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SOLUTION OF HW4 MINGFENG ZHAO October 09, 2011 Z 2 |(x − 1)(3x − 1)| dx. 1. [Page 83, Problem 16] Compute 0 Proof. First, we look at (x − 1)(3x − 1) ≥ 0, then we can get x ≥ 1, or x ≤ 1 . 3 Hence, we can get 2 Z |(x − 1)(3x − 1)| dx 0 1 3 Z 1 3 1 3 Z [3x2 − 4x + 1] dx − Z 2 1 1 [3x2 − 4x + 1] dx + 1 3 0 |(x − 1)(3x − 1)| dx (x − 1)(3x − 1) dx 1 3 1 3 Z Z (x − 1)(3x − 1) dx + 0 2 1 1 Z (x − 1)(3x − 1) dx − = = Z |(x − 1)(3x − 1)| dx + 0 = 1 Z |(x − 1)(3x − 1)| dx + = Z 2 [3x2 − 4x + 1] dx 1 1 1 2 [x3 − 2x2 + x]03 − [x3 − 2x2 + x] 1 + [x3 − 2x2 + x]1 3 = 1 2 1 1 2 1 − + − 0 − [1 − 2 + 1] + − + + [8 − 8 + 2] − [1 − 2 + 1] 27 9 3 27 9 3 = 2 4 2 − + +2 27 9 3 = 2 − 12 + 18 + 54 27 = 62 27 = 2 8 . 27 1 2 MINGFENG ZHAO Z −4 2. [Page 83, Problem 20] Compute (x + 4)10 dx. −2 Proof. Z = −4 1 · [x + 4]10+1 1 + 10 −2 = 1 · [−4 + 4]11 − [−2 + 4]11 11 −4 (x + 4)10 dx −2 = − 211 11 = − 2048 11 = −186 2 . 11 3. [Page 83, Problem 21] Find all values of c for which Z c x(1 − x) dx = 0. a. Z0 c |x(1 − x)| dx = 0. b. 0 Proof. a. Since we have Z 0 c x(1 − x) dx = 0 Z = c [x − x2 ] dx 0 c x3 x2 − 2 3 0 2 c c3 = − −0 2 3 = = c2 c3 − 2 3 = 3c2 − 2c3 6 = c2 [3 − 2c] . 6 SOLUTION OF HW4 3 So we get c2 [3 − 2c] = 0, therefore, we have c = 0, 3 . 2 b. Firstly, we look at x(1 − x) ≥ 0, then we get 0 ≤ x ≤ 1. So if c ≤ 0, then c Z 0 |x(1 − x)| dx = 0 c Z = − x(1 − x) dx 0 c Z = [x − x2 ] dx − 0 c x3 x2 − 2 3 0 2 c c3 − − −0 2 3 = = − = − c2 c3 + 2 3 = − 3c2 − 2c3 6 = − c2 [3 − 2c] . 6 Then we get c2 [3 − 2c] = 0, then c = 0 or c = 32 , but we know that c ≤ 0, then in this case, we have c = 0. If 0 < c ≤ 1, then we have Z 0 c |x(1 − x)| dx = 0 Z c x(1 − x) dx = 0 Z = 0 c [x − x2 ] dx 4 MINGFENG ZHAO c x2 x3 − 2 3 0 2 c c3 − −0 2 3 = = = c3 c2 + 2 3 = 3c2 − 2c3 6 = c2 [3 − 2c] . 6 Then we get c2 [3 − 2c] = 0, then c = 0 or c = 23 , but we know that 0 < c ≤ 1, then in this case, we do not have solution. If c > 1, then we know that Z c Z |x(1 − x)| dx 1 ≥ 0 |x(1 − x)| dx 0 Z 1 x(1 − x) dx = 0 Z = 1 [x − x2 ] dx 0 1 x2 x3 − = 2 3 0 1 1 = − −0 2 3 = 1 6 > 0. So in this case, we do not have solution. Z c In a summary, if |x(1 − x)| dx = 0, then 0 c = 0. SOLUTION OF HW4 5 Z 4. [Page 83, Problem 23] Find a quadratic polynomial P for which P (0) = P (1) = 0, and 1 P (x) dx = 0 1. Proof. Since P is a quadratic polynomial, then we can assume that there exists some a, b, c ∈ R such that Q(x) = ax2 + bx + c. Since 0 = P (0) = c, then c = 0. And by 0 = P (1) = a + b + c = a + b, then b = −a. On the other hand, 1 Z 1 = P (x) dx 0 1 Z [ax2 + bx + c] dx = 0 1 Z [ax2 − ax] dx = 0 = = = 1 ax2 ax3 − 3 2 0 ha 3 − ai −0 2 2a − 3a 6 a = − . 6 That is, we get 1 = − a6 , so a = −6, which implies that b = −a = 6, and P (x) = −6x2 + 6x. Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009 E-mail address: [email protected]