Problem. a) Compute the definite integral 4 + d This can be done by a u-substitution. Take u = +, so that du = d, which menas that 4 d = du. Notice that u() = and u() = 6, so our integral becomes 6 u du = ln u 6 = ln(6) ln() = ln(6/) = ln(). b) Compute the integral ln( ) d. This one looks ripe for integration by parts. According to LAITE, we should take u = ln( ). Then du = d = d. We need to use dv = d, so v =. Then ln( ) d = ( ln( ) ) ( ) ( ) d = ln( ) d = ln( ) 9 + C. c) Compute the integral π/6 sin + cos d First, let s do the indefinite integral. This is a substitution. Use u = + cos, so that du = sin d. Then sin d = du. Rewriting the integral in terms of u, we get u du = ln u = ln + cos + C. To evaluate the definite integral, we plug in the bounds π/6 ( cos t) sin t dt = ln + cos π/6 = ln (ln ln ) =. = (ln + ln + ) d) Compute (ln ) d Hint: One strategy is to make a u-substitution and then integrate by parts. Let s try z = ln, so that dz = d (I m calling the new variable z because I want to reserve u for the integration by parts). Then dz = d and so dz = d. But = (e z ) = e z, and so the integral becomes e z z dz.
Now we need to do this integral, which is going to be by parts, twice. Take u = z and dv = e z dz, so du = z dz and v = ez dz. Then we get e z z dz = z e z zez dz = z e z ze z dz Do it again, now with u = z and dv = e z dz, so du = dz and v = ez. Then we get z e z Now plug back in z = ln and we get (ln ) d = (This is a tough one.) ze z dz = ( ) z e z zez ez dz = ( z e z zez ) 4 ez + C ( = z z + ) e z + C. 4 ( (ln ) (ln ) + 4 ) + C
Problem. a) Sketch and compute the area of the region bounded by y = 4 and y =. Here s the region:.5..5 -.5 -. -.5.5..5 -.5 -. -.5 The area is going to be ( 4 ) ( ) d = ( 4 ) d The integrand is an even function (since it s the sum of a bunch of even functions), and so we can write this as ) ( 4 ) d = ( 5 5 ( = 5 ) ( ) = 5. b) Set up the integral for the arc length of the top edge of the region. (You don t need to actually evaluate it.) The upper edge is defined by y = 4, and the arc length is + f () d = + ( 4 ) d = + 66 d.
cos t Problem. Consider the function g() = dt. t a) Is g() an increasing or decreasing function in the range π? What about π π? If is in this range, then g () = cos is positive, and so the integral is increasing; thus g() is an increasing function. On the second range, g () = cos is negative, which means that g() is decreasing. b) Compute the derivative h () = d sin t dt. d t This is a matter for the fundamental theorem of calculus, which tells us that the answer is simply sin. c) Compute the derivative d sin t dt. d t This is equal to d sin t dt = d d t d g( ) = g ( )(), by the chain rule. We already found that g () = sin, and so the result is sin sin = 4
Problem 4. A parabolic antenna is formed by rotating the part of the graph of y = between = and = around the y-ais. a) Draw a sketch of the antenna, and compute its volume. To find the volume, we slice the paraboloid horizontally. The area of the slice at height y is given by A(y) = π y = πy for y 9, and so the volume is 9 πy dy = 8π. b) A rainstorm fills the antenna with water to a height of y = 5 m. Assuming a density of ρ = kg /m and gravitational force ρ = 9.8 m /s, set up the integral for the work done in pumping the water out over the top of the antenna. The water at height y needs to be lifted a distance of D(y) = 9 y to get it over the rim of the antenna. The work is then (using the notation from the book) b a ρga(y)d(y) dy = The units of the answer are N m. 5 9.8 πy(9 y) dy. 5
Problem 5. a) Compute the partial fraction epansion of. The first order of business is to factor it: = ( 5)(+). There are no repeated or non-reducible factors, so we know that the final answer should take the form: Clearing denominators yields = A 5 + B + = A( + ) + B( 5) Plugging in =, we get 4 = B( 5), so B = 4. Plugging in = 5, we get = A(7), 7 so A =. This gives 7 = 7( 5) + 4 7( + ) b) Evaluate d. Using our answer to the previous problem, we obtain 7 ln 5 + 4 ln + + C. 7 6
Problem 6. Sketch and find the area bounded by the curves y =, y =, and y =. This is easier if we integrate with respect to y. The area is ( ) y (y ( y)) dy = + y ( 8 = + 4 ) ( + ) = 4. Problem 7. a) Determine whether the improper integral is convergent or divergent. We have 4 d e 4d e. K = lim 4e K d = lim K K 4e = lim K 4e K + 4e = 4e. The limit eists, so it s convergent. b) Determine whether the improper integral 4 + + d is convergent or divergent. To compute the integral, we need to have a partial fractions epansion. The denominator factors as ( )( + ), so 4 + + = A + B +. Clearing denominators, 4 + = A( + ) + B( ). Plugging in = we get 6 = A() + B( ), so B =. Plugging in = we get 6 = A(), so A =. This means 4 + + = + +. Now for the actual integral. The integrand is not defined at =, so we need a limit 4 + d = lim + K K 4 + d = lim + K K + + d = lim ( ln + ln + ) K K = lim ( ln + ln(k + )) ( ln(k ) + ln(k + )) K The problem here is the term ln(k ), which is going to be close to ln when K is near : this causes the integral to be infinite. 7
Problem 8. Let f() =. Use the trapezoidal rule with n = to estimate d. We use the points,,,, so =. The formula says d (f() + f() + f() + f()) = + + 6 + 7 =.5 (The true value is 8/4 =.5, so this is a good approimation.) Problem 9. Suppose you re asked to estimate the volume of a football. You measure and find that a football is 8 cm long. You use a piece of string and measure the circumference at its widest point to be 5 cm. The circumference 7 cm from the end is 45 cm. Use Simpson s Rule to make your estimate. Notice that 7 cm is a quarter the length of the football field, so we can use Simpson s rule with n = 4. The circumferences at different points are: 7 4 8 circ 45 5 45 The volume is 45 A() d If the circumference is C, the radius is C/(π), and the area A() is C /(4π), and so the areas are 7 4 8 A() 45 /(4π) 5 /(4π) 45 /(4π) Then Simpson s rule says 45 A() d (A() + 4 A(7) + A(4) + 4 A() + A(8)) 7 = ( + 4 45 /(4π) + 5 /(4π) + 4 45 /(4π) + ) 7 = 766 6π 45.8. This is in cubic centimeters. It sounds like a lot, but that s roughly the same volume as a cube with side length 5 cm, so it s at least the right order of magnitude. 8