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Professor Readdy
Math 214
Spring 2017
Review
(1) Complete the square: Complete the squares of following polynomials. x2 + 2x − 3, x2 − 4x + 13
and 2s2 − 2s + 5.
(2) Finding rational roots: 2x3 + 3x2 − x − 1 = 0, 3x4 − x3 + 7x2 + 5 = 0, x5 + 9/2 · x4 − 11/2 ·
x3 − 19x2 − 25/2 · x − 15/2 = 0.
(3) Solving quadratic equations: x2 + 2x − 35 = 0, 2x2 + 5x − 12 = 0, x2 − 3x + 29/2 = 0.
(4) Factor polynomials: 2x2 + 7x − 15, x3 + 5x2 + 7x + 3, x4 − 5x3 + 6x2 + 4x − 8.
(5) Partial fractions: (x+5)/(x2 −2x−3), (7x2 −7x−2)/(x3 −x2 −x+1), (x3 −8x+11)/(x2 +2x−3).
p
(6) Simplify expressions: e4·ln(x) , ln(xy) − ln(x/y), 1 − cos2 (x), sin(x − π/6) + cos(x + π/3),
cos(x) − sin2 (x) cos(x), sin(2x)/ cos(x), ln(5ex ).
(7) Recursions:
– Solve an = 5an−1 with the initial condition a0 = 4.
– Solve bn = 3bn−2 with the initial conditions b0 = 4 and b1 = −1.
– Solve cn = 2n2 · cn−1 with the initial condition c0 = 5.
√
(8) Derivatives: Take the derivatives of 1/ x, (x2 + 2x + 3)/(x − 1), (x − 3)/(x2 − x − 6), 10x ,
√
2
tan(x), arctan(x), sin(5x2 ), ex , cos( x), sec(x) and 2x sin(x) + (2 − x2 ) · cos(x).
Z
Z
Z
Z
Z
dt
dx
dt
5/2
−2t
√
(9) Basic integrals: Compute x dx,
, e
dt,
,
.
2
2
1
+
x
t
1−t
Z
Z
Z
√
x
x · cos(x2 )
p
(10) Integration by substitution: Compute
dx,
x
·
x
−
1
dx,
dx.
x2 + 4
sin(x2 )
(11) The elementary differential equation: Find the function y(t) such that y 0 = −2 · y and
y(0) = e.
(12) Definite integration: What is wrong with the argument
Z 5
5
dx
= ln |x + 4| −5 = ln(9) − ln(1) = ln(9)?
−5 x + 4
(13) Taylor series:
– Find the coefficient of xn in the Taylor series of e5x .
– Find the coefficient of xn in the Taylor series of (1 + x)α .
– Find the coefficient of xn in the Taylor series of 1/(1 + 2x).
– Find the coefficient of xn in the Taylor series of (11x − 3)/(1 − 5x + 6x2 ).
– Find the coefficient of x4 in the Taylor series of ex+x
2 /2
.
– Find the coefficient of x6 in the Taylor series of cos(2x).
– Find the coefficient of x8 in the Taylor series of sin(x + x3 ).
(14) Taylor series and limits: Find the coefficients of 1, x, x2 and x3 in the Taylor series of
f (x) = ex − sin(x) − cos(x) − x2 . Use this to compute limx−→0 f (x)/x3 .
Z
Z
Z
2 x
(15) Integration by parts: Compute x e dx, 9x cos(3x) dx, arctan(x) dx.
Z
x+5
dx,
2
x − 2x − 3
Z
x+4
(16) Integration of rational functions: Compute
dx,
2
x + 4x + 13
Z
2x5 − 3x4 − 6x3 + 9x2 + 6x − 4
dx.
x4 − 2x2 + 1
Z 3
Z ∞
Z π
Z 4
Z 2p
dx
x2
2 dx,
(17) Definite integrals: Compute
x2 ln(x) dx,
4
−
x
x sin(x) dx,
,
dx,
x2 /9 + 1 0 x3 + 1
1
0
0
0
Z π
Z π/2
p
5
cos (x) dx,
cos(x) sin(x) dx.
0
0
(18) Partial derivatives: For f = x2 yexy compute
∂f ∂f ∂ 2 f
∂2f
∂x , ∂y , ∂x2 , ∂x∂y
and
∂2f
∂y∂x .
2
(19) Chain rule: View y and z as functions of the variable x and compute the derivative of y · ex·y ,
x2 /z and sin(x · y 2 · z 3 ).
Solutions: (1) (x + 1)2 − 4, (x − 2)2 + 9, 2((s − 1/2)2 + 9/4). (2) x = −1/2, none, x = −1/2, 3.
(3) x = 5, −7, x = 3/2, −4 and x = 3/2±7/2·i. (4) (x+5)·(2x−3), (x+3)·(x+1)2 , (x−2)3 ·(x+1).
(5) 2/(x−3)−1/(x+1), 4/(x−1)−1/(x−1)2 +3/(x+1), x−2+1/(x−1)−2/(x+3). (6) x4 , 2 ln(y),
| sin(x)|, 0, cos3 (x), 2 sin(x), x + ln(5). (7) an = 4 · 5n , b2k = 4 · 3k , b2k+1 = −3k , cn = 5 · 2n · (n!)2 .
If you can not do problems (1) through (7), please review/retake Precalculus. (8) −1/2 · x−3/2 ,
√
√
2
1 − 6/(x − 1)2 , −1/(x + 2)2 , ln(10) · 10x , sec2 (x), 1/(x2 + 1), 10x · cos(5x2 ), 2x · ex , − sin( x)/(2 x),
sec(x) · tan(x), x2 · sin(x). (9) 2/7 · x7/2 + C, arcsin(x) +pC, −1/2 · e2t + C, arctan(x) + C and ln |t| + C.
(10) 1/2 · ln(x + 4) + C, 2/15 · (3x + 2) · (x − 1)3/2 + C, sin(x2 ) + C. (11) e1−2t . (12) The integrand
has a pole at x = −4. If you can not do problems
(8) through (11), please review/retake Calculus 1.
(13) 5n /n!, α · (α − 1) · · · (α − n + 1)/n! = αn , (−2)n , 2 · 3n − 5 · 2n , 5/12, −26 /6! and 0 since it is
an odd function. (14) f (x) = x3 /3 + O(x4 ) and hence the limit is 1/3. (15) (x2 − 2x + 2) · ex + C,
cos(3x) + 3x sin(3x) + C, x arctan(x) − 1/2 ln(x2 + 1) + C. (16) 2 ln(x − 3) − ln(x + 1) + C, 1/2 · ln(x2 +
4x + 13) + 2/3 · arctan(x/3 + 2/3) + C, x2 − 3x − 1/(x − 1) − 2 ln(x + 1) + C. (17) 9 ln(3) − 26/9,
3π/2, ln(65)/3, π, π, 0, 2/3. If you can not do problems (13) through (17), please review/retake
Calculus 2. (18) (x2 y 2 + 2xy) · exy , (x3 y + x2 ) · exy , (x2 y 3 + 4xy 2 + 2y) · exy , (x3 y 2 + 4x2 y + 2x) · exy
2
dy
dy
dy
dz
dz
+ 2xy 2 dx
) · exy , (2xz − x2 dx
)/z 2 , (y 2 z 3 + 2xyz 3 dx
+ 3xy 2 z 2 dx
) · cos(xy 2 z 3 ). If you can
(19) (y 3 + dx
not do problems (18) and (19) please review/retake Calculus 3.