Solved: Area of a segment of a circle Use two approaches

What change of variables is suggested by an integral containing 2x2 - 9?

What change of variables is suggested by an integral containing 2x2 + 36?

What change of variables is suggested by an integral containing 2100 - x2?

If x = 4 tan u, express sin u in terms of x.

If x = 2 sin u, express cot u in terms of x.

If x = 8 sec u, express tan u in terms of x.

716. Sine substitution Evaluate the following integrals. L 5>2 0 dx 225 - x2

716. Sine substitution Evaluate the following integrals. L 3>2 0 dx 19 - x223>2

716. Sine substitution Evaluate the following integrals. L 10 5 2100 - x2 dx

716. Sine substitution Evaluate the following integrals. L 12 0 x2 24 - x2 dx

716. Sine substitution Evaluate the following integrals. L 1>2 0 x2 21 - x2 dx

716. Sine substitution Evaluate the following integrals. L 1 1>2 21 - x2 x2 dx

716. Sine substitution Evaluate the following integrals. L dx 116 - x221>2

716. Sine substitution Evaluate the following integrals. L 236 - t2 dt

716. Sine substitution Evaluate the following integrals. L 29 - x2 x dx

716. Sine substitution Evaluate the following integrals. L 136 - 9x22-3>2 dx

1746. Trigonometric substitutions Evaluate the following integrals. L 264 - x2 dx

1746. Trigonometric substitutions Evaluate the following integrals. L dx 2x2 - 49 , x 7 7

1746. Trigonometric substitutions Evaluate the following integrals. L dx 11 - x223>2

1746. Trigonometric substitutions Evaluate the following integrals. L dx 11 + x223>2

1746. Trigonometric substitutions Evaluate the following integrals. L dx x22x2 + 9

1746. Trigonometric substitutions Evaluate the following integrals. L dt t229 - t2

1746. Trigonometric substitutions Evaluate the following integrals. L dx 236 - x2

1746. Trigonometric substitutions Evaluate the following integrals. L dx 216 + 4x2

1746. Trigonometric substitutions Evaluate the following integrals. L dx 2x2 - 81 , x 7 9

1746. Trigonometric substitutions Evaluate the following integrals. L dx 21 - 2x2

1746. Trigonometric substitutions Evaluate the following integrals. L dx 11 + 4x223>2

1746. Trigonometric substitutions Evaluate the following integrals. L dx 1x2 - 3623>2, x 7 6

1746. Trigonometric substitutions Evaluate the following integrals. L x2 216 - x2 dx

1746. Trigonometric substitutions Evaluate the following integrals. L dx 181 + x222

1746. Trigonometric substitutions Evaluate the following integrals. L 2x2 - 9 x dx, x 7 3

1746. Trigonometric substitutions Evaluate the following integrals. L 29 - 4x2 dx

1746. Trigonometric substitutions Evaluate the following integrals. L x2 24 + x2 dx

1746. Trigonometric substitutions Evaluate the following integrals. L 24x2 - 1 x2 dx, x 7 1 2

1746. Trigonometric substitutions Evaluate the following integrals. L dx 23 - 2x - x2

1746. Trigonometric substitutions Evaluate the following integrals. L y4 1 + y2 dy

1746. Trigonometric substitutions Evaluate the following integrals. L 29x2 - 25 x3 dx, x 7 5 3

1746. Trigonometric substitutions Evaluate the following integrals. L 29 - x2 x2 dx

1746. Trigonometric substitutions Evaluate the following integrals. L x2 125 + x222 dx

1746. Trigonometric substitutions Evaluate the following integrals. L dx x229x2 - 1 , x 7 1 3

1746. Trigonometric substitutions Evaluate the following integrals. L x2 1100 - x223>2 dx

1746. Trigonometric substitutions Evaluate the following integrals. L dx x32x2 - 100 , x 7 10

1746. Trigonometric substitutions Evaluate the following integrals. L x3 181 - x222 dx

1746. Trigonometric substitutions Evaluate the following integrals. L dx x32x2 - 1 , x 7

1746. Trigonometric substitutions Evaluate the following integrals. L dx x1x2 - 123>2, x 7 1

1746. Trigonometric substitutions Evaluate the following integrals. L x3 1x2 - 1623>2 dx, x 6 -4

4756. Evaluating definite integrals Evaluate the following definite integrals. L 1 0 dx 2x2 + 16

4756. Evaluating definite integrals Evaluate the following definite integrals. L 16 812 dx 2x2 - 64

4756. Evaluating definite integrals Evaluate the following definite integrals. L 1 1>13 dx x221 + x2

4756. Evaluating definite integrals Evaluate the following definite integrals. L 22 1 dx x224 - x2

4756. Evaluating definite integrals Evaluate the following definite integrals. L 1>23 0 2x2 + 1 dx

4756. Evaluating definite integrals Evaluate the following definite integrals. L 2 12 2x2 - 1 x dx

4756. Evaluating definite integrals Evaluate the following definite integrals. L 1>3 0 dx 19x2 + 123>2

4756. Evaluating definite integrals Evaluate the following definite integrals. L 10 10>13 dy 2y2 - 25

4756. Evaluating definite integrals Evaluate the following definite integrals. L 4 4>13 dx x21x2 - 42

4756. Evaluating definite integrals Evaluate the following definite integrals. L 613 6 z2 1z2 + 3622 dz

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If x = 4 tan u, then csc u = 4>x. b. The integral 1 2 1 21 - x2 dx does not have a finite real value. c. The integral 1 2 1 2x2 - 1 dx does not have a finite real value. d. The integral L dx x2 + 4x + 9 cannot be evaluated using a trigonometric substitution.

5865. Completing the square Evaluate the following integrals. L dx x2 - 6x + 34

5865. Completing the square Evaluate the following integrals. L dx x2 + 6x + 18

5865. Completing the square Evaluate the following integrals. L du 2u2 - 12u + 36

5865. Completing the square Evaluate the following integrals. L x2 - 2x + 1 2x2 - 2x + 10 dx

5865. Completing the square Evaluate the following integrals. L x2 + 2x + 4 2x2 - 4x dx, x 7 4

5865. Completing the square Evaluate the following integrals. L x2 - 8x + 16 19 + 8x - x223>2 dx

5865. Completing the square Evaluate the following integrals. L 4 1 dt t2 - 2t + 10

5865. Completing the square Evaluate the following integrals. L 112 + 32>12122 1>2 dx 8x2 - 8x + 11

Area of an ellipse The upper half of the ellipse centered at the origin with axes of length 2a and 2b is described by y = b a 2a2 - x2 (see figure). Find the area of the ellipse in terms of a and b. 2a a x y 2b b y 5 oea2 2 x2 b a

Area of a segment of a circle Use two approaches to show that the area of a cap (or segment) of a circle of radius r subtended by an angle u (see figure) is given by Aseg = 1 2 r21u - sin u2. a. Find the area using geometry (no calculus). b. Find the area using calculus. _ r cap or segment

Area of a lune A lune is a crescent-shaped region bounded by the arcs of two circles. Let C1 be a circle of radius 4 centered at the origin. Let C2 be a circle of radius 3 centered at the point 12, 02. Find the area of the lune (shaded in the figure) that lies inside C1 and outside C2. (2, 0) 4 3 C2 C1

Area and volume Consider the function f 1x2 = 19 + x22-1>2 and the region R on the interval 30, 44 (see figure). a. Find the area of R. b. Find the volume of the solid generated when R is revolved about the x-axis. c. Find the volume of the solid generated when R is revolved about the y-axis. 0 R 1 2 3 4 0.4 0.3 0.2 0.1 y x y 5 (9 1 x2)21/2

Area of a region Graph the function f 1x2 = 116 + x22-3>2 and find the area of the region bounded by the curve and the x-axis on the interval 30, 34.

Arc length of a parabola Find the length of the curve y = ax2 from x = 0 to x = 10, where a 7 0 is a real number.

Computing areas On the interval 30, 24, the graphs of f 1x2 = x2>3 and g1x2 = x219 - x22-1>2 have similar shapes. a. Find the area of the region bounded by the graph of f and the x-axis on the interval 30, 24. b. Find the area of the region bounded by the graph of g and the x-axis on the interval 30, 24. c. Which region has greater area?

7375. Using the integral of sec3 u By reduction formula 4 in Section 7.3, L sec3 u du = 1 2 1sec u tan u + ln  sec u + tan u 2 + C. Graph the following functions and find the area under the curve on the given interval. f 1x2 = 19 - x22-2, 30, 32 4

7375. Using the integral of sec3 u By reduction formula 4 in Section 7.3, L sec3 u du = 1 2 1sec u tan u + ln  sec u + tan u 2 + C. Graph the following functions and find the area under the curve on the given interval. f 1x2 = 14 + x221>2, 30, 24

7375. Using the integral of sec3 u By reduction formula 4 in Section 7.3, L sec3 u du = 1 2 1sec u tan u + ln  sec u + tan u 2 + C. Graph the following functions and find the area under the curve on the given interval. f 1x2 = 1x2 - 2521>2, 35, 104

7677. Asymmetric integrands Evaluate the following integrals. Consider completing the square. L dx 21x - 1213 - x2

7677. Asymmetric integrands Evaluate the following integrals. Consider completing the square. L 4 2 + 12 dx 21x - 121x - 32

Clever substitution Evaluate L dx 1 + sin x + cos x using the substitution x = 2 tan-1 u. The identities sin x = 2 sin x 2 cos x 2 and cos x = cos2 x 2 - sin2 x 2 are helpful.

A torus (doughnut) Find the volume of the solid torus formed when the circle of radius 4 centered at 10, 62 is revolved about the x-axis.

Bagel wars Bob and Bruce bake bagels (shaped like tori). They both make bagels that have an inner radius of 0.5 in and an outer radius of 2.5 in. Bob plans to increase the volume of his bagels by decreasing the inner radius by 20% (leaving the outer radius unchanged). Bruce plans to increase the volume of his bagels by increasing the outer radius by 20% (leaving the inner radius unchanged). Whose new bagels will have the greater volume? Does this result depend on the size of the original bagels? Explain. T

Electric field due to a line of charge A total charge of Q is distributed uniformly on a line segment of length 2L along the y-axis (see figure). The x-component of the electric field at a point 1a, 02 is given by Ex1a2 = kQa 2L L L -L dy 1a2 + y223>2 , where k is a physical constant and a 7 0. a. Confirm that Ex1a2 = kQ a2a2 + L2 . b. Letting r = Q>2L be the charge density on the line segment, show that if LS _, then Ex1a2 = 2kr>a. (See the Guided Project Electric field integrals for a derivation of this and other similar integrals.) inner radius outer radius O y 2L L y Current I (a, 0) a r x _ O x y 2L L Line of charge (a, 0)

Magnetic field due to current in a straight wire A long, straight wire of length 2L on the y-axis carries a current I. According to the Biot-Savart Law, the magnitude of the magnetic field due to the current at a point 1a, 02 is given by B1a2 = m0 I 4p L L -L sin u r2 dy, where m0 is a physical constant, a 7 0, and u, r, and y are related as shown in the figure. a. Show that the magnitude of the magnetic field at 1a, 02 is B1a2 = m0 IL 2p a2a2 + L2 . b. What is the magnitude of the magnetic field at 1a, 02 due to an infinitely long wire 1LS _2?

Fastest descent time The cycloid is the curve traced by a point on the rim of a rolling wheel. Imagine a wire shaped like an inverted cycloid (see figure). A bead sliding down this wire without friction has some remarkable properties. Among all wire shapes, the cycloid is the shape that produces the fastest descent time (see the Guided Project The amazing cycloid for more about the brachistochrone property). It can be shown that the descent time between any two points 0 a 6 b p on the curve is descent time = L b a A 1 - cos t g1cos a - cos t2 dt, where g is the acceleration due to gravity, t = 0 corresponds to the top of the wire, and t = p corresponds to the lowest point on the wire. a. Find the descent time on the interval 3a, b4 by making the substitution u = cos t. b. Show that when b = p, the descent time is the same for all values of a; that is, the descent time to the bottom of the wire is the same for all starting points.

Maximum path length of a projectile (Adapted from Putnam Exam 1940) A projectile is launched from the ground with an initial speed V at an angle u from the horizontal. Assume that the x-axis is the horizontal ground and y is the height above the ground. Neglecting air resistance and letting g be the acceleration due to gravity, it can be shown that the trajectory of the projectile is given by y = - 1 2 kx2 + ymax, where k = g 1V cos u22 and ymax = 1V sin u22 2g . a. Note that the high point of the trajectory occurs at 10, ymax2. If the projectile is on the ground at 1-a, 02 and 1a, 02, what is a? b. Show that the length of the trajectory (arc length) is 21 a 0 21 + k2x2 dx. c. Evaluate the arc length integral and express your result in terms of V, g, and u. d. For a fixed value of V and g, show that the launch angle u that maximizes the length of the trajectory satisfies 1sin u2 ln 1sec u + tan u2 = 1. e. Use a graphing utility to approximate the optimal launch angle.

Show that L dx x2x2 - 1 = e sec-1 x + C = tan-1 2x2 - 1 + C if x 7 1 -sec-1 x + C = -tan-1 2x2 - 1 + C if x 6 -1.

Evaluate for L 2x2 - 1 x3 dx, for x 7 1 and for x 6 -1.

Graph the function f 1x2 = 2x2 - 9 x and consider the region bounded by the curve and the x-axis on 3-6, -34. Then evaluate L -3 -6 2x2 - 9 x dx. Be sure the result is consistent with the graph.

Graph the function f 1x2 = 1 x2x2 - 36 on its domain. Then find the area of the region R1 bounded by the curve and the x-axis on 3-12, -12>134 and the area of the region R2 bounded by the curve and the x-axis on 312>13, 124. Be sure your results are consistent with the graph.

Visual proof Let F 1x2 = 1 x 02a2 - t2 dt. The figure shows that F 1x2 = area of sector OAB + area of triangle OBC. a. Use the figure to prove that F 1x2 = a2 sin-11x>a2 2 + x2a2 - x2 2 . b. Conclude that L 2a2 - x2 dx = a2 sin-11x>a2 2 + x2a2 - x2 2 + C. (Source: The College Mathematics Journal 34, 3, May 2003)