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MATH 101 Quiz #2 (v.M2) Last Name: Friday, January 29 First Name: Grade: Student-No: Section: Very short answer question Z 3 cos x − x) dx. Simplify as far as possible. 1. 1 mark Evaluate 0 Answer: sin 3 − 9 2 Solution: By FTC2, Z 0 3 cos x − x) dx = 3 x2 9 9 sin x − − sin 0 − 0 = sin 3 − . = sin 3 − 2 0 2 2 Marking scheme: 1 for a correct answer in the box Short answer questions—you must show your work Z 2. 2 marks Evaluate (9 − x2 )x √ dx. x2 − 9 Solution: We use the substitution u = x2 − 9, for which du = 2x dx: Z Z 9 − x2 1 x2 − 9 √ x dx = − √ (2x) dx 2 x2 − 9 x2 − 9 Z 1 u √ du =− 2 u Z √ 1 u du =− 2 1 u3/2 =− +C 2 3/2 1 = − (x2 − 9)3/2 + C 3 Marking scheme: • 1 mark for substituting u = x2 − 9 • 1 mark for the correct answer (including restoring the x–dependence). 3. 2 marks Find the area to the right of the y–axis and to the left of the curve x = y − 2y 2 . A “calculator-ready” answer is acceptable. Solution: A point (x, y) on the curve x = y − 2y 2 = y(1 − 2y) has x = 0 for y = 0, 1/2, has x > 0 for 0 < y < 1/2 (then both factors y and 1 − 2y are positive) and has x < 0 for y < 0 and y > 1/2 (then one of the factors y and 1 − 2y is positive and the other is negative). This leads to the figure below. So, using horizontal slices, y (0, 1/2) x = y − 2y 2 1/2 Z y 2 2y 3 (y − 2y ) dy = − 2 3 2 area = 0 1/2 1 1 1 − = 8 12 24 = 0 (0, 0) x Marking scheme: • 1 mark for any correct integral. If they use vertical slices, the resulting integral is √ √ Z 1/8 Z 1/8 √ 1 + 1 − 8x 1 − 1 − 8x 1 − 8x − dx dx = 4 4 2 0 0 • 1 mark for the correct answer. Long answer question—you must show your work Z x2 4. 5 marks Consider the function F (x) = −t e Z 0 0 2 e−t dt. dt + −x 0 (a) Find F (x). (b) Find the value of x for which F (x) takes its minimum value. Solution: (a) Write 2 F (x) = G(x ) − H(−x) Z with G(y) = y e −t Z dt, H(y) = 0 y 2 e−t dt 0 By the Fundamental Theorem of Calculus, G0 (y) = e−y H 0 (y) = e−y 2 Hence, by the chain rule, 2 2 F 0 (x) = 2xG0 (x2 ) − (−1)H 0 (−x) = 2xe−(x ) + e−(−x) = (2x + 1)e−x 2 (b) Observe that F 0 (x) < 0 for x < −1/2 and F 0 (x) > 0 for x > −1/2. Hence F (x) is decreasing for x < −1/2 and increasing for x > −1/2, and F (x) must take its minimum value when x = −1/2 Marking scheme: • 4 marks for part (a), – with 2 marks for 2xG0 (x2 ) (including 1 mark for use of the chain rule), R0 R −x 2 2 – 1 mark for −x e−t dt = − 0 e−t dt and – 1 mark for −H 0 (−x) 2 2 If the student writes down F 0 (x) = 2xe−x + e−x without any justification, they still get 4 marks. • 1 mark for part (b), even if they just solve F 0 (x) = 0, without justifying why it is a minimum.