advertisement

MATH 172.504 Examination 1 Solutions February 16, 2012 Z 1. 2 6x Evaluate p 2x2 + 1 dx. 0 (A) 26 (B) 27 (C) 28 (D) 29 (E) 30 Solution: A We make the substitution u = 2x2 + 1, du = 4x dx, and find (remembering the substitute the limits as well) Z 2 Z p 2 6x 2x + 1 dx = 0 2. 1 9 6√ 3 h 2 3/2 i9 u du = u = 27 − 1 = 26. 4 2 3 1 Consider the region in the first quadrant bounded by y = x2 , y = 9, and x = 0. Which integral below gives the volume of the solid obtained by rotating this region about the x-axis? Z 9 (A) π (9 − x2 ) dx 0 Z 3 x(9 − x2 ) dx (B) 2π 0 Z (C) 2π Z (x4 − 9) dx 0 3 (D) π Z 9 (81 − x4 ) dx 0 3 (E) π y 2 dy 0 Solution: D We use washers, noting that the horizontal line y = 9 sits above the parabola y = x2 on the interval 0 ≤ x ≤ 3. So the outer radius at a point x is 9, and the inner radius is x2 . 3. Our ideal spring follows Hooke’s Law. It requires 5 Newtons of force to be held 10 cm from rest. How much work (in Joules) is done in stretching the spring from rest to 20 cm? (A) 0 (B) 1 (C) 10 (D) 100 (E) 1000 1 Solution: B We first need to calculate the spring constant. We are given that the force at 0.1 meters (= 10 cm) is F (0.1) = 5. Hooke’s Law asserts that F (x) = kx for a constant k, so k(0.1) = 5, which implies that k = 50. Therefore to calculate the work we evaluate Z 0.2 h i0.2 = 25(0.2)2 = 1. 50x dx = 25x2 W = 0 0 Z 4. Let F (x) = xp t3 + 1 dt. Which of the following statements about F (x) are true? 1 I. F (1) = 0. II. F 0 (2) = 2. III. F 0 (2) = 3. (A) I only (B) II only (C) III only (D) I and II only (E) I and III only Solution: E Since F (1) is equal to a definite integral over an interval of length 0, its value must be zero. Therefore, I is true . Now the Fundamental Theorem of Calculus says that √ √ F 0 (x) = x3 + 1, and so F 0 (2) = 9 = 3. Thus III is true but II is false . 5. What is the average value of f (x) = cos(3x) on the interval 0 ≤ x ≤ π2 ? 2 (A) − 3π 3 (B) − 2π 1 (C) 3π 1 (D) 2 (E) 0 Solution: A We need to calculate the integral π/2 Z 2 π/2 2 1 2 3π 2 cos(3x) dx = sin(3x) = sin =− . π 0 π 3 3π 2 3π 0 6. Let R be the region in the first quadrant of the xy-plane bounded above by the union of √ the curve y = 2 x and the line y = −x + 8 and bounded below by the x-axis. Find the area of R by setting up the integral with respect to y. √ Solution: When we consider the region in question, we have the parabola y = 2 x to the left of the line y = −x + 8. The two curves intersect in the first quadrant at (4, 4). Now converting to y, the parabola is x = y 2 /4 and the line is x = 8 − y, and so we find the area by integrating 4 Z 4 y2 y3 16 56 y2 A= (8 − y) − dy = 8y − − = 32 − 8 − = . 4 2 12 0 3 3 0 2 7. Evaluate the following integrals. Z π/3 4 cos2 θ sin θ dθ (a) 0 Z √ (b) x2 x + 1 dx e2x dx e2x + 1 Solution: (a) We substitute u = cos θ, du = − sin θ dθ, giving Z (c) Z π/3 1/2 Z 2 4 cos θ sin θ dθ = − 0 1 4 3 4u du = u 3 2 1 1/2 4 4 = − 3 3 1 7 = . 8 6 (b) We substitute u = x + 1, du = dx, x = u − 1, and find Z Z Z √ √ 2√ 2 x x + 1 dx = (u − 1) u du = (u2 − 2u + 1) u du Z = (u5/2 − 2u3/2 + u1/2 ) du 2 4 2 = u7/2 − u5/2 + u2/3 + C 7 5 3 2 4 2 = (x + 1)7/2 − (x + 1)5/2 + (x + 1)3/2 + C . 7 5 3 (c) We substitute u = e2x + 1, du = 2e2x dx, and find Z Z 1 1 1 e2x dx = du = ln |u| + C = 2x e +1 2 u 2 8. 1 2 ln(e2x + 1) + C . Let R be the region in the first quadrant of the xy-plane bounded by the curves πx y = cos , y = 2 − 2x2 , and the y-axis. 2 Set up integrals with respect to x (do not evaluate!) that express the volumes of the solids obtained by rotating R about the following axes of rotation. (It may help to start by sketching the region R.) (a) The x-axis. (b) The y-axis. (c) The line y = −1. Solution: In the first quadrant the parabola sits above the cosine curve over the interval [0, 1]. (a) Here we use washers: Z V =π 1h (2 − 2x2 )2 − cos2 0 3 πx i 2 dx (b) Here we use cylindrical shells: Z V = 2π 0 1 h πx i x (2 − 2x2 ) − cos dx 2 (c) Here we use washers again: Z V =π 0 9. 1 πx 2 (2 − 2x2 + 1)2 − cos dx +1 2 [8 points] Let S be the 3-dimensional solid whose base is the region in the first quadrant of the xy-plane bounded by y = cos x, the x-axis, and the y-axis, and whose cross-sections perpendicular to the x-axis are squares. Write down an integral for the volume of S. (You do not need to evaluate the integral.) Solution: The region sits below y = cos x over the interval 0 ≤ x ≤ π2 . At a point x in the interval, the cross-section of S is then a square of side-length cos x. Therefore, Z V = 0 4 π/2 cos2 x dx .