Calculus I Exam 2 Review Solutions 2002

Review: Exam 2 1. Goals for this portion of the course: • Be able to compute the area between curves, the volume of solids of revolution, and understand the mean value of a function. We had three basic volumes: Disk, Washer and Shell. • Stimulate mental connections between geometry and calculus- Given a picture, write down the formula for the given area or volume of a solid. Apply algebra if needed to obtain the required information (i.e., points of intersection). Also, be able to go from the integral of an area/volume to a geometric picture. • Be able to finish the integration to obtain a final solution. • Understand and apply u, du substitution to find antiderivatives (and evaluate the integral). 2. Material to be covered: 6.1, 6.2, 6.3, 6.5 plus review of u, du substitution. Review Questions: 1. Find the area of the region bounded between the curves: (a) y = ex − 1, y = x2 − x, x = 1 Top function: ex − 1, bottom function: x2 − x, so the area is 1 − (x2 − x) dx which is e − (11/6) Z 1 ex − 0 (b) x − 2y + 7 = 0, y 2 − 6y − x = 0 The rightmost function is the line, leftmost is the parabola, and integrate in y: Z 7 (2y − 7) − (y 2 − 6y) dy = 36 1 2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. You may use any method. √ (a) y = x − 1, y = 0, x = 3, about the x-axis. √ Use disks with radius x − 1: Z 3 √ π( x − 1)2 dx = 2π 1 (b) y = e−2x , y = 1 + x, x = 1, about the x-axis. One point of intersection between y = e−2x and y = 1 + x is at (0, 1) since that is the y−intercept of the line, and we know that e0 = 1. The lines y = 1 + x and x = 1 intersect at (1, 2), and e−2x intersects x = 1 at (1, e−2 ). The area is the small region between these points of intersection. 1 Using washers, the inside radius is e−2x and the outside radius is 1 + x. Therefore, the volume is: Z 1 Z 1   π (1 + x)2 − (e−2x )2 dx = π 1 + 2x + x2 − e−4x dx = 0 0  x3 1 x+x + + e−4x 3 4 1 2 =π  0 25 1 + 12 4e4  (c) y = x3 , y = 8, x = 0, about the y−axis. Use disks and integrate with respect to y: Z 8 96π (y 1/3 )2 dy = 5 0 3. If f is continuous, and Z 3 f (x) dx = 8, so that f takes on the value 4 at 1 least once on the interval [1, 3]. ***TYPO: “so that” should be “show that” From the Mean Value Theorem for integrals, there is a c between a and b so that: Z b 1 1 f (c) = f (x) dx = ·8=4 b−a a 3−1 4. Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. You may use any method. (a) y = x3 , y = x2 , about y = 1 Washers: Z 1   π (1 − x3 )2 − (1 − x2 )2 dx 0 (b) y = x3 , y = 8, x = 0 about x = 2. Shells: Z 2 2π(8 − x3 )(2 − x) dx 0 5. Each integral represents the volume of a solid. Describe the solid (with words and/or pictures). Z π/2 (a) 2πx cos(x) dx 0 Rotate the region under cos(x) (and above the x-axis) for 0 ≤ x ≤ π/2 about the y − axis 2 (b) Z π/2 2π cos2 (x) dx 0 √ Rotate the region under 2 cos(x) (above the x-axis) with 0 ≤ x ≤ π/2. Z 1  √  (c) π (2 − x2 )2 − (2 − x)2 dy 0 ***TYPO: The dy should be dx √ The solid is found by rotating the region between y = x2 and y = x (0 ≤ x ≤ 1) about the line x = 2. √ 6. Find the average value of f (x) = x2 1 + x2 on the interval [0, 2]. √ √ ***TYPO: 1 + x2 should be 1 + x3 favg = 1 2 Z 0 2 x2 p 1 + x3 dx 3 Use substitution with u = 1 + x , du = 3x2 dx Z 1 9√ 26 favg = u du = 6 1 9 7. Let R be the region in the first quadrant bounded by the curves y = x3 and y = 2x − x2 . Calculate the following quantities: (a) The area of R. 1 Z 2x − x2 − x3 dx = 0 5 12 (b) The volume obtained by revolving R about the x−axis. A cross section is a washer with inner radius x3 and outer radius 2x − x2 Z 1   41π π (2x − x2 )2 − x6 dx = 105 0 (c) The volume obtained by revolving R about the y−axis. Using shells, the radius is x and the height is 2x − x2 − x3 , so we get: Z 1 13π 2πx(2x − x2 − x3 ) dx = 30 0 8. Find the numbers b so that the average value of f (x) = 2 + 6x − 3x2 on the interval [0, b] is equal to 3. The requirement is that 1 b−0 Z b 2 + 6x − 3x2 dx = 3 0 3 Write out the left hand side of that equation to get: √ 3± 5 2 + 3b − b2 = 3 ⇒ b = 2 both of which are valid, since both are positive. 9. Let R1 be the region bounded by y = x2 , y = 0, and x = b, with b > 0. Let R2 be the region bounded by y = x2 , x = 0, and y = b2 (Same b). (a) Is there a b so that R1 and R2 have the same area? R b2 √ Rb The area of R1 is 0 x2 dx = 13 b3 . The area of R2 is 0 y dy = 23 b3 . Therefore, no value of b > 0 will make these equal. (b) Is there a value of b so that R1 sweeps out the same volume if R1 is rotated about the x−axis versus rotating R1 about the y−axis? Rb The first volume is found with disks: 0 π(x2 )2 dx = 15 πb5 . The second volume can be found with shells: Z b 1 2πx(x2 ) dx = πb4 2 0 So setting the volumes equal: 1 5 1 5 πb = πb4 ⇒ 2b = 5 ⇒ b = 5 2 2 (c) Is there a value of b so that R1 and R2 sweep out the same volume when rotated about the y−axis? The volume of R1 about the y−axis was computed already as 12 πb4 . The volume of R2 about the y−axis is given by (disks): Z 0 b2 1 √ π( y)2 dy = πb4 2 Therefore, the two volumes are always equal (for all values of b)! 10. Evaluate: Z √ (a) x x − 1 dx Let u = x − 1, so du = dx. Upon substitution, we see an extra x, so we can substitute u + 1 = x, and obtain: Z Z √ 2 2 (u + 1) u du = u3/2 + u1/2 du = u5/2 + u3/2 + C = 5 3 2 2 (x − 1)5/2 + (x − 1)3/2 + C 5 3 4 (b) Z (c) Z x+1 dx Let u = x2 + 2x, and get: x2 + 2x Z 1 1 1 1 du = ln |u| + C = ln |x2 + 2x| + C 2 u 2 2 x dx Let u = x + 2 (Note that x = u − 2): x+2 Z Z 8 4 u−2 √ du = u3/4 − 2u−1/4 du = u7/4 − u3/4 + C = 4 7 3 u √ 4 4 8 (x + 2)7/4 − (x + 2)3/4 + C 7 3 (d) x2 dx Let u = 1 − x. (Note that x = 1 − u so that x2 = 1−x (1 − u)2 ), and: Z Z (1 − u)2 √ du = − u−1/2 − 2u1/2 + u3/2 du = − u Z √ 4 2 4 2 −2u1/2 + u3/2 + u5/2 +C = −2(1−x)1/2 + (1−x)3/2 + (1−x)5/2 +C 3 5 3 5 5