Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
MAT 266 Test 1 SOLUTIONS, FORM A 1. [20 points] Perform the appropriate trigonometric substitution on the following integral. Do NOT try to find the resulting integral. Z √ 2 x − 9 + 5x √ dx x e − x2 − 9 p √ Solution: The presence ofpthe x2 − 9 formula suggests that you let x = 9 · sec θ = 3 sec θ. Then dx = 3 sec θ tan θ dθ and x2 − 9 = 3 tan θ. Substituting these into the formula produces: Z 3 tan θ + 5 · 3 sec θ · 3 sec θ tan θ dθ e3 sec θ − 3 tan θ Grading (roughly): +9 points for the substitutions for x, dx, and tion. Grading for common mistakes: −5 points for not replacing dx. √ ·; +11 points for substitu- 2. [15 points] Find the FORM of the Partial Fraction Decomposition of the rational function below. You do not need to determine the coefficients. (x + 4)(x − 2)2 (x2 + 5x + 8) x3 (x + 3)(x2 − 5x + 8)2 B C D Ex + F Gx + H A + 2+ 3+ + 2 + 2 . Note that x2 − 5x + 8 is a quadx x x x + 3 x − 5x + 8 (x − 5x + 8)2 ratic polynomial with no real roots. Grading: Done on a 0−3−5−7−10−12−15-point basis (these were the grades usually given); generally, you went from 15 to 12 if you made one mistake, 12 to 10 if you made another, etc. Solution: 2 MAT 266 Test 1 SOLUTIONS, FORM A 3. Find the following integrals. Z a. [15 points] x2 e5x dx Solution: This is done using Integration by Parts twice: Z 2 5x x e e5x dx = x · − 5 2 = = x2 e5x 2 − 5 5 Z e5x x2 e5x 2 2x · dx = − 5 5 5 x2 e5x 2 − 5 5 x· e5x − 5 Z xe5x e5x − 5 25 1· xe e5x dx 5 # v 0 = e5x e5x u0 = 2x v = 5 # " u = x v 0 = e5x e5x u0 = 1 v = 5 " Z 5x dx u = x2 +C Grading for common mistakes: +5 points (total) for starting with u = e5x and v 0 = x2 ; +7 points (total) for only one integration-by-parts. Z 1 b. [15 points] dx x2 − 4x Solution: This is a rational function, so do it with partial fractions: 1 1 A B = = + x2 − 4x x(x − 4) x x−4 1 = A(x − 4) + Bx 1 = −4 · A 1=4·B ⇒ ⇒ A=− B= 1 4 1 4 (x = 0) (x = 4) Then Z 1 dx = x2 − 4x Z 1 1 1 1 1 1 − · + · dx = − ln x + ln(x − 4) + C. 4 x 4 x−4 4 4 Grading for common mistakes: +5 points (total) for ln(x2 −4x); −3 points for factoring x2 −4x incorrectly. Z c. [15 points] sin5 x cos2 x dx Solution: This integral has trig functions, so it is a trig integral. This one can be done by setting up the substitution u = cos x (since sin x has an odd power, and cos x has an even power); the rest of the substitution is du = − sin x dx, and sin2 x = 1 − cos2 x. Z Z Z sin5 x cos2 x dx = sin4 x cos2 x · sin x dx = (1 − cos2 x)2 cos2 x · sin x dx Z Z 2 2 2 = (1 − u ) · u · (−1) du = −u2 + 2u4 − u6 du = −u3 2u5 u7 − cos3 x 2 cos5 x cos7 x + − +C = + − + C. 3 5 7 3 5 7 Grading for common mistakes: +7 points (total) for using power-reducing formulas. 3 MAT 266 Test 1 SOLUTIONS, FORM A Z 5 ln(2 + sin x) dx cannot be determined. Approxi- 4. [20 points] The exact value of the definite integral 2 mate this integral using the Trapezoid Rule and Simpson’s Rule for n = 4. (You will only need to write out the formulas; make sure you identify which is which.) Solution: Before providing these formulas, a few numbers need to be calculated: ∆x = b−a 5−2 = = 0.75, n 4 and x0 , x1 , x2 , x3 , x4 is the sequence 2, 2.75, 3.5, 4.25, 5. The Trapezoid Rule states that Z 5 ln(2 + sin x) dx ≈ 2 = ∆x [f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + f (x4 )] 2 0.75 · [ln(2 + sin 2) + 2 ln(2 + sin 2.75) + 2 ln(2 + sin 3.5) + 2 ln(2 + sin 4.25) + ln(2 + sin 5)] 2 ≈ 1.516527504. Simpson’s Rule states that Z 5 ln(2 + sin x) dx ≈ 2 = ∆x [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 )] 3 0.75 · [ln(2 + sin 2) + 4 ln(2 + sin 2.75) + 2 ln(2 + sin 3.5) + 4 ln(2 + sin 4.25) + ln(2 + sin 5)] 3 ≈ 1.496386232. Grading: +2 points for ∆x, +4 points for x0 , x1 , x2 , x3 , x4 , +5 points for each basic formula, +2 points for each substitution. Grading for common mistakes: −1 points for a calculation done in degree mode; −3 points for keeping f in the final answer. 4 MAT 266 Test 1 SOLUTIONS, FORM B 1. [20 points] Perform the appropriate trigonometric substitution on the following integral. Do NOT try to find the resulting integral. p Z sin 4 − x2 + 5 √ dx 1 − x 4 − x2 p √ 2 Solution: The presence p of the 4 − x formula suggests that you let x = 4 · sin θ = 2 sin θ. Then dx = 2 cos θ dθ and 4 − x2 = 2 cos θ. Substituting these into the formula produces: Z sin (2 cos θ) + 5 · 2 cos θ dθ 1 − 2 sin θ · 2 cos θ Grading (roughly): +9 points for the substitutions for x, dx, and tion. Grading for common mistakes: −5 points for not replacing dx. √ ·; +11 points for substitu- 2. [15 points] Find the FORM of the Partial Fraction Decomposition of the rational function below. You do not need to determine the coefficients. (x + 5)2 x(x2 − 3x + 9)2 (x + 1)3 (x − 5)(x2 + 3x + 9)2 Solution: A B Ex + F Gx + H C D + + + + + . (Note that x + 1 (x + 1)2 (x + 1)3 x − 5 x2 + 3x + 9 (x2 + 3x + 9)2 x2 + 3x + 9 is a quadratic polynomial with no real roots.) Grading: Done on a 0−3−5−7−10−12−15-point basis (these were the grades usually given); generally, you went from 15 to 12 if you made one mistake, 12 to 10 if you made another, etc. 2 MAT 266 Test 1 SOLUTIONS, FORM B 3. Find the following integrals. Z 1 a. [15 points] dx x2 + 3x Solution: This is a rational function, so do it with partial fractions: x2 1 1 A B = = + + 3x x(x + 3) x x+3 1 = A(x + 3) + Bx 1=3·A ⇒ 1 = −3 · B A= ⇒ 1 3 (x = 0) B=− 1 3 (x = −3) Then Z 1 dx = 2 x + 3x Z 1 1 1 1 1 1 · − · dx = ln x − ln(x + 3) + C. 3 x 3 x+3 3 3 Grading for common mistakes: +5 points (total) for ln(x2 + 3x). Z b. [15 points] tan3 x sec80 x dx Solution: This integral has trig functions, so it is a trig integral. This one can be done either by setting up the substitution u = tan x or u = sec x; after the substitution u = sec x, the new integral is easier to do. Note that, since u = sec x, du = sec x tan x dx, and tan2 x = sec2 x − 1. Z Z Z tan3 x sec80 x dx = tan2 x sec79 x · sec x tan x dx = (sec2 x − 1) sec79 x · sec x tan x dx Z = Z c. [15 points] (u2 − 1)u79 du = Z u81 − u79 du = u82 u80 sec82 x sec80 x − +C = − +C 82 80 82 80 x2 e−4x dx Solution: This is done using Integration by Parts twice: Z 2 −4x x e e−4x − dx = x · −4 2 = = e−4x −x2 e−4x 1 2x · dx = + −4 4 2 Z −x2 e−4x 1 + 4 2 1 −x2 e−4x + 4 2 x· e−4x − −4 Z 1· −xe−4x e−4x − 4 16 e−4x dx −4 # v 0 = e−4x −4x e u0 = 2x v = −4 # " u = x v 0 = e−4x e−4x u0 = 1 v = −4 " Z xe −4x dx u = x2 +C Grading for common mistakes: +5 points (total) for letting u = e−4x and v 0 = x2 . 3 MAT 266 Test 1 SOLUTIONS, FORM B Z 5 tan(5 + ln x) dx cannot be determined. Approxi- 4. [20 points] The exact value of the definite integral 3 mate this integral using the Trapezoid Rule and Simpson’s Rule for n = 4. (You will only need to write out the formulas; make sure you identify which is which.) Solution: Before providing these formulas, a few numbers need to be calculated: ∆x = b−a 5−3 = = 0.5, n 4 and x0 , x1 , x2 , x3 , x4 is the sequence 3, 3.5, 4, 4.5, 5. The Trapezoid Rule states that Z 5 tan(5 + ln x) dx ≈ 3 = ∆x [f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + f (x4 )] 2 0.5 · [tan(5 + ln 3) + 2 tan(5 + ln 3.5) + 2 tan(5 + ln 4) + 2 tan(5 + ln 4.5) + tan(5 + ln 5)] 2 ≈ 0.1867116031. Simpson’s Rule states that Z 5 tan(5 + ln x) dx ≈ 3 = ∆x [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 )] 3 0.5 · [tan(5 + ln 3) + 4 tan(5 + ln 3.5) + 2 tan(5 + ln 4) + 4 tan(5 + ln 4.5) + tan(5 + ln 5)] 3 ≈ 0.1892353202. Grading: +2 points for ∆x, +4 points for x0 , x1 , x2 , x3 , x4 , +5 points for each basic formula, +2 points for each substitution. Grading for common mistakes: −1 points for a calculation done in degree mode; −3 points for keeping f in the final answer. 4