advertisement

MATH 101 Quiz #1 (v.T2) Last Name: Thursday, January 14 First Name: Grade: Student-No: Section: Very short answer question Z 2 Z Z 2f (x) + 3g(x) dx. Please 0 0 0 2 g(x) dx = −4, calculate f (x) dx = 3 and 1. 1 mark If 2 write your answer in the box. Simplify your answer completely. Answer: −6 Marking scheme: 1 for a correct answer in the box Solution: The operation of integration is linear, so that: Z 2 Z 2 Z 2 [2f (x) + 3g(x)] dx = 2f (x) dx + 3g(x) dx 0 0 0 Z 2 Z 2 g(x) dx = (2 × 3) + (3 × (−4)) = −6 f (x) dx + 3 = 2 0 0 Short answer questions—you must show your work Z 3 f (x) dx, where 2. 2 marks Use elementary geometry to calculate 0 ( x, if x ≤ 2, f (x) = 2, if x > 2. Solution: If we sketch the function, we have a trapezoid:R a linear increase from x = 0 to 3 x = 2, followed by a constant. Using the interpretation of 0 f (x) dx as the area under the function f (x) between 0 and 3, we can break this trapezoid into: • R2 • R3 0 2 f (x) dx: a right-angled triangle of height 2 and base 2. Area = 2. f (x) dx: a rectangle of height 2 and base 1. Area = 2. Summing up: R3 0 f (x) dx = 4. Marking scheme: • 1 mark for splitting the integral into 2, or otherwise identifying simpler geometric structures, or using the formula for the area of a trapezoid. • 1 mark for calculating the correct answer. 3. 2 marks The value of the following limit is equal to the area below a graph of y = f (x), integrated over the interval [0, b]: 2 n X 4i 4 log 2 + . lim n→∞ n n i=1 Find f (x) and b. (Do not evaluate the limit.) Please write your answers in the boxes. Answer: b = 4 Answer: f (x) = log2 (2 + x) Solution: We identify this as the right Riemann sum, with interval ∆x = 4/n and xi = 4i/n. Therefore, b = 4 and f (x) = log2 (2 + x). Marking scheme: 1 mark for each answer Long answer question—you must show your work 4. 5 marks Consider the integral: Z 3 (7 + x3 ) dx. (∗) 0 (a) (1 mark) Approximate this integral using the left Riemann sum with n = 3 intervals. Solution: The left Riemann sum is defined as Ln = n−1 X f (xi )∆x. i=0 We subdivide into 3 intervals, (i.e. ∆x = 1), defining for i = 0, . . . , 2, the values xi = 0, 1, 2. The function f (x) = 7 + x3 has the values: f (xi ) = 7, 8, 15, from which we evaluate L3 = 30. Marking scheme: 1 mark for a correct answer (b) (4 marks) Write down the expression for the right Riemann sum with n intervals and calculate the sum. Now take the limit n → ∞ in your expression for the Riemann sum, to evaluate the integral (∗) exactly. You can leave your answer in “calculator-ready” form. Hint: you may use the identity n X i=1 i3 = n4 + 2n3 + n2 . 4 Solution: We divide into n intervals ∆x = 3/n and set xi = 3i/n. The right Riemann sum is therefore: n X (3i)3 3 7+ 3 . Rn = n n i=1 To calculate the sum: Rn = ! n 21 X 1 + n i=1 = 21 + n 81 X 3 i n4 i=1 ! 81 81 n4 + 2n3 + n2 = 21 + (1 + 2/n + 1/n2 ) × 4 n 4 4 To evaluate the limit exactly, we take n → ∞. The expressions involving 1/n vanish leaving: Z 3 81 1 = 41 . (7 + x3 ) dx = lim Rn = 21 + n→∞ 4 4 0 Marking scheme: • 1 mark for writing Rn . • 2 marks for calculating the sum using the formula in the hint, or otherwise. • 1 mark for taking the limit and getting the answer.