MIDTERM 1 REVIEW First, a (No Promise of Being Exhaustive) List Of

MIDTERM 1 REVIEW First, a (No Promise of Being Exhaustive) List Of

MIDTERM 1 REVIEW MICHAEL First, a (no promise of being exhaustive) list of topics. (1) Being familiar with homework problems is a good idea. (2) Anything from last quarter is fair game, of course. (3) Definition of an integral. Note you should definitely know the geometric pictures, but knowing the technicalities of partitions, upper sums, and lower sums is also good. (4) Elementary properties of integrable functions: basic lower and upper bounds, additivity, linearity, integrable on [a; b] implies integrable on a subset of [a; b]... (5) Continuous implies integrable. (6) Basic examples of non-integrable functions (most importantly, not bounded implies not integrable). R x (7) Definition of antiderivative; definition of F (x) = c f(t) dt and basic properties of this function. (8) Theorem: If f is continuous on [a; b], and F and G are antiderivatives of f, then there exists a constant c such that f(x) = g(x) + c. (9) First Fundamental Theorem of Calculus: if c 2 [a; b], f is continuous at d 2 (a; b), and F (x) = R x 0 c f(t) dt, then F (d) = f(d). Also know its more useful statement: if c 2 [a; b], f continuous R x 0 on [a; b], then F (x) = c f(t) dt is continuous on [a; b], differentiable on (a; b), and F (x) = f(x). Understanding the proof geometrically is good. (10) Using the above two points to prove the Second Fundamental Theorem of Calculus for continuous functions. R tan(x) (11) Applying the First Fundamental Theorem to differentiate, for example, cos(x) sin(t) dt on (−π=2; π=2). (12) The Second Fundamental Theorem of Calculus: if f is integrable on [a; b], and F is an antiderivative R b for f on [a; b], then a f(x) dx = F (b) − F (a). Knowing that the hypotheses here are necessary is also good. (13) More advanced properties: f < g =) R f < R g, j R fj ≤ R jfj, f and g integralbe implies f · g integralbe, etc. (14) Definition of indefinite integrals, and inherent qualms therein (e.g., f need not be integrable for R f to exist; if f is integrable, then R f need not exist; \+C" does not really mean plust a constant, with like f(x) = 1=x). (15) The integrals we should know, e.g., xα for α 6= −1, and those coming from the derivatives of the trig functions. 0 R b 0 (16) The statement of u-substitution: if f is continuous, ' differentiable, and ' integrable, then a f('(x))' (x) dx = R '(b) '(a) f(x) dx. Also how this is derived simply from integrating the chain rule. (17) Using u-substitution in the forward direction, to solve things like R sin(x) cos(x) dx, R (x2+2x5)=(x3+ x6) dx, or R sec3(x) tan(x) dx. p p (18) Using u-substitution in the reverse direction, to solve things like R 1 − x2 dx, R p9 − x dx. (19) Using u-substitution justp to make a simple change of coordinates to make integrals easier, like R (x + 3)(x − 1)5 dx, R x 4 − x dx, R (x + 7)=(3x + 4)3 dx. Last modified: January 24, 2020, Date of test: January 27, 2020. 1 2 MICHAEL (20) The Mean Value Theorem for Integrals: if f is continuous on [a; b], g is integrable on [a; b], g(x) ≥ 0, R b R b then there exists c 2 (a; b) such that a f(x)g(x) dx = f(c) a g(x) dx. (21) Solving for g-weighted averages, I guess: this is really just solving integrals, but the terminology needs to be known. (22) Symmetry tricks are good: integral of an odd function about a symmetric domain is zero, trying a substitution that flips bounds of integration on a definite integral, etc. A lot of other \tricks" that people talk about are just guessing convenient u's, for doing the \reverse direction" version of u-substitution. (23) Using integrals to set-up area problems, e.g., finding the area bounded by some subset of curves in the plane. Now the next problems are not comprehensive. If you want more problems, you should be able to solve anything from Chapter 2 of the textbook should be fine. Problems: (1) For the following problems, answer true or false. If true, provide a brief reason. If false, provide a counterexample. R b R b (i) Suppose a f(x) dx = 0. Then a jf(x)j dx = 0. R x (ii) If u is once differentiable, then 0 u(t) dt is twice differentiable. R 1 R 1 2 (iii) If 0 u du = 0, then 0 u = 0. R 1 2 R 1 (iv) If u is continuous then 0 u dx = 0, then 0 u dx = 0. (v) If R 1 jf(x)j dx ≤ 2, then R 1 f(x) dx ≤ 2. 0 0 (vi) If f is continuous and g > 0, then Z b R b f a f dx dx = b : a g R a g dx (vii) If f and g are two continuous functions defined on [a; b], then b b R f dx R g(x) dx R fg a a = : b − a b − a b − a R f (viii) If f and g are continuous and g 6= 0, then g dx exists. R b (ix) If a f(x) dx = 0, then there exists c such that f(c) = 0. (x) Z π=2 p Z π=2 cos x dx < cos x: 0 0 (xi) Suppose f is a continuous function defined on [0; 1) that is positive. If x > y > 0, then Z y Z x f(t) dt < f(t) dt: 0 0 R b R b (xii) If a f(t) dt ≤ a g(t) dt, then there exists c 2 (a; b) such that f(c) < g(c). (xiii) Suppose that f; g are continuous defined on [a; b] and that for c 2 [a; b], Z c Z c f(t) dt < g(t) dt: a a Then f(x) ≤ g(x) for all x 2 [a; b]. MIDTERM 1 REVIEW 3 R b R a (xiv) If a f(t) dt = c, then b f(t) dt = −c. (xv) −0 = 0. (xvi) If u:[a; b] ! [c; d] and f :[c; d] ! R are continuous, then Z d Z b f(t) dt = f(u(t))u0(t) dt: c a Z 16 1 (xvii) 3 dt < 1. 0 1 + x R 1 2020 (xviii) 0 sin (x) dx < 1 R x (xix) If f(x) > 0, then 0 f(t) dt has no local extrema. (xx) If f(t) > 0 and g(x) > 0 are continuous, then Z g(x) f(t) dt 0 has no local extrema. (xxi) If f(x) > 0, continuous, and f is defined on [0; 1), then sZ x f(t) dt 0 has no critical points. R cos(x) (xxii) Suppose that f is continuous. Then F (x) = 0 f(t) dt is periodic. (xxiii) Suppose that f and g are continuous functions defined on [a; b]. Then there exists c such that R b f(t) dt R b g(t) dt f(c) = a and g(c) = a : b − a b − a (xxiv) Suppose f > 0 on [0; 1) then 9x > 0, such that Z x f(t) dt = π 0 R x R x p (xxv) If 0 f(t) dt has a horizontal asymptote, then 0 f(t) dt has a horizontal asymptote. R b R b 2 (xxvi) Suppose a jf(x)j dx = 0. Then a f(x) dx = 0. R x (xxvii) There exists a continuous function g such that 0 g(t) dt has a cusp at x = 1. (xxviii) 10 10 Z X X Z sin(ix) dx = sin(ix) dx: i=1 i=1 (xxix) If jfj ≤ 1, then Z Z gf dx ≤ g dx (xxx) If F; G are two antiderivatives of 1=x on its domain of definition, then F − G is constant. R x R x 0 (xxxi) Let H(x) = 0 f(t) dt 0 g(t) dt. Then H (x) = f(x)g(x). R 3 0 R 3 (xxxii) Let F (x) = 2 x cos(t) dt. Then F (x) = 2 cos(t) dt. (xxxiii) Suppose that f is continous function defined on [0; 1]. Let P1 = f0; 1=2; 1g and let P2 = f0; 1=3; 2=3; 1g. Then for all f Lf (P1) ≤ Lf (P2): (xxxiv) If p(x) is a polynomial defined on R and F is an antiderivative of p(x), then F is a polynomial. 4 MICHAEL R π (xxxv) If f(x) = f(x + π) then 0 f(t) dt = 0. R x (xxxvi) If F (−x) = −F (x), then 0 F (t) dt is an odd function. (xxxvii) If f is continuous, Z 1 Z 1 f(x) dx = f(1 − x) dx: 0 0 0 R x (xxxviii) If F = f, then 0 f = F (x) + c for some constant c. (xxxix) If m > n then Z 1 Z 1 xm dx > xn dx: 0 0 (xl) Z b Z b Z b jf + gj ≤ jfj + jgj a a a (xli) If f is defined on R and 1 ≤ f ≤ 2, then there exists c such that Z c f(t) dt = 2020: 0 R c (xlii) If 0 < f < 2, then there exists c such that 0 f(t) dt = 2020.

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