Math 2211: Calculus Post Exam 3 Review PDF

MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 Study Guide for Post Exam 3 in MATH 2211 Section 5.2 1. Definition: If f is a function defined for a ≤ x ≤...

MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 Study Guide for Post Exam 3 in MATH 2211 Section 5.2 1. Definition: If f is a function defined for a ≤ x ≤ b, we divide the interval [a, b] into n subintervals of equal width with ∆x = (b − a)/n. We let a = x0 , x1 ,... , xn = b be the endpoints of these subintervals and we let x∗1 , x∗2 ,... , x∗n be any sample points in these subintervals, so x∗i lies in the [xi−1 , xi ]. Then the definite integral of f from a to b is Z b n X f (x) dx = lim f (x∗i )∆x a n→∞ i=1 provided the limit exists and gives the same value for all possible choices of sample points. If it does exist, we say that f is integrable on [a, b]. 2. If the function f (x) has positive and negative values, the the definite integral from a to b is considered the net area, i.e. is the difference A1 − A2 where A1 is the area of the regions above the x-axis and A2 is the area of the regions below the x-axis. Z b (a) Net Area = f (x) dx = A1 − A2. a Z b (b) Total Area = |f (x)| dx = A1 + A2. a 3. Theorem: If f is continuous on [a, b] or if f has only a finite number of jump discontinuities, then f is integrable on [a, b]; i.e. the definite integral exists. 4. Theorem: If f is integrable on [a, b], then Z b n X f (x) dx = lim f (xi )∆x a n→∞ i=1 where ∆x = (b − a)/n and xi = a + ∆x. Z b n X 5. Midpoint Rule: f (x) dx ≈ f (xi )∆x = ∆x(f (xi ) + · · · + f (xn )) where ∆x = (b − a)/n a i=1 1 and xi = (xi−1 + xi ) is the midpoint of [xi−1 , xi ]. 2 – 1– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 6. Properties of Integrals: Z a (a) f (x) dx = 0 a Z a Z b (b) f (x) dx = − f (x) dx b a Z b (c) c dx = c(b − a) for c a constant. a Z b Z b Z b (d) [f (x) ± g(x)] dx = f (x) dx ± g(x) dx a a a Z b Z b (e) cf (x) dx = c f (x) dx a a Z b Z c Z b (f) f (x) dx = f (x) dx + f (x) dx a a c Z b (g) If f (x) ≥ 0 for a ≤ x ≤ b, then f (x) dx ≥ 0. a Z b Z b (h) If f (x) ≥ g(x) for a ≤ x ≤ b, then f (x) dx ≥ g(x) dx. a a Z b (i) If m ≤ f (x) ≤ M for a ≤ x ≤ b, then m(b − a) ≤ f (x) dx ≤ M (b − a). a Problems to be able to Solve: 1. Given a function and interval, approximate the definite integral using the left or right Riemann sum or midpoint rule. 2. Use integral properties to find areas. 3. Use geometry to find areas. 4. Solve any question similar to lecture, homework, or recitation. Section 5.3 1. The Fundamental Theorem of Calculus, Part I: If f is continuous on [a, b], then the function defined by Z x g(x) = f (t) dt, a ≤ x ≤ b a is continuous on [a, b] and differentiable on (a, b), and g ′ (x) = f (x). – 2– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 2. Note: We must use chain rule if the upper limit is not an x but rather a function. Z x d 3. Note: The first part can also be written as f (t) dt = f (x) which says that if we dx a integrate a continuous function f and differentiate, we end up at the same function. 4. The Fundamental Theorem of Calculus, Part II: If f is continuous on [a, b], then Z b f (x) dx = F (b) − F (a) a where F (x) is any antiderivative of f (x), that is F ′ (x) = f (x). 5. Note: We must be careful with the continuous assumption of f (x). Z x 6. Note: The second part can be written as F ′ (t) dt = F (x) − F (a) which says that if we a differentiate F (x) then integrate the result, we end at the same function F (x) plus a constant. Problems to be able to Solve: 1. Use the Fundamental Theorems in practice. 2. Evaluate definite integrals using antiderivatives from section 4.9. 3. Find the area under a curve. 4. Solve any question similar to lecture, homework, or recitation. Section 5.4 1. The indefinite integral is given by Z f (x) dx = F (x) + C where F ′ (x) = f (x), meaning that F (x) is any antiderivative of f (x). (Do not forget the +C!). 2. Remember that we can often use algebra to simplify our integrals into simpler integrands that are on our charts. 3. Net Change Theorem: The integral of a rate of change is the net change Z b F ′ (x) dx = F (b) − F (a). a – 3– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 4. Common Applications of the Net Change Theorem: (a) V (t) is the volume of water in a reservoir at time t, then Z b V ′ (t) dt = V (b) − V (a) a is the change in the amount of water between time a and b, here V ′ (t) is the rate at which water flows into the reservoir. dn (b) is the rate of growth of a population t, then dt Z b dn dt = n(b) − n(a) a dt is the net change in population during the time period a to b. (c) C(x) is the cost of producing x units of an item, then Z b C ′ (x) dx = C(b) − C(a) a is the increase in cost when production is increased from a units to b units. (d) v(t) is the velocity of an object moving in a straight line at time t, then Z b v ′ (t) dt = s(b) − s(a) a is the net change of position between time a and b, also called displacement. The integral Z b |v ′ (t)| dt = total distance travelled from time a to b a Note: Idea for total distance traveled is to split your integral up into when v(t) > 0 and v(t) < 0 and add up the absolute value of what each integral equals. Problems to be able to Solve: 1. Find indefinite integrals. 2. Solve any question involving the above Net Change Theorem. 3. Solve any question similar to lecture, homework, or recitation. – 4– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 Section 5.5 1. Substitution Rule: If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then Z Z ′ f (g(x))g (x) dx = f (u) du = F (u) + C = F (g(x)) + C. (Note that we are using the convention that F ′ (x) = f (x) where F is any antiderivative of f.) Note: We typically want to pick u = g(x) in such a way that du = g ′ (x)dx is also in our integrand up to a constant. Some tricks we saw were letting u be the inner function of a composition function or the power of ef (x) for example. 2. Substitution Rule for Definite Integrals: If g ′ is continuous on [a, b] and f is continuous on the range of u = g(x), then Z b Z g(b) ′ f (g(x))g (x) dx = f (u) du. a g(a) 3. Note: You can use this method for these problems, or solve the indefinite integral and use the solution to it to solve the definite integral. 4. Exponential functions can be integrated using the following formulas: Z ex dx = ex + C ax Z ax dx = +C ln a Note: For substitutions involving exponential functions, a good starting point for picking our u is by looking at the function in the power of e or a when it s more complicated than just x. 5. The following formula can be used to evaluate integrals involving logarithms: Z x−1 dx = ln |x| + C Note: For substitutions involving logarithms or resulting in logarithms, a good starting point for picking our u is by looking at the function in the denominator of our integrand. – 5– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 6. The following integration formulas yield inverse trigonometric solutions: Z   du u (a) √ = arcsin +C a2 − u2 |a| Z du 1 u (b) = arctan +C a2 + u 2 a a Z   du 1 u (c) √ = arcsec +C u a2 − u 2 |a| |a| 7. Recall that a function is even means that f (−x) = f (x) and that a function is odd means that f (−x) = −f (x), it then follows that if f (x) is continuous on [−a, a] and Z a Z a (a) If f (x) is even, then f (x) dx = 2 f (x) dx. −a 0 Z a (b) If f (x) is odd, then f (x) dx = 0. −a Problems to be able to Solve: 1. Use substitution rules to solve indefinite or definite integrals. 2. Use of the below integrals to solve problems. 3. Use even or odd properties to evaluate a function’s integral. 4. Solve any question similar to lecture, homework, or recitation. – 6– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 Table of Helpful Integrals: Z Z 1. k dx = kx + C 11. csc x cot x dx = − csc x + C xn+1 Z Z 2. xn dx = +C 12. tan x dx = − ln | cos x| + C n+1 Z 1 Z 3. dx = ln |x| + C x 13. cot x dx = ln | sin x| + C Z 4. ex dx = ex + C Z 14. sec x dx = ln | sec x + tan x| + C Z x a 5. ax dx = +C ln a Z Z 15. csc x dx = ln | csc x − cot x| + C 6. sin x dx = − cos x + C Z 1 Z 16. dx = arctan x + C 1 + x2 7. cos x dx = sin x + C Z 1 1 x Z 17. dx = arctan + C 8. 2 sec x dx = tan x + C a2 +x 2 a a Z Z 1 2 18. √ dx = arcsin x + C 9. csc x dx = − cot x + C 1 − x2 Z 1 x Z 10. sec x tan x dx = sec x + C 19. √ dx = arcsin + C a2 −x 2 a – 7– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 Practice Problems Section 5.2: Z 5 Z 9 Z 13 1. Given that f (x) dx = 6, f (x) dx = 9, and f (x) dx = 7, find each of the following: 4 5 4 Z 9 Z 13 Z 4 Z 4 (a) f (x) dx (b) f (x) dx (c) f (x) dx (d) f (x) dx 4 9 4 13 Z 9 Z 9 Z 13 2. Given that g(x) dx = −8, g(x) dx = −7, and g(x) dx = 6, find each of the −2 0 −2 following: Z 0 Z 13 Z 13 Z −2 (a) g(x) dx (b) g(x) dx (c) g(x) dx (d) g(x) dx −2 9 0 0 Z 4 Z 4 3. Given that g(x) dx = 6, and f (x) dx = −3, find each of the following: 1 1 Z 4 Z 4 (a) [2f (x) − g(x)] dx (c) [3g(x) + 1] dx 1 1 Z 4 Z 4 (b) [f (x) + 2g(x)] dx (d) [2g(x) − 3f (x)] dx 1 1 Section 5.3: 4. Using Part II, evaluate each definite integral below: Z 4 Z 3   2 4 (a) (3x + 1) dx (f) 4x + dx −2 1 x √ Z 5 Z 3 3 2 8 (b) x (x − 2) dx (g) dx 0 1 1 + x2 Z 2 2 5 Z (c) ( x−3/2 + 2x1/2 + 6x) dx (h) 7x dx 1 2 0 Z π Z 5 (d) (3 cos x − 8 sin x) dx (i) 6 dx 0 1 Z π/6 Z 7 5 (e) sec x(cos x − tan x) dx (j) dx −π/3 1 x – 8– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 Z x 5. Let F (x) = (3t2 + 3) dt. Find each of the following: 0 (a) F (2) (b) F ′ (x) (c) F ′ (2) (d) F ′′ (x) (e) F ′′ (2) Z 2   2 6. Let H(x) = dt. Find each of the following: x t (a) H(3) (b) H ′ (x) (c) H ′ (3) (d) H ′′ (x) (e) H ′′ (3) Z 1 7. Let H(x) = sin(π − t) dt. Find each of the following: x (a) H ′ (x) (b) H ′ (π/3) (c) H ′′ (π/3) (d) H ′′′ (x) (e) H ′′′ (π/3) Section 5.4: 8. Evaluate the indefinite integrals: Z Z   18 (a) (x4 + 3x) dx (f) 3x + 2 dx x Z Z 4 (b) x6 (x2 − 2x) dx (g) √ dx 1 − x2 Z   7 −3/5 Z 5/4 (c) x + x + 6x dx (h) 8x dx 3 Z Z (d) (−2 cos x + tan x) dx (i) 7 dx Z Z 8 (e) cos x(− sec x + 8) dx (j) dx x 9. The velocity of a particle moving in a straight line is given by v(t) = t2 − 6t + 8 for any time t. Find the displacement and total distance travelled by the particle from t = 1 to t = 6. 10. The velocity of a particle moving in a straight line is given by v(t) = t2 − 8t + 15 for any time t. Find the displacement and total distance travelled by the particle from t = 1 to t = 5. 11. Water flows into a storage tank at r(t) = 10 + 6t liters per minute, where 0 ≤ t ≤ 20. Find the amount of water that flows into the tank during the first 15 minutes. 12. The marginal cost of producing x wooden birds is given by C ′ (x) = 6x + 200. Find the increase in cost if production is increased from 5 birds to 8 birds. – 9– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 Section 5.5: 13. Evaluate using substitution method: Z x2 Z (a) 3x(x2 + 2)7 dx (f) dx (3x3 + 1)3 Z Z (b) (5x − 2)3 dx (g) sec(2x) tan(2x) dx Z Z 4 (c) x2 (3 − x3 )4 dx (h) 3x3 ex dx Z Z (d) (2 cos x)(6 sin x) dx (i) cos(6x) dx Z √ Z 2ex (e) 2x 4x2 + 4 dx (j) dx 2 + 3ex 14. Evaluate each definite integral using substitution method: Z 4p Z 1/2 x (a) 8y + 1 dy (d) √ dx 0 0 1 − x2 Z π/2 Z π/3 2 sec2 x (b) 4 cos x sin x dx (e) dx 0 0 3 − tan x Z 1 Z 1/3 6x (c) 2 dx (f) 4 tan(πx) dx 0 3 + 2x 0 15. Evaluate each definite integral using substitution method: Z Z 1 1 (a) dx (d) √ dx 2 + x2 6 − x2 Z 1 Z 1 (b) dx (e) √ dx 3 + 4x 2 64 − x2 Z 8 Z √2 1 1 (c) 2 dx (f) √ dx 0 144 + x 0 9 − x2 16. Using symmetry properties, evaluate the functions. Z 1 Z π 3 (c) 2 sin(πx) dx (a) (x + x) dx −1 −π Z 1 Z 1 2 7x2 − 8 + tan(3x) dx
(b) (x + 2) dx (d) −1 −1 – 10– MATH 2211: Calculus of One Variable I Study Guide for Post Exam 3 17. Evaluate each definite integral using substitution method: e − e−x Z Z x (a) e2x dx (d) dx ex + e−x √ Z Z −x (b) 3 dx (e) e2x 1 − e2x dx π/3 sin x − cos x eln(1−t) Z Z (c) dx (f) dt 0 sin x + cos x 1−t 18. Evaluate each definite integral using substitution method: Z cos x Z arcsin(x) (a) 2 dx (d) √ dx 4 + sin x 1 − x2 ex Z Z tan(arcsin(x)) (b) 2x dx (e) √ dx 1+e 1 − x2 y2 t arctan(t2 ) Z Z (c) dy (f) dt 2 + y6 1 + t4 – 11–