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Tài liệu này giả định rằng bạn đã có một khóa học về giải tích thông qua tích phân giải tích và muốn dạy kèm (hoặc gia sư tại nhà) học sinh đang học tích phân tích lần đầu tiên. Chúng tôi tập trung vào việc phát triển kỹ năng và trực giác để chuẩn bị tinh thần để đối mặt với một nhóm nhỏ sinh viên đang mắc kẹt hoặc tệ hơn là chưa, không bị mắc kẹt và đầy đủ các câu hỏi. Chúng tôi bắt đầu với việc xem xét những điều cơ bản. Trong của chúng tôi thảo luận, chúng tôi lưu ý đến các hàm có giá trị thực được giảm trừ cho tất cả các số thực hoặc các khoảng số thực giống như những số bạn đã học trong giải tích cho đến điểm này.
Integral Calculus - Exercises 6.1 Antidifferentiation The Indefinite Integral In problems through 7, find the indicated integral R√ xdx Solution Z Z √ √ xdx = x dx = x + C = x x + C 3 R x 3e dx Solution Z Z x 3e dx = ex dx = 3ex + C R √ (3x2 − 5x + 2)dx Solution Z Z Z √ √ Z √ 2 (3x − 5x + 2)dx = x dx − xdx + dx = R³ √ √ = · x3 − · x x + 2x + C = 3 √ = x3 − x 5x + 2x + C ´ − x22 + √3x dx Solution ả Z Z Z Z 1 −2 − 2+√ dx − x dx + x− dx = dx = 2x x x x 1 = ln |x| − · (−1)x−1 + · 2x + C = √ ln |x| = + + x + C x 2x 40 INTEGRAL CALCULUS - EXERCISES 41 R¡ ¢ 2ex + x6 + ln dx Solution ả Z Z Z Z x x 2e + + ln dx = e dx + dx + ln dx = x x = 2ex + ln |x| + (ln 2)x + C R x2 +3x−2 √ dx x Solution Z Z Z Z 1 x + 3x − √ dx = x dx + x dx − x− dx = x = x + · x − · 2x + C = 3 = x + 2x − 4x + C = √ √ 2√ = x x + 2x x − x + C R ¡ ¢ (x3 − 2x2 ) x1 − dx Solution ả Z Z (x − 2x ) − dx = (x2 − 5x3 − 2x + 10x2 )dx = x Z = (−5x3 + 11x2 − 2x)dx = 1 = −5 · x4 + 11 · x3 − · x2 + C = 11 = − x + x − x2 + C Find the function f whose tangent has slope x3 − x22 + for each value of x and whose graph passes through the point (1, 3) Solution The slope of the tangent is the derivative of f Thus f (x) = x3 − +2 x2 and so f(x) is the indefinite integral ¶ Z µ Z x − + dx = f (x) = f (x)dx = x = x + + 2x + C x INTEGRAL CALCULUS - EXERCISES 42 Using the fact that the graph of f passes through the point (1, 3) you get = + + + C or C = − 4 Therefore, the desired function is f (x) = 14 x4 + x2 + 2x − 54 It is estimated that t years from now the population of a certain lakeside community will be changing at the rate of 0.6t2 + 0.2t + 0.5 thousand people per year Environmentalists have found that the level of pollution in the lake increases at the rate of approximately units per 1000 people By how much will the pollution in the lake increase during the next years? Solution Let P (t) denote the population of the community t years from now Then the rate of change of the population with respect to time is the derivative dP = P (t) = 0.6t2 + 0.2t + 0.5 dt It follows that the population function P (t) is an antiderivative of 0.6t2 + 0.2t + 0.5 That is, Z Z P (t) = P (t)dt = (0.6t2 + 0.2t + 0.5)dt = = 0.2t3 + 0.1t2 + 0.5t + C for some constant C During the next years, the population will grow on behalf of P (2) − P (0) = 0.2 · 23 + 0.1 · 22 + 0.5 · + C − C = = 1.6 + 0.4 + = thousand people Hence, the pollution in the lake will increase on behalf of · = 15 units 10 An object is moving so that its speed after t minutes is v(t) = 1+4t+3t2 meters per minute How far does the object travel during 3rd minute? Solution Let s(t) denote the displacement of the car after t minutes Since v(t) = ds = s0 (t) it follows that dt Z Z s(t) = v(t)dt = (1 + 4t + 3t2 )dt = t + 2t2 + t3 + C During the 3rd minute, the object travels s(3) − s(2) = + · + 27 + C − − · − − C = = 30 meters INTEGRAL CALCULUS - EXERCISES 43 Homework In problems through 13, find the indicated integral Check your answers by differentiation R R R x5 dx R x dx dx 5dx R¡ √ R x2 ¢ 2 x − + (x − 3x + 6)dx dx x x √ ´ R ³√ R ¡ ex √ ¢ + x x dx x − 2√x + dx √ ´ R³1 R x2 +2x+1 − 2x32 + e2 + 2x dx 10 dx 3x ¢ R 3¡ R√x x(x − 1)dx 11 R x 2x + x dx 12 13 x(2x + 1) dx 14 Find the function whose tangent has slope 4x + for each value of x and whose graph passes through the point (1, 2) 15 Find the function whose tangent has slope 3x2 + 6x − for each value of x and whose graph passes through the point (0, 6) 16 Find a function whose graph has a relative minimum when x = and a relative maximum when x = 17 It is estimated that t months from now the population of a certain town will be changing at the rate of + 5t people per month If the current population is 10000, what will the population be months from now? 18 An environmental study of a certain community suggests that t years from now the level of carbon monoxide in the air will be changing at the rate of 0.1t + 0.1 parts per million per year If the current level of carbon monoxide in the air is 3.4 parts per million, what will the level be years from now? 19 After its brakes are applied, a certain car decelerates at the constant rate of meters per second per second If the car is traveling at 108 kilometers per hour when the brakes are applied, how far does it travel before coming to a complete stop? (Note: 108 kmph is the same as 30 mps.) 20 Suppose a certain car supplies a constant deceleration of A meters per second per second If it is traveling at 90 kilometers per hour (25 meters per second) when the brakes are applied, its stopping distance is 50 meters (a) What is A? INTEGRAL CALCULUS - EXERCISES 44 (b) What would the stopping distance have been if the car had been traveling at only 54 kilometers per hour when the brakes were applied? (c) At what speed is the car traveling when the brakes are applied if the stopping distance is 56 meters? Results 16 x6 + C − x1 + C 5 23 x − 95 x + 6x + C 12 ex + 25 x + C 3 13 ln |x| + 2x + e2 x + 13 x + C 11 25 x5 + 13 x3 + C 13 x4 + 43 x3 + 12 x2 + C 14 f (x) = 2x2 + x − 15 f (x) = x3 + 3x2 − 2x + 16 f (x) = 13 x3 − 52 x2 + 4x; not unique 17 10128 18 4.15 parts per million 19 75 meters 20 (a) A = 6.25 (b) 42 meters (c) 120.37 kilometers per hour 47 x + C 5x + C 2xp2 + x12 + ln |x| + C √ √ 25 (x3 )x − x + 2x + C 10 x − x1 + ln x + C 12 27 x − 23 x + C INTEGRAL CALCULUS - EXERCISES 6.2 45 Integration by Substitution In problems through 8, find the indicated integral R (2x + 6)5 dx Solution Substituting u = 2x + and 12 du = dx, you get Z Z 1 (2x + 6) dx = u5 du = u6 + C = (2x + 6)6 + C 12 12 R [(x − 1)5 + 3(x − 1)2 + 5] dx Solution Substituting u = x − and du = dx, you get Z Z ¤ £ (x − 1) + 3(x − 1) + dx = (u5 + 3u2 + 5)du = u + u3 + 5u + C = (x − 1)6 + (x − 1)3 + 5(x − 1) + C = = Since, for a constant C, C − is again a constant, you can write Z Ê Ô (x 1)5 + 3(x − 1)2 + dx = (x − 1)6 + (x − 1)3 + 5x + C R xex dx Solution Substituting u = x2 and 12 du = xdx, you get Z Z 1 x2 xe dx = eu du = eu + C = ex + C 2 R x5 e1−x dx Solution Substituting u = − x6 and − 16 du = x5 dx, you get Z Z 1 1−x6 xe dx = − eu du = − eu + C = − e1−x + C 6 R 2x4 dx x5 +1 Solution Substituting u = x5 + and 25 du = 2x4 dx, you get Z 2x4 dx = x +1 Z ¯ 2 ¯ du = ln |u| + C = ln ¯x5 + 1¯ + C u 5 INTEGRAL CALCULUS - EXERCISES R 46 √10x −5x dx x4 −x2 +6 Solution Substituting u = x4 − x2 + and 52 du = (10x3 − 5x)dx, you get Z Z Z 1 10x3 − 5x 5 − 12 + C = √ √ du = dx = u du = · 2u 2 u x4 − x2 + √ = x4 − x2 + + C R dx x ln x Solution Substituting u = ln x and du = x1 dx, you get Z Z 1 dx = du = ln |u| + C = ln |ln x| + C x ln x u R ln x2 dx x Solution Substituting u = ln x and du = x1 dx, you get Z ln x2 dx = x Z ln x dx = x Z udu = · u2 + C = (ln x)2 + C Use an appropriate change of variables to find the integral Z (x + 1)(x − 2)9 dx Solution Substituting u = x − 2, u + = x + and du = dx, you get Z Z Z 9 (x + 1)(x − 2) dx = (u + 3)u du = (u10 + 3u9 )du = 11 u + u10 + C = 11 10 (x − 2)11 + (x − 2)10 + C = 11 10 = 10 Use an appropriate change of variables to find the integral Z √ (2x + 3) 2x − 1dx Solution Substituting u = 2x − 1, u + = 2x + and 12 du = dx, you INTEGRAL CALCULUS - EXERCISES 47 get Z Z Z Z √ √ 1 (2x + 3) 2x − 1dx = (u + 4) udu = u du + u du = 2 2 = · u2 + · u2 + C = u2 + u2 + C = 5 = (2x − 1) + (2x − 1) + C = √ √ = (2x − 1) 2x − + (2x − 1) 2x + C = ả √ x− + +C = = (2x 1) 2x 5 ả √ 17 = x+ (2x − 1) 2x − + C 25 Homework In problems through 18, find the indicated integral and check your answer by differentiation R R√ R e5x dx R 4x − 1dx dx e1−x dx R R 3x+5x2 −1 2xe dx x(x2 + 1)5 dx R √ R 3 3x R x2 x + 8dx Rx (x + 1)2 dx 10 (x + 1)(x + 2x + 5)12 dx +5)2 dx (x R R +12x3 +6 11 R (3x2 − 1)ex −x dx 12 R x53x dx +5x4 +10x+12 3x−3 6x−3 13 14 dx −2x+6)2 dx R 4x21−4x+1 R (x ln 5x 15 dx 16 dx √ x) R 2xxln(x2 +1) R x(ln e√ x 17 dx 18 dx x2 +1 x In problems 19 through 23, use an appropriate change of variables to find the indicated R x integral R √ dx 20 R x x + 1dx 19 R x−1 x x+3 21 22 dx dx (x−4)2 R (x−5) x 23 dx 2x+1 √ 24 Find the function whose tangent has slope x x2 + for each value of x and whose graph passes through the point (2, 10) 2x 25 Find the function whose tangent has slope 1−3x for each value of x and whose graph passes through the point (0, 5) INTEGRAL CALCULUS - EXERCISES 48 26 A tree has been transplanted and after x years is growing at the rate of + (x+1) meters per year After two years it has reached a height of five meters How tall was it when it was transplanted? 27 It is projected that t years from now the population of a certain country will be changing at the rate of e0.02t million per year If the current population is 50 million, what will the population be 10 years from now? Results 15 e5x + C 13 ln |3x + 5| + C ex −1 + C √ (x2 + 8) x2 + + C − 3(x31+5) + C 11 ex −x + C 13 − 2(x2 −2x+6) +C 15 ln 5x + C 17 12 ln2 (x2 + 1) + C 19 x + ln |x − 1| + C 1 21 − (x−5) +C − 4(x−5)4 2x+1 23 x − ln |2x + 1| + C √ 16 (4x − 1) 4x − + C −e1−x + C 12 (x2 + 1)6 + C 21 (x3 + 1) + C 10 26 (x2 + 2x + 5)13 + C 12 35 ln |x5 + 5x4 + 10x + 12| + C 14 32 ln |2x − 1| + C 16 − ln +C √x x 18 2e + C √ √ 20 25 (x + 1)2 x + − 23 (x + 1) x + + C 22 − x−4 + ln |x − 4| + C √ 24 f (x) = 13 (x2 + 5) x2 + + 25 f (x) = − 13 ln |1 − 3x2 | + 26 meters 27 61 million INTEGRAL CALCULUS - EXERCISES 6.3 49 Integration by Parts In problems through 9, use integration by parts to find the given integral R xe0.1x dx Solution Since the factor e0.1x is easy to integrate and the factor x is simplified by differentiation, try integration by parts with g(x) = e0.1x Then, Z G(x) = and e0.1x dx = 10e0.1x f (x) = x and f (x) = and so Z Z 0.1x 0.1x xe dx = 10xe − 10 e0.1x dx = 10xe0.1x − 100e0.1x + C = R = 10(x − 10)e0.1x + C (3 − 2x)e−x dx Solution Since the factor e−x is easy to integrate and the factor 3−2x is simplified by differentiation, try integration by parts with g(x) = e−x Then, G(x) = Z and e−x dx = −e−x f (x) = − 2x and f (x) = −2 and so Z Z −x −x (3 − 2x)e dx = (3 − 2x)(−e ) − e−x dx = = (2x − 3)e−x + 2e−x + C = (2x − 1)e−x + C R x ln x2 dx Solution In this case, the factor x is easy to integrate, while the factor ln x2 is simplified by differentiation This suggests that you try integration by parts with g(x) = x and f (x) = ln x2 Then, G(x) = Z xdx = x2 and f (x) = 2x = x x INTEGRAL CALCULUS - EXERCISES 6.5 59 The Definite Integral In problems through evaluate the given definite integral R2 ln (et − e−t ) dt Solution Z ¡ t ¢ ¡ ¢¯2 1 e − e−t dt = et + e−t ¯ln = e2 + e−2 − eln − e− ln = ln 2 = e2 + e−2 − eln − eln = e2 + e−2 − = e2 + e−2 − −2 = R0 (2x + 6)4 dx Solution Substitute u = 2x + Then 12 du = dx, u(−3) = 0, and u(0) = Hence, ¯6 Z Z ¯¯ 65 3888 (2x + 6) dx = u du = u¯ = −0 = 10 10 −3 Solution Substitute u = x3 + Then 13 du = x2 dx, u(1) = 2, and u(2) = Hence, ¯9 Z Z x2 −2 ¯¯ 1 dx = u du = − ¯ = − + = 3 3u 27 54 (x + 1) −3 R2 x2 dx (x3 +1)2 R e2 e dx x ln x Solution Substitute u = ln x Then du = x1 dx, u(e) = 1, and u(e2 ) = Hence, Z e e2 dx = x ln x Z du = ln |u||21 = ln − ln = ln u R e2 t ln 2tdt Solution Since the factor t is easy to integrate and the factor ln 2t is simplified by differentiation, try integration by parts with g(t) = t and f (t) = ln 2t INTEGRAL CALCULUS - EXERCISES Then, G(t) = and so Z e t ln 2tdt = Z 60 tdt = t2 and f (t) = t ¯e ¯e Z e ¯2 2 2 ¯¯ ¯ tdt = t ln 2t − t ¯ = t ln 2t¯ − 2 12 1 2 1 1 1 e ln e − e2 − ln + = e2 − e2 + = = 16 16 16 16 1 = e + = (e2 + 1) 16 16 16 R1 x2 e2x dx R R Solution Apply the reduction formula xn eax dx = a1 xn eax − na xn−1 eax dx twice to get ¯1 Z Z ¯ x2 e2x dx = x2 e2x ¯¯ − xe2x dx = 0 ¯01 ¯ Z 2x ¯¯ 2x ¯¯1 1 2x = x e ¯ − xe ¯ + e dx = 2 0 ả à ¶ ¯ 1 2x 2x 2x ¯¯1 2x ¯¯1 x e − xe + e = ¯ = 2x − 2x + e = 2 0 ả 1 1 1 = − + e − e = e − = (e2 − 1) 2 4 4 R5 5−t te− 20 dt Solution Integrate by parts with g(t) = e− Then, G(t) = Z 5−t 5−t 20 and 5−t e− 20 dt = 20e− 20 f(t) = t and f (t) = and so Z Z ¯ ³ ´¯5 5−t ¯5 ¯ − 5−t − − 5−t − 5−t − 5−t 20 20 20 20 20 te dt = 20te e dt = 20te − 400e ¯ − 20 ¯ = 0 0 ¯ 5−t ¯5 = 20(t − 20)e− 20 ¯ = 20 · (−15)e0 − 20 · (−20)e− = ³ ´ − 14 − 14 = −300 + 400e = 100 4e − INTEGRAL CALCULUS - EXERCISES (a) Show that Rb Rc b f (x)dx = Rc f (x)dx R1 (b) Use the formula in part (a) to evaluate −1 |x| dx R4 (c) Evaluate (1 + |x − 3|)2 dx Solution (a) By the Newton-Leibniz formula, you have Z b Z c f(x)dx + f(x)dx = F (b) − F (a) + F (c) − F (b) = a b Z c = F (c) − F (a) = f (x)dx a f (x)dx + 61 a a (b) Since |x| = −x for x ≤ and |x| = x for x ≥ 0, you have to break the given integral into two integrals ¯0 Z Z ¯¯ 1 |x| dx = (−x)dx = − x ¯ = + = −1 2 −1 −1 and Z Thus, Z |x| dx = −1 |x| dx = Z Z 0 −1 ¯1 ¯¯ 1 xdx = x ¯ = − = 2 |x| dx + Z |x| dx = 1 + = 2 (c) Since |x − 3| = −x + for x ≤ and |x − 3| = x − for x ≥ 3, you get Z Z Z 2 (1 + |x − 3|) dx = [1 + (−x + 3)] dx + [1 + (x − 3)]2 dx = Z0 Z = (−x + 4)2 dx + (x − 2)2 dx = ¯3 ¯4 ¯ ¯ 1 3¯ 3¯ = − (−x + 4) ¯ + (x − 2) ¯ = 3 64 70 = − + + − = 3 3 (a) Show that if F is an antiderivative of f , then Z b f (−x)dx = −F (−b) + F (−a) a INTEGRAL CALCULUS - EXERCISES 62 (b) A function f is said to be even if f (−x) = f (x) [For example, f (x) = x2 is even.] Use problem and part (a) to show that if f is even, then Z a Z a f (x)dx = f (x)dx −a (c) Use part (b) to evaluate R1 −1 |x| dx and R2 x2 dx −2 (d) A function f is said to be odd if f(−x) = −f (x) Use problem and part (a) to show that if f is odd, then Z a f (x)dx = −a R 12 (e) Evaluate −12 x3 dx Solution (a) Substitute u = −x Then du = −dx, u(a) = −a and u(b) = −b Hence, Z b Z −b f(−x)dx = − f (u)du = −F (u)|−b −a = −F (−b) + F (−a) a −a (b) Since f(−x) = f (x), you can write Z a Z Z f (x)dx = f (−x)dx + −a −a a f (x)dx By the part (a), you have Z f(−x)dx = −F (0) + F (−(−a)) = F (a) − F (0) = −a Z a = f (x)dx Hence, Z a f (x)dx = −a Z a f (x)dx (c) Since f(x) = |x| is an even function, you have Z Z Z ¯1 |x| dx = |x| dx = xdx = x2 ¯0 = − = −1 0 Analogously, Z −2 x dx = Z ¯2 ¯¯ 16 x dx = x ¯ = 3 INTEGRAL CALCULUS - EXERCISES 63 (d) Since f(−x) = −f (x), you can write Z a Z Z f (x)dx = − f (−x)dx + −a −a a f(x)dx = = F (0) − F (a) + F (a) − F (0) = (e) Since f(x) = x3 is an odd function, you have Z 12 x3 dx = −12 10 It is estimated that t days from now a farmer’s crop will be increasing at the rate of 0.3t2 + 0.6t + bushels per day By how much will the value of the crop increase during the next days if the market price remains fixed at euros per bushel? Solution Let Q(t) denote the farmer’s crop t days from now Then the rate of change of the crop with respect to time is dQ = 0.3t2 + 0.6t + 1, dt and the amount by which the crop will increase during the next days is the definite integral Z ¢ ¡ ¢¯5 ¡ Q(5) − Q(0) = 0.3t2 + 0.6t + dx = 0.1t3 + 0.3t2 + t ¯0 = = 12.5 + 7.5 + = 25 bushels Hence, the value of the market price will increase by 25 · = 75 euros Homework In problems 17, R through (x − 3x + 1) dx R05 (2 + 2t + 3t2 ) dt ¢ R3¡ + x1 + x12 dx R6 x (x − 1)dx R04 √ dt R 6t+1 6x 11 dx R02 x +1 13 (t + 1)(t − 2)9 dt R12 15 xe−x dx R−2 10 17 (20 + t)e−0.1t dt evaluate R the 5given 2definite integral ³(3x − 3x´ + 2x − 1) dx R−1 √ t − √1t dt R −1 t+1 dt t3 R−3 (2x − 4)5 dx √ R1 10 (t + t) t4 + 2t2 + 1dt R0e+1 x 12 dx R2e2 x−1 14 ln tdt R1e2 (ln x)2 16 dx x INTEGRAL CALCULUS - EXERCISES 64 18 A study indicates that x months from now the population of a certain town will be increasing at the rate of 5+3x people per month By how much will the population of the town increase over the next months? 19 It is estimated that the demand for oil is increasing exponentially at the rate of 10 percent per year If the demand for oil is currently 30 billion barrels per year, how much oil will be consumed during the next 10 years? 20 An object is moving so that its speed after t minutes is + 2t + 3t2 meters per minute How far does the object travel during the 2nd minute? Results 20 83 + ln 43 23 13 − 110 17 152.85 − 72 29 10 76 14 e2 + 18 98 people 144 252 11 ln 15 −3e−2 − e2 19 515.48 billion barrels 40 − 16 12 e 16 83 20 15 meters INTEGRAL CALCULUS - EXERCISES 6.6 65 Area and Integration In problems through find the area of the region R R is the triangle with vertices (−4, 0), (2, 0) and (2, 6) Solution From the corresponding graph (Figure 6.1) you see that the region in question is bellow the line y = x + above the x axis, and extends from x = −4 to x = y y=x+4 -4 -2 x Figure 6.1 Hence, A= Z −4 (x + 4)dx = ả2 x + 4x ¯¯ = (2 + 8) − (8 − 16) = 18 −4 R is the region bounded by the curve y = ex , the lines x = and x = ln 12 , and the x axis Solution Since ln 12 = ln − ln = − ln ' −0.7, from the corresponding graph (Figure 6.2) you see that the region in question is bellow the line y = ex above the x axis, and extends from x = ln 12 to x = y y=ex -1 -ln2 Figure 6.2 x INTEGRAL CALCULUS - EXERCISES 66 Hence, A= Z ln ex dx = ex |0ln = e0 − eln = − 2 1 = 2 R is the region in the first quadrant that lies below the curve y = x2 +4 and is bounded by this curve, the line y = −x + 10, and the coordinate axis Solution First sketch the region as shown in Figure 6.3 Note that the curve y = x2 + and the line y = −x + 10 intersect in the first quadrant at the point (2, 8), since x = is the only positive solution of the equation x2 + = −x + 10, i.e x2 + x − = Also note that the line y = −x + 10 intersects the x axis at the point (10, 0) y y=x2+4 10 y=-x+10 -2 10 x Figure 6.3 Observe that to the left of x = 2, R is bounded above by the curve y = x2 + 4, while to the right of x = 2, it is bounded by the line y = −x + 10 This suggests that you break R into two subregions, R1 and R2 , as shown in Figure 6.3, and apply the integral formula for area to each subregion separately In particular, ả2 Z 32 (x + 4)dx = A1 = x + 4x ¯¯ = + = 3 0 and A2 = Z 10 (−x + 10)dx = Therefore, ả10 − x + 10x ¯¯ = −50 + 100 + − 20 = 32 2 A = A1 + A2 = 32 128 + 32 = 3 INTEGRAL CALCULUS - EXERCISES 67 R is the region bounded by the curves y = x2 + and y = −x2 , the line x = 3, and the y axis Solution Sketch the region as shown in Figure 6.4 y 15 y=x2+5 10 -2 -5 -1 x y=-x2 Figure 6.4 Notice that the region in question is bounded above by the curve y = x2 + and below by the curve y = −x2 and extends from x = to x = Hence, ả3 Z Z ¯ 2 A= [(x +5)−(−x )]dx = (2x +5)dx = x + 5x ¯¯ = 18+15 = 33 0 R is the region bounded by the curves y = x2 − 2x and y = −x2 + Solution First make a sketch of the region as shown in Figure 6.5 and find the points of intersection of the two curves by solving the equation x2 − 2x = −x2 + i.e 2x2 − 2x − = x = −1 and x = to get The corresponding points (−1, 3) and (2, 0) are the points of intersection y y=-x2+4 -3 -2 y=x2-2x -1 -2 Figure 6.5 x INTEGRAL CALCULUS - EXERCISES 68 Notice that for −1 ≤ x ≤ 2, the graph of y = −x2 + lies above that of y = x2 − 2x Hence, Z Z 2 A = [(−x + 4) − (x − 2x)]dx = (−2x2 + 2x + 4)dx = 1 ả2 16 = x3 + x2 + 4x ¯¯ = − + + − − + = 3 −1 √ R is the region bounded by the curves y = x2 and y = x Solution Sketch the region as shown in Figure 6.6 Find the points of intersection by solving the equations of the two curves simultaneously to get √ √ √ x2 = x x2 − x = x(x − 1) = and x=0 x = The corresponding points (0, 0) and (1, 1) are the points of intersection y y=x2 y= x -2 -1 x Figure 6.6 √ Notice that for ≤ x ≤ 1, the graph of y = x lies above that of y = x2 Hence, ả1 Z 3 ¯¯ 2 x − x ¯ = − = A= ( x − x )dx = 3 3 0 (a) R is the region to the right of the y axis that is bounded above by the curve y = − x2 and below the line y = (b) R is the region to the right of the y axis that lies below the line y = and is bounded by the curve y = − x2 , the line y = 3, and the coordinate axes INTEGRAL CALCULUS - EXERCISES 69 Solution Note that the curve y = − x2 and the line y = intersect to the right of the y axis at the point (1, 3), since x = is the positive solution of the equation − x2 = 3, i.e x2 = (a) Sketch the region as shown in Figure 6.7 y y=4-x2 y=3 -3 -2 -1 x Figure 6.7 Notice that for ≤ x ≤ 1, the graph of y = − x2 lies above that of y = Hence, ả1 Z Z 1 ¯¯ 2 A= (4−x −3)dx = (1−x )dx = x − x ¯ = 1− = 3 0 (b) Sketch the region as shown in Figure 6.8 y y=4-x2 y=3 -3 -2 -1 x Figure 6.8 Observe that to the left of x = 1, R is bounded above by the curve y = 3, while to the right of x = 1, it is bounded by the line y = − x2 This suggests that you break R into two subregions, R1 and R2 , as shown in Figure 6.8, and apply the integral formula INTEGRAL CALCULUS - EXERCISES 70 for area to each subregion separately In particular, Z A1 = 3dx = 3x|10 = and A2 = Z 2 (4 − x )dx = µ so ¶¯2 ¯¯ 4x − x ¯ = − − + = , 3 3 14 = 3 A = A1 + A2 = + R is the region bounded by the curve y = x12 and the lines y = x and y = x8 Solution First make a sketch of the region as shown in Figure 6.9 and find the points of intersection of the curve and the lines by solving the equations =x x2 x i.e x3 = = x and and x3 = to get x=1 and y y= x12 x = y=x y= -2 -1 x x Figure 6.9 Then break R into two subregions, R1 that extends from x = to x = and R2 that extends from x = to x = 2, as shown in Figure 6.9 Hence, the area of the region R1 is ¯1 Z 1³ Z x´ 7 ¯¯ A1 = x− dx = xdx = x¯ = 16 16 0 INTEGRAL CALCULUS - EXERCISES 71 and the area of the region R2 is ả ả2 Z 2µ 1 ¯¯ x 1 − dx = − − x ¯ = − − + + = A2 = x x 16 16 16 1 Thus, the area of the region R is the sum A = A1 + A2 = 12 = 16 R is the region bounded by the curves y = x3 − 2x2 + and y = x2 + 4x − Solution First make a rough sketch of the two curves as shown in Figure 6.10 You find the points of intersection by solving the equations of the two curves simultaneously x3 − 2x2 + = x2 + 4x − x3 − 3x2 − 4x + 12 = x2 (x − 3) − 4(x − 3) = (x − 3)(x − 2)(x + 2) = to get x = −2, and x=2 y 20 x = y=x3-2x2+5 10 y=x +4x-7 -5 -4 -2 x -10 Figure 6.10 The region whose area you wish to compute lies between x = −2 and x = 3, but since the two curves cross at x = 2, neither curve is always above the other between x = −2 and x = However, since the curve y = x3 − 2x2 + is above y = x2 + 4x − between x = −2 and x = 2, and since y = x2 + 4x − is above y = x3 − 2x2 + between x = and x = 3, it follows that the area of the region between x = −2 and x = 2, is INTEGRAL CALCULUS - EXERCISES A1 = = Z Z −2 −2 µ £¡ Ă 72  à ÂÔ x3 2x2 + − x2 + 4x − dx = ¢ x3 − 3x2 − 4x + 12 dx = ¶¯2 ¯ = x − x − 2x + 12x ¯¯ = −2 = − − + 24 − − + + 24 = 32 and the area of the region between x = and x = 3, is Z  à ÂÔ ÊĂ A2 = x + 4x − − x3 − 2x2 + dx = Z ¢ ¡ = −x + 3x2 + 4x 12 dx = ả3 ¯ = − x + x + 2x − 12x ¯¯ = 81 = − + 27 + 18 − 36 + − − + 24 = 81 = − + 21 = 4 Thus, the total area is the sum A = A1 + A2 = 32 + 131 = 4 Homework In problems through 20 find the area of the region R R is the triangle bounded by the line y = − 3x and the coordinate axes R is the rectangle with vertices (1, 0), (−2, 0), (−2, 5) and (1, 5) R is the trapezoid bounded by the lines y = x + and x = and the coordinate axes √ R is the region bounded by the curve y = x, the line x = 4, and the x axis INTEGRAL CALCULUS - EXERCISES 73 R is the region bounded by the curve y = 4x3 , the line x = 2, and the x axis R is the region bounded by the curve y = − x2 and the x axis R is the region bounded by the curve y = −x2 − 6x − and the x axis R is the region in the first quadrant bounded by the curve y = − x2 and the lines y = 3x and y = √ R is the region bounded by the curve y = x and the lines y = − x and y = 10 R is the region in the first quadrant that lies under the curve y = 16 x and that is bounded by this curve and the lines y = x, y = 0, and x = 11 R is the region bounded by the curve y = x2 −2x and the x axis (Hint: Reflect the region across the x axis and integrate the corresponding function.) 12 R is the region bounded by the curves y = x2 + and y = − x2 between x = −2 and x = 13 R is the region bounded by the curve y = ex and the lines y = and x = 14 R is the region bounded by the curve y = x2 and the line y = x 15 R is the region bounded by the curve y = x2 and the line y = 16 R is the region bounded by the curves y = x3 − 6x2 and y = −x2 17 R is the region bounded by the line y = x and the curve y = x3 18 R is the region in the first quadrant bounded by the curve y = x2 + and the lines y = 11 − 8x and y = 11 19 R is the region bounded by the curves y = x2 − 3x + and y = −x2 + 2x + 20 R is the region bounded by the curves y = x3 − x and y = −x2 + x Results 83 43 11 43 16 625 12 15 32 12 12 17 12 14 19 13 e − 18 40 14 19 16 6 √ 11 33 16 10 8(1 + ln 4) 15 32 20 37 12