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MATH 101 Quiz #2 (v.M1) Last Name: Friday, January 29 First Name: Grade: Student-No: Section: Very short answer question Z 2 x3 + sin x) dx. Simplify as far as possible. 1. 1 mark Evaluate 0 Answer: 5 − cos 2 Solution: By FTC2, Z 2 3 x + sin x) dx = 0 2 24 x4 − cos x = − cos 2 + cos 0 = 4 − cos 2 + 1 = 5 − cos 2. 4 4 0 Marking scheme: 1 for a correct answer in the box Short answer questions—you must show your work Z 2. 2 marks Evaluate (x2 − 4)x √ dx. 4 − x2 Solution: We use the substitution u = 4 − x2 , for which du = −2x dx: Z Z x2 − 4 1 4 − x2 √ √ x dx = (−2x) dx 2 4 − x2 4 − x2 Z 1 u √ du = 2 u Z 1 √ = u du 2 1 u3/2 = +C 2 3/2 = 1 (4 − x2 )3/2 + C 3 Marking scheme: • 1 mark for substituting u = 4 − x2 • 1 mark for the correct answer (including restoring the x–dependence). 3. 2 marks Find the area to the left of the y–axis and to the right of the curve x = y 2 + y. A “calculator-ready” answer is acceptable. Solution: A point (x, y) on the curve x = y 2 + y = y(y + 1) has x = 0 for y = −1, 0, has x < 0 for −1 < y < 0 (the factors y and y + 1 have opposite signs) and has x > 0 for y < −1 and y > 0 (the factors y and y + 1 are either both positive or both negative). This y leads to the figure below. So, using horizontal slices, (0, 0) x Z 0 3 0 y 1 y2 1 1 0 − (y + y) dy = − + =− + = 3 2 −1 3 2 6 2 area = −1 (0, −1) x = y + y2 Marking scheme: • 1 mark for any correct integral. If they use vertical slices, the resulting integral is √ √ Z 0 Z 0 √ −1 + 1 + 4x −1 − 1 + 4x − 1 + 4x dx dx = 2 2 −1/4 −1/4 • 1 mark for the correct answer. Long answer question—you must show your work Z x2 4. 5 marks Consider the function F (x) = 0 Z t2 0 4 et dt. e 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 Z t2 e dt, H(x) = 0 x 4 et dt 0 By the Fundamental Theorem of Calculus, G0 (y) = ey 2 H 0 (x) = ex 4 Hence, by the chain rule, 4 4 F 0 (x) = 2xG0 (x2 ) − H 0 (x) = 2xex − ex = (2x − 1)ex 4 (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 4 Rx 4 – 1 mark for x et dt = − 0 et dt and – 1 mark for H 0 (x) 4 4 If the student writes down F 0 (x) = 2xex − ex 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.