1/3 1 −2/3 x |x=64 3 Anti-derivatives � F (x) = f (x)dx means that F is the antiderivative of f . Other ways of saying this are: F � (x) = f (x) or, dF = f (x)dx Examples: � sin xdx = − cos x + c where c is any constant. 1. � 2. � 3. � 4. xn+1 + c for n �= −1. n+1 dx = ln |x| + c x (This takes care of the exceptional case n = −1 in 2.) sec2 xdx = tan x + c � dx 1 √ = sin−1 x + c (where sin−1 x denotes “inverse sin” or arcsin, and not ) sin x 1 − x2 � dx = tan−1 (x) + c 1 + x2 5. 6. xn dx = Proof of Property 2: The absolute value |x| gives the correct answer for both positive and negative x. We will double check this now for the case x < 0: ln |x| d ln(−x) dx d ln(−x) dx = ln(−x) � � du d ln(u) where u = −x. = dx du 1 1 1 (−1) = (−1) = = u −x x 2 Lecture 15 18.01 Fall 2006 Uniqueness of the antiderivative up to an additive constant. If F � (x) = f (x), and G� (x) = f (x), then G(x) = F (x) + c for some constant factor c. Proof: (G − F )� = f − f = 0 Recall that we proved as a corollary of the Mean Value Theorem that if a function has a derivative zero then it is constant. Hence G(x) − F (x) = c (for some constant c). That is, G(x) = F (x) + c. Method of substitution. � Example 1. x3 (x4 + 2)5 dx Substitution: u = x4 + 2, du = 4x3 dx, (x4 + 2)5 = u5 , x3 dx = 1 du 4 Hence, � � Example 2. 1 x (x + 2) dx = 4 3 4 5 � u5 du = u6 u6 1 4 = +c= (x + 2)6 + c 4(6) 24 24 x √ dx 1 + x2 Another way to find an anti-derivative is “advanced guessing.” First write � � x √ dx = x(1 + x2 )−1/2 dx 1 + x2 Guess: (1 + x2 )1/2 . Check this. d 1 (1 + x2 )1/2 = (1 + x2 )−1/2 (2x) = x(1 + x2 )−1/2 dx 2 Therefore, � � Example 3. x(1 + x2 )−1/2 dx = (1 + x2 )1/2 + c e6x dx Guess: e6x . Check this: d 6x e = 6e6x dx Therefore, � e6x dx = 3 1 6x e +c 6 Lecture 15 18.01 Fall 2006 � Example 4. 2 xe−x dx 2 Guess: e−x Again, take the derivative to check: 2 d −x2 e = (−2x)(e−x ) dx Therefore, � � Example 5. sin x cos xdx = 2 2 1 xe−x dx = − e−x + c 2 1 sin2 x + c 2 Another, equally acceptable answer is � 1 sin x cos xdx = − cos2 x + c 2 This seems like a contradiction, so let’s check our answers: d sin2 x = (2 sin x)(cos x) dx and d cos2 x = (2 cos x)(− sin x) dx So both of these are correct. Here’s how we resolve this apparent paradox: the difference between the two answers is a constant. 1 1 1 1 sin2 x − (− cos2 x) = (sin2 x + cos2 x) = 2 2 2 2 So, 1 1 1 1 1 sin2 x − = (sin2 x − 1) = (− cos2 x) = − cos2 x 2 2 2 2 2 The two answers are, in fact, equivalent. The constant c is shifted by 12 from one answer to the other. � dx Example 6. (We will assume x > 0.) x ln x 1 Let u = ln x. This means du = dx. Substitute these into the integral to get x � � dx 1 = du = ln u + c = ln(ln(x)) + c x ln x u 4