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sample exam problems

 

(c) 1997-2008 Dan Bach and B & L Math Enterprises; all rights reserved. download for personal use only.

 

calculus II (dvc math193) / set #1: answers below

 

#1 - 6 : Integrate, show steps, and simplify :

Q1. ln(x^2) dx

Q2.cos(ln x) dx

Q3.sec^4(5t) dt

Q4.[ 1 /(x^2 + 2 x + 2) ] dx

Q5.[ (x + 3) / (x^2 + 2x) ] dx

Q6. Use a table:x^3 sin(x^2) dx

Q7. Use Left, Right, Midpt, Trap, Simpson

(n = 2) to estimate ln 5 using an integral.

 

Q8. Convergent or Divergent? (Eval if conv.)

x e^(-2x) from 0 to oo (zero to infinity)

Q9. Find the length of y = cosh x from x = -1 to x = 1.

Def'n of hyperbolic cosine: cosh x = [e^x + e^(-x)] / 2.

 

Q10. Give (and explain) an example of an infinite

surface area containing a finite volume (!)

 

Q11. Solve the diff. eqn. y' = 3x / y with initial condition

y(1) = 5 , using separation of variables, and

draw a slope field and solution curve.

 

Q12. If y = A sin 3x + B cos 3x , what values of A and B will

make the initial conditions y(0) = 15 and y'(0) = 30 ?

 

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answers to sample problems (calc II, set #1): back to set #1

 

A1. Use ln(x^2) = 2 ln x ; int by parts: u = ln x ; dv = dx ;

du = dx / x ; v = x ; answer is 2 x ln x -- 2 x + C

 

A2. u = ln x ; x = e^u ; dx = e^u du ; cos(u) e^u du ;

int. by parts twice ; ans. is (x/2)[cos(ln x) + sin(ln x)] + C

 

A3. sec^4(5t) = sec^2(5t) sec^2(5t) = sec^2(5t)[1 + tan^2(5t)]

u = tan(5t) ; integrate (1 + u^2) du. (where did sec^2(5t) go?)

 

 

A4. u = x+1 ; x^2 + 2 x + 2 = u^2 + 1 ; du = dx ; then sub:

u = tan(t) ; du = sec^2(t) dt ; integ. =(sec t) dt

ans: ln | sec t + tan t | = ln | x+1 +(x^2 + 2 x + 2)| + C

 

A5. (x + 3) / [ x(x+2)] = A/x + B/(x+2) ; x + 3 = A(x+2) + Bx ;

A+B = 1 , 2A = 3 ; A = 3/2 ; B = -1/2 ; (3/2) ln | x | -- (1/2) ln | x+2 | + C

 

A6. By Formula 82, it's [ sin(x^2) -- x^2 cos(x^2)] / 2 + C

 

A7. ln 5 =dx / x from 1 to 5 ; 2 rects ; x0=1 , x1=3 , x2=5.

delta x = (5 - 1)/2 ; y0=1/1 , y1=1/3 , y2=1/5 ;

L2 = 2(1/1 + 1/3) = 8/3 ~ 2.6667 ;

R2 = 2(1/3 + 1/5) = 16/15 ~ 1.0667 ;

M2 = 2(1/2 + 1/4) = 3/2 = 1.5000 ; (m1=2 ; m2=4)

T2 = (2/2)(1/1 + 2*1/3 + 1/5) = 28/15 ~ 1.8667 ;

S2 = (2/3)(1/1 + 4*1/3 + 1/5) = 73/45 ~ 1.6222 ;

Actual A = ln 5 ~ 1.6094 ; errors . . . ?

 

A8.x e^(-2x) = (-1/2) x e^(-2x) + (1/2) e^(-2x) dx ;

take lim b -> oo from 0 to b, use L'H, ans = 1/4 (conv.)

 

A9. This is one where the 1 + (dy/dx)^2 looks just like

the (dy/dx)^2 expression except for the + or - 1/2.

The integrand simplifies to [e^x + e^(-x)] / 2 , and

so this curve has arclength equal to its area !!

A = L = e -- 1/e ; I'm pretty sure ; is it right?

 

A10. Gabriel's Horn, 8.2 #27 (revolve y = 1/x), done in class!

 

A11. y dy = 3x dx ; y^2 / 2 = 3 x^2 / 2 + C/2 ; y = +/-(3 x^2 + C)

If x=1, y=5 so C = 22 ; solution curve is a hyperbola y =(3 x^2 + 22)

 

A12. y(0) = A sin 0 + B cos 0 = B = 15 ;

y'(0) = 3A cos 0 - 3B sin 0 = 30 ; A = 10 .

 

 

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sample exam problems

calculus II (dvc math193) / set #2: answers below

 

Q13. Write an expression for the nth term of the series

1/2 + 2/4 + 3/8 + 4/16 + . . .

Q14. Prove using the integral test that the series in #1 converges.

Q15. Find the radius of convergence of the power series

f(x) = (n / 2^n) x^n and discuss the endpoints.

Q16. True or false? (Give a reason or an example.)

a) If a series converges, then the terms go to zero.

b) If the terms go to 0, the sequence of partial sums converges.

c) Every continuous function equals its Taylor series.

Q17. Give an example of a "conditionally convergent" series.

Q18. Find the Taylor poly of degree 6, about x = a, for each:

i) f(x) = sin(x^2) , a = 0 ; ii) f(x) = x ln(x) , a = 1

Q19. How many terms of the usual series for e^x would you

need to use to approximate e to two decimal places?

Use Taylor's inequality and explain steps clearly.

Q20. If g(x) = 1 / (1 -- 2x) , what is g(50)(0)? (the fiftieth deriv.)

Q21. a) Verify that y = 3 cos(5x) is a solution of the initial

value problem y''(0) = -25 y , y(0) = 3 , y'(0) = 0.

b) Guess a solution to y''(0) = 25 y , y(0) = 2 , y'(0) > 0.

Q22. a) Sketch the slope field for y' = x -- y on a grid.

b) Draw the solution curve y = f(x) thru the pt (0, 1)

c) Use Euler's method to estimate f(1) with h = 0.25.

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answers to sample problems (calc II, set #2)

back to set #2 | back to math193

 

A13. an = n / (2^n) ; S = 2 as seen by geom series method

A14. Let f(x) = x / (2^x) = x 2^(-x) ;

x 2^(-x) dx = (-x/ln 2) 2^(-x) + (1/ln 2)2^(-x) dx

( u = x , du = dx ; dv = 2^(-x) dx , v = (-1/ln 2) 2^(-x) . . . )

= (-x/ln 2) 2^(-x) + (-1/ln 2)^2 2^(-x) ; converges (by L'H) as x --> oo .

A15. Ratio test says lim | an+1 / an | |x| = lim ((n+1)/2^(n+1)) / (n / 2^n) |x|

= lim ((n+1)/n) (1/2) |x| = |x|/2 < 1 if |x| < 2 ; so R = 2.

If x = +/-2 the sum is (+/-1)^n * n which both diverge.

A16. a) This one is True by the 'Divergence Test'.

b) False, for example the alternating harmonic series.

c) False, e.g. f(x) = e^(-1 / x^2) as seen in class and text.

A17. Example: the alternating harmonic series (-1)^n / n

A18. i) sin(x^2) ~ (x^2) -- (x^2)^3 / 3! = x^2 -- x^6 / 6

ii) x ln(x) ~ (1 + (x-1))*[ (x-1) - (x-1)^2 / 2 + . . . - (x-1)^6 / 6 ]

= (x-1) + (x-1)^2 / 2 . . . you do the algebra!

A19. Taylor's inequality says |Rn| < Mn x^(n+1) / (n+1)!

For e = e^(1/2) we have |x| < 0.5 so Mn < e^0.5 < 2 ;

find first partial sum Tn(0.5) whose Rn is small enough.

A20. a) If you're busy taking 50 derivatives, stop!

g(x) = 1 / (1 -- 2x) = (2x)^n =2^n x^n ;

so an = g(n)(0)/n! = 2^n ; g(50)(0) = 2^50 * 50! .

A21. a) y = 3 cos(5x) , y' = -15 sin(5x) , y'' = -75 cos(5x) = -25 y.

y(0) = 3 cos(0) = 3 ; y'(0) = -15 sin(0) = 0.

b) Try y = A e^kx ; y' = A k e^kx ; y'' A k^2 e^kx = 25 y

means k = 5 or -5 ; y(0) = A e^0 = 2 means A = 2 ;

y'(0) > 0 means k = 5 not -5 ; solution is y = 2 e^5x .

A22. a) b)

c) Let (xn, yn) be end of nth segment; mn = y'(xn, yn)

Then mn = xn -- yn and yn+1 = yn + h mn

x0 = 0.00 ; y0 = 1 ; m0 = 0 - 1 = -1

x1 = 0.25 ; y1 = 1 + 0.25(-1) = 0.75 ;

x2 = 0.50 ; y2 = 0.75 + 0.25(-0.5) = 0.625 ;

x3 = 0.75 ; y1 = 0.625 + 0.25(-0.125) = 0.59375 ;

x4 = 1.00 ; y1 = 0.59375 + 0.25(0.15625) = 0.63281

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You're welcome, and good luck! - Dan