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Integral Calculus of One Variable Functions Northwestern University, Summer 2019

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Integral Calculus of One Variable Functions Northwestern University, Summer 2019 Math 224: Integral Calculus of One Variable Functions Northwestern University, Summer 2019 Shuyi Weng Last Update: August 12, 2019 Contents Lecture 1: Antiderivatives and the Area Problem 1 Lecture 2: Definite Integrals 11 Lecture 3: The Fundamental Theorem of Calculus 19 Lecture 4: Substitution Rule and Integration by Parts 26 Lecture 5: Trigonometric Integrals 32 Lecture 6: Partial Fractions 39 Lecture 7: Improper Integrals 47 Lecture 8: Areas, Volumes, and Arc Lengths 54 Lecture 9: Sequences and Limits 62 Lecture 10: Series 71 Lecture 11: Convergence Tests 79 Lecture 12: Power Series 85 Lecture 13: Taylor and Maclaurin Series 92 Lecture 1: Antiderivatives and the Area Problem Today: Introduction, Review of Math 220, Antiderivatives, The Area Problem Welcome to Math 224! You have studied the idea of differentiation in your previous calculus course(s). We will steer our focus to integration in this course. Let's start by reviewing some important ideas of differential calculus. Review: Differential Calculus The central idea in differential calculus is, of course, differentiation. Definition. Given a function f(x), we define the derivative of f(x) by d f(x + h) − f(x) f 0(x) = f(x) = lim ; dx h!0 h given any value of x for which this limit exists. If the limit exists for all x in the domain of definition of f(x), we say that f(x) is differentiable. The following theorem includes some of the elementary differentiation rules. Theorem 1. Some elementary differentiation rules: i. (Constant Function) If c is a constant, then d (c) = 0: dx ii. (Power Rule) If n is any real number, then d (xn) = nxn−1: dx iii. (Constant Multiple) If c is a constant and f(x) is a differentiable function, then d d [cf(x)] = c f(x): dx dx iv. (Sum and Difference) If f(x) and g(x) are differentiable functions, then d d d [f(x) ± g(x)] = f(x) ± f(x): dx dx dx v. (Sine and Cosine) d d (sin x) = cos x; (cos x) = − sin x: dx dx 1 vi. (Exponential and Logarithm) d d 1 (ex) = ex; (log jxj) = : dx dx x vii. (Product Rule) If f(x) and g(x) are differentiable functions, then d d d [f(x)g(x)] = f(x) [g(x)] + g(x) [f(x)]: dx dx dx viii. (Quotient Rule) If f(x) and g(x) are differentiable functions, then d d d hf(x)i g(x) [f(x)] − f(x) [g(x)] = dx dx dx g(x) [g(x)]2 ix. (Chain Rule) If f and g are both differentiable and F = f ◦g is the composition defined by F (x) = f(g(x)), then F is differentiable, and its derivative is given by F 0(x) = f 0(g(x)) · g0(x): Exercise. Find the derivative of tan x. Use the quotient rule d d h sin x i (tan x) = dx dx cos x cos x · cos x − sin x · (− sin x) = (cos x)2 cos2 x + sin2 x = cos2 x 1 = : cos2 x Exercise. Find the derivative of xx. Use the chain rule and the product rule d d d (xx) = [(eln x)x] = (ex ln x) dx dx dx d = ex ln x · (x ln x) dx d d = xx · x (ln x) + ln x (x) dx dx 1 = xx · x · + ln x · 1 x = xx(1 + ln x): An important tool in differential calculus is graph sketching. We will continue to use graphs of functions in this course to understand integrals and related concepts. Here is a basic example. 2 Exercise. Sketch the graph of sin(x). sin(x) 1 x 3π π π 3π -2π - -π - π 2π 2 2 2 2 -1 The curve shows the evaluation of sin(x) at each value of x, and the slope of the tangent line at a point on the curve represents the derivative of sin(x) at this point. Antiderivatives In some cases, we would like to know the original function based on a known (or measured) derivative. For example, if the velocity of a car over a period of time is known, we might wish to know its position at a given time. The problem is to find a function F whose derivative is a known function f. If such a function F exists, it is called an antiderivative of f. Definition. A function F is called an antiderivative of f on an interval I if F 0(x) = f(x) for all x 2 I. The following theorem characterizes all antiderivatives of a function, given that an antideriva- tive exists. Theorem 2. If F is an antiderivative of f on an interval I, then all antiderivatives of f are of the form F (x) + C; where C is an arbitrary constant. Going back to the function sin(x), we know that its derivative is cos(x). So an antiderivative of cos(x) is sin(x). However, since constant functions have zero derivatives, we know that sin(x) + 1 also has derivative cos(x), and in fact, sin(x) + C for any constant C would have 3 derivative cos(x). The following figure shows some of these antiderivatives of cos(x). sin(x) 2 1 x 3π π π 3π -2π - -π - π 2π 2 2 2 2 -1 -2 The following table gives some of the most elementary antiderivatives Function Antiderivative Function Antiderivative cf(x) cF (x) + C ex ex + C f(x) + g(x) F (x) + G(x) + C cos(x) sin(x) + C xn+1 xn (n 6= −1) + C sin(x) − cos(x) + C n + 1 x−1 ln jxj + C sec2(x) tan(x) + C Exercise. If f(x) is a polynomial of degree n, we can write 2 n f(x) = a0 + a1x + a2x + ··· + anx ; where a0; : : : ; an are constants. Find all antiderivatives of f. Use power rule and linearity, we get a x2 a x3 a xn+1 F (x) = a x + 1 + 2 + ··· + n + C 0 2 3 n + 1 4 The Area Problem We start our discussion by considering the area of regular shapes. Question. What is the area of each of the following three shapes? 2 2 2 1 1 1 -2 -1 1 2 -1 1 2 -1 1 2 The area of the rectangle is (length) × (width) = 4 × 2 = 8: The area of the trapezoid is (base 1 + base 2) × (height) (1 + 2) × 3 9 = = : 2 2 2 The area of the triangle is (base) × (height) 3 × 2 = = 3: 2 2 Things get much more complicated in some cases. For example, we do not yet have the proper tools to find the area of shapes with curved boundary. Question. What is the area of the shaded region? (1,1) y=x2 0 1 The shaded region has a curved side, namely the graph of the function f(x) = x2. It is not clear what the exact area of the region is. However, we can do some estimates. First of all, the area of the region must be between 0 and 1, because the region is contained in the square of side length 1. 5 We then divide the region in two parts by drawing a vertical line at x = 1=2. The area of the region is the sum of the area of these two parts. Call these two parts S1 and S2. S2 S1 0 /2 11 Now we estimate the area of S1 and S2, respectively. The area of S1 is bounded below by zero, and is bounded above by the area of the rectangle whose base is the same as S1 and whose height is the same as the right edge of S1. The area of S2, however, is bounded below by the area of the rectangle whose base is the same as S2 and whose height is the same as the left edge of S2. An upper bound for the area of S2 would be the area of the rectangle whose base is the same as S2 and whose height is the same as the right edge of S2. 0 /2 11 0 /2 11 Thus we get a lower bound 1 1 1 Area = S + S > 0 + · = ; 1 2 2 4 8 and an upper bound 1 1 1 5 Area = S + S < · + · 1 = : 1 2 2 4 2 8 So we know that the area of the shaded region must be between 1=8 and 5=8. This is a better estimate than 0 and 1. In order to find an even better estimate, we divide the region further into four vertical strips, 6 and repeat the same process. 0 1 0 1 0 1 The lower and upper bounds in this case can be computed by 1 12 1 12 1 32 7 Lower Bound = 0 + · + · + · = = 0:21875 4 4 4 2 4 4 32 1 12 1 12 1 32 1 15 Upper Bound = · + · + · + · 12 = = 0:46875 4 4 4 2 4 4 4 32 Once again, this gives a better estimate. We can repeat this process with a larger number n of strips. The following table shows the lower bound Ln and the upper bound Un for some large numbers n. Number of Strips Lower Bound Upper Bound 10 0:285 0:385 20 0:30875 0:35875 50 0:3234 0:3434 100 0:32835 0:33835 1000 0:3328335 0:3338335 It appears that the area of the shaded region is somewhere near 1=3. Proposition 3. The sum of the areas of the upper approximating rectangles for the shaded region approaches 1=3, that is 1 lim Un = : n!1 3 Proof.
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