Math 226 Calculus Spring 2016 Exam 2V1

Transcription

1 Math 6 Calculus Spring 6 Exam V () (35 Points) Evaluate the following integrals. (a) (7 Points) tan 5 (x) sec 3 (x) dx (b) (8 Points) cos 4 (x) dx

2 Math 6 Calculus Spring 6 Exam V () (Continued) Evaluate the following integrals. x 3 + x + 3x + 5 (c) ( Points) x (x + ) dx (d) (8 Points) 3 3/ x 9 x dx

3 Math Calculus Spring 6 Exam V () (5 Points) Circle the correct form for the partial fraction decomposition of the rational function x5 + 3x 4 x 3 + 5x x + (x 4 )(x + ) : a) Ax + B x 4 + Cx + D (x + ) c) Ax + B x + + e) Ax + B x 4 + C (x ) + Ex + F (x + ) 3 C x + + D (x + ) b) A x + B (x ) + C x + + d) Ax + B x + Cx + D x + + f) A x + B x + + E x + + D (x + ) + E F (x + ) (x + ) 3 C (x + ) + D (x + ) 3 + Ex + F x +. (3) ( Points) Does x dx converge or diverge? Why? Evaluate if it converges. + x6 (4) ( Points) Does dx converge or diverge? Why? Evaluate if it converges. x4

4 Math 6 Calculus Spring 6 Exam V (5) ( Points) Use the Comparison Theorem to determine whether the following improper integral converges or diverges. DO NOT COMPUTE THE EXACT VALUE OF THE INTEGRAL, but show all work needed for the Comparison Theorem. x x3 x dx

5 Math 6 Calculus Spring 6 Exam V (6) ( Points) Let the curve C be y = f(x) = x 3 for x. (a) (5 Points) Set up the integral for the arc length of that curve. DO NOT TRY TO EVALUATE OR SIMPLIFY IT. (b) (5 Points) Find the area of the surface made by revolving that curve about the y-axis. DO EVALUATE THIS INTEGRAL.

6 Math 6 Calculus Spring 6 Exam V (7) ( points) The curve defined by parametric equations x = t and y = t 3 3t has been discussed in the textbook. Answer each of the following questions about it. In parts (b), (c) and (d), just set up the appropriate integral but DO NOT EVALUATE OR SIMPLIFY IT. WRITE IT AS AN INTEGRAL INVOLVING EXACTLY ONE VARIABLE, t. (a) (3 Points) Find the equation of the tangent line to that curve at t =. SHOW YOUR WORK. (b) ( Points) Write the integral for the arclength of the part of the curve where t 3, (c) ( Points) Set up the integral for the area between that curve and the x-axis for t 3. (d) (3 Points) Set up the integral for the surface area when the curve for t 3 is rotated about the x-axis.

7 Math 6 Calculus Spring 6 Exam V Solutions () (35 Points) Evaluate the following integrals. (a) (7 Points) tan 5 (x) sec 3 (x) dx = tan 4 (x) sec (x) tan(x) sec(x) dx Let u = sec(x) so du = tan(x) sec(x)dx, and tan (x) = sec (x) = u, so the integral becomes tan 5 (x) sec 3 (x) dx = tan 4 (x) sec (x) tan(x) sec(x) dx = (u ) u du = (u 6 u 4 + u ) du = u7 7 u5 5 + u3 3 + C = sec7 (x) + sec5 (x) + sec3 (x) + C ( ) + cos(4x) (b) (8 Points) cos 4 (x) dx = dx = ( + cos(4x) + cos (4x) ) dx = x sin(4x) + ( ) + cos(8x) dx 8 4 = x 4 + sin(4x) + ( x + sin(8x) ) = 3x sin(4x) + sin(8x) + C x 3 + x + 3x + 5 (c) ( Points) x (x + ) (partial fractions) where dx = ( A x + B x + Cx + D ) x + dx x 3 +x +3x+5 = (A)x(x +)+B(x +)+(Cx+D)x = (A+C)x 3 +(B+D)x +Ax+B gives the equations A + C =, B + D =, A = 3, B = 5 so A = 3, B = 5, C =, D = 3. Then the integral is x 3 + x + 3x + 5 x (x + ) dx = 3 5 x dx + x dx + x x + dx + = 3 ln x 5 x ln x + 3 tan (x) + C = ln x 3 x + 5 x 3 tan (x) + C either expression is correct. 3 x + dx

8 Math 6 Calculus Spring 6 Exam V Solutions (d) (8 Points) 3 3/ x 9 x dx Since cos (u) = sin (u), use the trig substitution x = 3 sin(u) so 9 x = 9 9 sin (u) = 3 cos(u) and dx = 3 cos(u)du. The limits of integration also change: x = u = and x = 3 3/ u = π 3 from the usual degree triangle. Get = 9 3 3/ π/3 x π/3 dx = 9 sin (u)3 cos(u) du 9 x 3 cos(u) ( cos(u)) du = 9 [ u sin(u) ] π/3 = 9 π/3 = 9 π/3 sin (u) du = 9 [ π 3 sin(π/3) ] = 3π cos(u) du () (5 Points) Since (x 4 ) = (x + )(x ) = (x + )(x + )(x ) and each of these factors is irreducible, the denominator factors into irreducibles as (x )(x+) 3 (x +), so the correct form for the partial fraction is: f) A x + B x + + C (x + ) + D (x + ) 3 + Ex + F x + (3) ( Points) Does x dx converge or diverge? Why? Evaluate if it converges. + x6 This integral is improper since the upper bound is infinity. We find the indefinite x integral + x 6 dx = 3 tan (x 3 ) + C so the improper integral is the converging limit [ lim tan (t 3 ) tan ( 3 ) ] = [ π t 3 3 π ] = π 4.

9 (4) ( Points) Does Math 6 Calculus Spring 6 Exam V Solutions dx converge or diverge? Why? Evaluate if it converges. x4 This integral is improper since the denominator of the integrand is zero at x =. We must break it up into x 4 dx + dx. If either integral diverges, then so does the x4 original. The indefinite integral is 3x + C. Both integrals diverge. The first improper 3 integral diverges since it is defined to be t lim t ( dx = lim x4 t 3t 3 ) 3( ) 3 =. The second improper integral diverges since it is defined to be lim s + s ( dx = lim x4 s + 3() 3 ) 3(s) 3 =. It is enough to show that one of these improper integrals diverges. (5) ( Points) Use the Comparison Theorem to determine whether the following improper integral converges or diverges. DO NOT COMPUTE THE EXACT VALUE OF THE INTEGRAL, but show all work needed for the Comparison Theorem. and then x x3 x dx For x we have < x 3 x < x 3 so x 3 x < x 3 = x x so x x3 x > We know that dx diverges for p, so xp Theorem, the given integral diverges. x x x = >. x x3 x > x x dx diverges. By the Comparison x/

10 (6) ( Points) Let the curve C be y = f(x) = x 3 for x. (a) (5 Points) Set up the integral for the arc length of that curve. DO NOT TRY TO EVALUATE OR SIMPLIFY IT. We have f (x) = 3x so + (f (x)) = + 9x 4 and the integral for the arclength is + (f (x)) dx = + 9x4 dx (b) (5 Points) Find the area of the surface made by revolving that curve about the y-axis. That surface area equals πx + (f (x)) dx = π x + (3x ) dx = π x + 9x 4 dx After the substitution u = 3x with du = 6xdx, the above integral equals π + u du 3 and then use the trig sub u = tan(θ) so du = sec (θ)dθ and u = θ = and u = 3 θ = tan (3) = α. This gives (using the integration formula from the cover page of the exam) π α sec 3 (θ)dθ = π [sec(θ) tan(θ) + ln sec(θ) + tan(θ) ] 3 3 = π [sec(α) tan(α) + ln sec(α) + tan(α) sec() tan() ln sec() + tan() ]. 6 Since the triangle corresponding to 3 = tan(α) has side 3 opposite angle α, adjacent side and hypotenuse, we have sec(α) =, and sec() = and tan() =, so the final answer is π 6 [3 + ln 3 + ]. α 3

11 Math 6 Calculus Spring 6 Exam V Solutions (7) ( points) The curve defined by parametric equations x = t and y = t 3 3t has been discussed in the textbook. Answer each of the following questions about it. In parts (b), (c) and (d), just set up the appropriate integral but DO NOT EVALUATE OR SIMPLIFY IT. WRITE IT AS AN INTEGRAL INVOLVING EXACTLY ONE VARIABLE, t. (a) (3 Points) Find the equation of the tangent line to that curve at t =. SHOW YOUR WORK. At t = we have the point (x, y) = (4, ) and the slope of the tangent line is dy dx = dy/dt dx/dt = 3(t ) t which equals 3(3) () = 9 at t =. so the equation of the tangent line to the 4 curve at that point is y = 9 4 (x 4). (b) ( Points) Write the integral for the arclength of the part of the curve where t 3, The arclength integral is 3 3 (dx/dt) + (dy/dt) dt = (t) + (3t 3) dt. (c) ( Points) Set up the integral for the area between that curve and the x-axis for t 3. The integral giving that area is 3 ydx = 3 (t 3 3t) tdt. (d) (3 Points) Set up the integral for the surface area when the curve for t 3 is rotated about the x-axis. The surface area of the curve rotated about the x-axis is 3 π(t 3 3t) (t) + (3t 3) dt