Common Mistakes in Second Semester Calculus

COMMON MISTAKES IN SECOND SEMESTER CALCULUS Many of the mistakes most frequently made by students in second semester calculus are illustrated in the problems given below. Most of the solutions presented are taken directly from student exam papers. The logical or technical flaw underlying each mistake is examined, the erroneous solution is followed with a correct one, and tips may be given for how to avoid the error in similar problems. The hope is that by seeing the common errors made by others, you will make fewer of these errors yourself. In addition, if you have made a certain mistake on some calculus problem, you may find a careful analysis of a similar problem below. What follows is by no means a comprehensive list, but rather a collection of a few of the most frequently made mistakes. ∫5 Problem #1: Evaluate the integral Solution: ∫ x +1 dx . 5 x+2 + C. x+2 5 x +1 dx = x n +1 + C , n ≠ − 1 ) cannot be n +1 applied here because the variable is in the exponent, not in the base. The Error: The power rule for integrals ( A Correct Solution: ∫5 x +1 dx = ∫ 5 ⋅ 5 x dx = 5 (using the formula Problem #2: Evaluate the integral Solution: ∫ 1 x−6 ∫ x n dx = ∫ ∫ 1 x−6 ∫ 5 x dx = 5 ⋅ a x dx = 5x 5 x +1 +C = +C ln 5 ln 5 ax + C ). ln a dx . dx = ln( x − 6 ) + C . The Error: This is an example of one of the very most common integration errors made: The integration formula 1 ∫ x dx = ln x + C does not allow one to write = ln(expression) + C in general. 1 ∫ expression dx A Correct Solution: Use the method of substitution: Let w = x − 6. Then dw = dx and we can write ∫ 1 x−6 dx = ∫ 1 w dw = Problem #3: Evaluate the integral Solution: 3 ∫ −3 2 9 − x dx = = 3 ∫w 3 ∫ −3 −1 / 2 9 − x 2 dx . ∫ −3 (9 − x ) 2 1/ 2 (9 − 3 2 ) 3 / 2 3/ 2 dw = 2w1/2 + C = 2 x − 6 + C . − (9 − x 2 ) 3 / 2 dx = 3/ 2 3 −3 (9 − (−3) 2 ) 3 / 2 = 0. 3/ 2 The Error: It is incorrect to apply the power rule for integrals here. To see that this is (9 − x 2 )3 / 2 wrong, note that the derivative of is not 9 − x 2 , because the chain rule 3/ 2 comes into play: d  (9 − x 2 ) 3 / 2 dx  3/ 2 2 1/ 2   = (3 / 2)(9 − x ) ⋅ (−2 x) = − 2 x(9 − x 2 )1 / 2 ≠  3/ 2  9 − x2 . In general, when the integrand is a composite function with the outer function being a power, the power rule for integrals should not be applied. This is because the derivative of the inner function is not accounted for in this approach. Here are several more examples of this error: 1 ∫ sin x dx ≠ 4 sin x + C 1 4 4 −3 ( x + 2 x ) dx ≠ − ( x + 2 x) − 2 ∫ 2 3 ∫ (e 4 +C x ∫ 1 x (e + 1) 3 + C 3 1 ln 2 x dx ≠ ln 3 x + C 3 + 1) 2 dx ≠ A Correct Solution: The best way to evaluate the integral 3 ∫ −3 9 − x 2 dx is to recognize what it represents geometrically. The graph of the function y = 9 − x 2 is shown below. Notice that just by making this graph you can immediately tell that the answer of 0 above cannot be right since the area under a curve lying above the x-axis must be positive. zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz To get the right answer, note that y = 9 − x 2 is the upper half of a semicircle of radius 3, thus the integral given above is the area of that semicircle, π(3)2/2 = 9π/2. -3 3 (You could also solve the integral with the rationalizing substitution x = 3 sin t if you have learned that method.) Problem #4: Evaluate the integral dx ∫ x2 + 4 . Solution: Let w = x 2 + 4. Then dw = 2 x dx => dx = dw/2x, so dx ∫ x2 + 4 ∫ = dw / 2 x w = 1 2x ∫ dw 1 = = ln w + C w 2x = = 1 ln x 2 + 4 + C . 2x The Error: There are two mistakes here: 1 from the integral is not an allowable step. . Only constants 2x can be pulled outside an integral, not variables. When performing a substitution, you should convert all instances of the original variable (here, x) into the substitution variable (here, w) and leave them inside the integral. Error #1: Pulling out Unfortunately, while that is the correct technique, if we try it above we obtain zzzzz ∫ = dw / 2 x w ∫ 2w = dw w−4 ∫ dw / 2 w − 4 w (x was replaced by solving w = x 2 + 4 for x) , which is a worse integral than the one we started with. And that zzzzzz brings us to the second error: Error #2: The substitution w = denominator should not be used when the integrand has the form 1/quadratic. dx ∫ Recall that A Correct Solution: = arctan x. We can convert the given x 2 +1 integral to this form as follows: dx ∫ x2 + 4 dx ∫ 4( x 2 / 4 + 1) = = dx ∫ 4( ( x / 2) 2 + 1) ; now let w = x/2, => dw = dx/2, to obtain 2dw ∫ 4(w 2 + 1) 1 2 = dw ∫ w2 +1 1 1  x arctan w + C = arctan   + C . 2 2 2 = Remark: Letting w = x 2 + 4 would have worked if the integral had contained an x in the numerator to accommodate dw = 2xdx. Problem #5: Evaluate the integral Solution: ∫ 1 1 −1 x4 ∫ dx = 1 −1 x −4 ∫ 1 1 −1 x4 dx . x −3 dx = −3 1 −1 2  1   −1  =  −  = − . 3  −3  −3 The Error: Notice that the integrand is always positive, so the definite integral cannot possibly come out negative. The error is failing to recognize that this an improper integral, since 1/x4 blows up at x = 0. A Correct Solution: ∫ 1 1 −1 x 4 1 1 dx = ∫ 0 x 4 dx = = 1 1 ∫ 0 x4 lim a → 0+ ∫ dx + 1 a ∫ 0 1 −1 x4 dx . Now x −3 dx = lim+ a→ 0 − 3 x4 1 1 a  1  1   lim    −    . a → 0+   − 3   − 3a 3   Since a is in the denominator and approaching 0, the limit does not exist. Hence the 1 1 dx diverges, and therefore the original integral diverges as well. integral 0 x4 ∫ Problem #6: Evaluate the integral Solution: Letting w = 1 + x2 gives therefore ∫ ∞ x 1+ x 2 −∞ dx = lim b→ ∞ ∫ ∫ ∞ −∞ ∫ x 1+ x 2 x 1+ x b −b 2 dx . ∫ dx = 1 2 dw 1 1 = ln w = ln(1 + x 2 ) ; w 2 2 dx = 1 ln(1 + x 2 ) b→ ∞ 2 −b b x 1+ x 2 lim 1  1 = lim  ln(1 + b 2 ) − ln(1 + (−b) 2 ) = lim [0] = 0. b→ ∞  2 b→ ∞ 2  The Error: An improper integral of the form ∫ ∞ f ( x) dx must be broken up into two −∞ integrals, since it is improper at both of the limits of integration. A Correct Solution: ∫ ∞ −∞ x 1+ x 2 dx = ∫ 0 x −∞ 1+ x 2 dx + ∫ ∞ x 1+ x 2 0 dx . For the second integral we have ∫ ∞ 0 x 1+ x 2 dx = lim b→ ∞ ∫ b 0 x 1+ x 2 b dx = 1 ln(1 + x 2 ) = b→ ∞ 2 0 lim 1   1 1 lim  ln(1 + b 2 ) − ln(1 + 0 2 ) = lim  ln(1 + b 2 ) − 0 = ∞. b→ ∞  2 b→ ∞  2 2   Thus the second integral is divergent, so the original integral diverges as well. ∞ Problem #7: Determine whether the series n+2 converges or diverges. n n =1 ∑ n+2 2  = lim 1 +  = 1 + 0 = 1. The series converges to 1. n→∞ n n→∞  n Solution: lim The Error: The solver has confused the convergence/divergence of the sequence of terms in the series with the convergence/divergence of the series itself. A Correct Solution: Each term in the series is greater than 1, thus the series is greater than the series 1 + 1 + 1 + … , which diverges to infinity. Hence the given series diverges to infinity also. ∞ Problem #8: Determine whether the series ∑ n =1 Solution: lim n→∞ n+2 n2 = lim n / n2 + 2 / n2 n→∞ n2 / n2 n+2 n2 converges or diverges. 1/ n + 2 / n2 0+0 = = 0. n→∞ 1 1 = lim Since the terms of the series tend to 0, the series converges. The Error: The solver is confused about the following fact (sometimes known as the divergence test): If the sequence of terms in a series does not tend to 0, the series diverges. But this does not mean that if the terms do converge to 0, then the series converges. We cannot tell what a series does just from knowing that its terms converge 1 1 to 0. For example, the series diverges, although converges while the series n n2 1 1 the two sequences of terms and both converge to 0. 2 n n ∑ ∑ ∞ A Correct Solution: The series ∑ n+2 n =1 ∞ series 1 ∑n since for every n, n+2 n2 > is larger, term-by-term, than the harmonic n = 1 . Since the harmonic series is n n2 n2 known to diverge, so does the given series, by comparison. n =1 ∞ Problem #9: Determine whether the series ∑ 2n 2 − 1 n =1 ∞ Solution: ∑ ∞ 2n 2 − 1 n =1 n ∑ n3 < 3 2n 2 n3 converges or diverges. ∞ (term-by-term) = 2 n =1 ∞ Since the harmonic series 1 ∑n . n =1 1 ∑ n diverges, so does the given series, by comparison. n =1 The Error: In order to use comparison to show that a series diverges, you must show that it is greater than another series that is known to diverge. In the solution above, the solver showed that the given series was smaller than a known divergent series. That tells nothing about whether the given series converges or diverges. ∞ A Correct Solution: ∑ 2n 2 − 1 n =1 n 3 ∞ = ∑ n2 + n2 −1 n n =1 3 ∞ > n2 ∑ n3 ∞ (term-by-term) = n =1 1 ∑n n =1 Since the given series is greater than the harmonic series, which diverges, so does the given series by comparison. ∞ Problem #10: Determine whether the series 1 ∑ n(ln n) 3 converges or diverges. n =10 ∞ Solution: 1 ∑ n(ln n) 3 n =10 ∞ < 1 ∑ n2 (term-by-term) n =10 Sine the series on the right is a convergent p-series (p > 1), the series on the left converges by comparison. The Error: The inequality is not true: (ln n)3 grows very slowly as n → ∞, and after a 1 1 1 = 2 and > while remains less than n from then on. Thus eventually, 3 n⋅ n n n(ln n) so the comparison given above is not valid. A Correct Solution: We can compare the given series to the improper ∞ 1 integral dx . Using the substitution w = ln x, we can show that this integral 10 x(ln x)3 converges. By comparison, so does the series. ∫ Problem #11: Find the second degree Taylor polynomial P2(x) that approximates the function f(x) = cos x + sin(2x) near x = 0. Solution: f(0) = cos 0 + sin(2⋅0) = 1, and the first two derivatives of f are f′ (x) = − sin x + 2cos(2x), f′′ (x) = − cos x − 4sin(2x). Substituting these values into the Taylor polynomial formula P2(x) = f(0) + f′ (0)x + 1 2! f′′ (0) x2 gives P2(x) = 1 + (− sin x + 2cos(2x)) x + 1 (− cos x − 4sin(2x)) x2 . 2 The Error: The solver used the derivative functions f′ (x), f′′ (x) rather than their values f′ (0), f′′ (0) in the Taylor polynomial formula. Note that the solver's answer is not even a polynomial, as it has trigonometric factors. A Correct Solution: Substituting zero for x in the derivatives found above gives f′ (0) = − sin 0 + 2cos(2⋅0) = 2 and f′′ (0) = − cos 0 − 4sin(2⋅0)= −1, thus P2(x) = 1 + 2x + 1 2! (−1)x2 = 1 + 2x − x2/2 .