1Area of a triangle
A = ½ bh
2Area of a trapezoid
A = ½ (b₁ + b₂) h
4circumfrence of a circle
C = 2πr or C = πd
5Lateral surface area of a cylinder
S = 2πrh
6Total surface area of a cylinder
S = 2πrh + 2πr²
7Surface area of a sphere
S = 4πr²
9Volume of a pyramid
V = ⅓Bh
10Volume of a cylinder
V = πr²h
11Volume of a cone
V = ⅓ πr²h
12Volume of a sphere
V = ⁴⁄₃ πr³
13Double angle idenitity for cos²(θ)
½( 1 + cos(2θ) )
14Double angle identity for sin²(θ)
½( 1 - cos(2θ) )
24( sin(θ) ) / ( cos(θ) )
tan(θ)
25( cos(θ) ) / ( sin(θ) )
cot(θ)
27cos(2θ) (in terms of cosθ and sinθ)
cos²(θ) - sin²(θ)
28cos(2θ) (in terms of cosθ)
2cos²(θ) - 1
29cos(2θ) (in terms of sinθ)
1 - 2sin²(θ)
34log (A × B)
log(A) + log(B)
35log (ᴬ/ᵦ)
log(A) - log(B)
37logₐ(b) expressed in terms of ln
ln(b) / ln(a)
41Pythagorean Theorem
a² + b² = c²
42Distance between (x₁,y₁) and (x₂,y₂)
√( (x₂ - x₁)² + (y₁ - y₂)² )
43Slope of the line through (x₁,y₁) and (x₂,y₂)
(y₁ - y₂) / (x₂ - x₁)
44Point-slope equation of a line
y - y₁ = m(x - x₁)
45Slope-intercept equation of a line
y = mx + b
46Standard form of the equation of a line
Ax + By = C
47Circle with center at point (h,k) and radius r
(x - h)² + (y - k)² = r²
48Hyperbola with the axes as asymptotes
xy = k
49Hyperbola
Ax² - By² = C
(x² / a²) - (y² / b²) = 1
50Ellipse
Ax² + By² = C
(x² / a²) + (y² / b²) = 1
51Parabola (vertical axis of symmetry)
y = ax² + bx + c
52Parabola (horizontal axis of symmetry)
x = ay² + by + c
53Domain of a function
The set of all possible values of x for a function
54Range of a function
The set of all possible values of y for a function
55A zero of a function
A value of x where the graph of a function intersects the x-axis
56Function symmetric across the y-axis
f(-x) = f(x)
57Even function
f(-x) = f(x)
58Function symmetric through the origin
f(-x) = -f(x)
59Odd function
f(-x) = -f(x)
60Inverse of a function
(f ○ f⁻¹)(x) = x
*(f ○ f⁻¹)(x) is the same as f( f⁻¹(x) )*
61To graph the inverse of a function
reflect the graph of the function across the line y = x
62To find the equation of the inverse of a function
interchange x and y, then solve the equation for y
63if point (a,b) lies on the inverse function f⁻¹, then...
point (b,a) lies on function f
64Inverse of y = eˣ
y = ln(x)
65Inverse of y = ln(x)
y = eˣ
66lim( k × f(x) )
k × lim( f(x) )
67lim( f(x) ± g(x) )
lim( f(x) ) ± lim( g(x) )
Quadradic formula
68(roots of y = ax² + bx + c)
x = ( -b ± √(b² - 4ac) ) / 2a
69lim( f(x) × g(x) )
lim( f(x) ) × lim( g(x) )
70lim( f(x) / g(x) )
lim( f(x) ) / lim( g(x) ), if lim( g(x) ) ≠ 0
71limit of [ sin(x)/x ] as x→0
1
72limit of [ sin(x)/x ] as x→∞
0
73limit of [ sin(1/x) ] as x→0
does not exist
74limit of [ ( cos(x) - 1 ) / x ] as x→0
0
75limit of [ (1 + 1/n)ⁿ ] as n→∞
e
76If limit of f(x) as x→a⁻ ≠ limit of f(x) as x→a⁺, then...
limit of f(x) as x→a does not exist
77f is continuous at x = a
• f(a) exists
• limit of f(x) as x→a exists
• limit of f(x) as x→a = f(a)
78f has an infinite discontinuity at x = a
limit of f(x) as x→a⁻ or limit of f(x) as x→a⁺ is ± ∞
79f has a removable discontinuity at x = a
limit of f(x) as x→a exists, but does not equal f(a)
Squeeze Theorem
If f(x) ≤ g(x) ≤ h(x) and
80limit of f(x) as x→c = limit of h(x) as x→c = L, then...
limit of g(x) as x→c = L
(Squeeze Theorem)
81f has a jump discontinuity at x = a
limit of f(x) as x→a⁻ and limit of F(x) as x→a⁺ both exist, but are not equal.
Intermediate Value Theorem
82If f is continuous on [a, b] and if k is between f(a) and f(b), then...
there exists a c in [a, b] such that f(c) = k
(Intermediate Value Theorem)
Extreme Value Theorem
83If f is continuous on the closed interval [a, b], then...
f has a maximum and a minimum value in the interval [a,b]
(Extreme Value Theorem)
84The graph of f has a horizontal asymptote at y = k
limit of f(x) as x→±∞ = k
85The graph of f has a vertical asymptote at y = a
limit of f(x) as x→a = ±∞
86Difference quotient for f at x = a
(f(a + h) - f(a) )/ h
87Symmetric difference quotient for f at x = a
(f(a + h) - f(a-h) )/ 2h
Definition of derivative
88f'(x) =
limit of [ f(x + h) - f(x) / h ] as h→0
Definition of derivative
89f'(c) =
limit of [ f(x ) - f(c) / (x -c) ] as x→c
90If f is differentiable at x = a...
then f is continuous at x = a
Rolle's Theorem
91If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then...
for some c in (a,b), f'(c) = 0
(Rolle's Theorem)
Mean Value Theorem
92If f is continuous on [a, b] and differentiable on (a, b), then...
for some c in (a,b), f'(c) = ( f(b) - f(a) ) / (b - a)
(Mean Value Theorem)
Mean Value Theorem
(as rates of change)
93If f is continuous on [a, b] and differentiable on (a, b), then...
at some point in the interval the instantaneous rate of change equals the average rate of change over the interval
Mean Value Theorem (geometrically)
94If f is continuous on [a, b] and differentiable on (a, b), then...
the line tangent to the curve at some point in the interval is parallel to the secant line across the interval
95If f is differentiable at x = a then...
limit of [ ( f(a + h) - f(a) ) / h ] as h → 0 exists
97(d/dx) (f ± g) =
f' ± g'
98(d/dx) (f × g) =
f × g' + g × f'
99(d/dx) (f)
(g × f' - f × g') / g²
100(d/dx) f( g(x) ) =
f'( g(x) ) × g'(x)
101(d/dx) f⁻¹(x) =
1 / f'(y)
102(dy/du) × (du/dx)
dy/dx, by the Chain Rule
108(d/dx) sec(x)
sec(x) tan(x)
109(d/dx) csc(x)
-csc(x) cot(x)
110(d/dx) arcsin(x) =
1 / √(1 - x²)
111(d/dx) arccos(x) =
-1 / √(1 - x²)
112(d/dx) arctan(x) =
1 / (1 + x²)
113(d/dx) arccot(x) =
-1 / (1 + x²)
114(d/dx) arcsec(x) =
1 / ∣x∣√(x² - 1)
115(d/dx) arccsc(x) =
-1 / ∣x∣√(x² - 1)
118(d/dx) logₐ(x)
1 / ( (x)ln(a) )
120Average rate of change of f on [a, b]
( f(b) - f(a) ) / (b-a)
121Instantaneous rate of change of f at x = a
f'(a)
122f'(a) = 0
The graph of f has a horizontal tangent at x = a
123dx/dy = 0
The graph of f has a vertical tangent
124Critical points of f
Points in the domain of f
• that are endpoints of the domain
• where f' does not exist
• where f' = 0
125f'(x) > 0 for a < x < b
f is increasing on the interval a < x < b
126f'(x) < 0 for a < x < b
f is decreasing on the interval a < x < b
127f"(x) > 0 for a < x < b
The graph of f is concave upward on the interval a < x < b.
128f'(x) is increasing for a < x < b
The graph of f is concave upward on the interval a < x < b.
129f"(x) < 0 for a < x < b
The graph of f is concave downward on the interval a < x < b.
130f'(x) is decreasing for a < x < b
The graph of f is concave downward on the interval a < x < b.
The 2nd Derivative Test
131f'(a) = 0 and f"(a) < 0
f has a maximum at x = a (2nd Derivative Test)
The 2nd Derivative Test
132f'(a) = 0 and f"(a) > 0
f has a minimum at x = a (2nd Derivative Test)
133The graph of f changes concavity
The graph of f has a point of inflection.
134At x = a, f is continuous and changes from increasing to decreasing
f has a maximum at x = a
135At x = a, f is continuous and changes from decreasing to increasing.
f has a minimum at x = a
136Velocity
ds/dt, where s(t) = position
138Acceleration (in terms of velocity)
dV/dt, where v(t) = velocity
139Acceleration (in terms of position)
d²s/dt², where s(t) = position
140f has a vertical tangent at x = a
f is continuous at x = a and ∣f'(x)∣ → ∞ as x → a
141Slope of the line tangent to f at x = a
f'(a)
142Increasing speed
Velocity and acceleration have the same signs.
143Decreasing speed
Velocity and acceleration have the opposite signs.
144Linear approximation to f at x = a
f(x) ≈ f(a) + f'(a)(x - a)
145If the graph of f is concave downward on [a, b], then the linear approximation to f at x = a ________ f(b).
overestimates
146If the graph of f is concave upward on [a, b], then the linear approximation to f at x = a ________ f(b).
underestimates
147The normal line to a curve at x = a is...
perpendicular to the line tangent to the curve at x = a
148An object in motion along a line reverses direction when...
the sign of the object's velocity changes
149An object is at rest when...
v(t) = 0
150Velocity is constant when...
acceleration is 0
151f has a global (absolute) maximum at x = a
f(a) is greater than or equal to every other y-value of f
152f has a global (absolute) minimum at x = a
f(a) is less than or equal to every other y-value of f
153f has a local (relative) maximum at x = a
f(a) is greater than or equal to every other y-value in some interval containing x = a
154f has a local (relative) minimum at x = a
f(a) is less than or equal to every other y-value in some interval containing x = a
A Riemann Sum equivalent to the definite integral ∫ [ f(x) ]dx is...
155*Lower and upper bounds are [ a ] and [ b ], respectively.*
Limit of [ Σf(xᵢ) Δx ] as Δx → 0 for a ≤ x ≤ b
If f is continuous and bounded on an interval containing x = a and F(x) = ∫ [ f(t) ]dt, then...
156*Lower and upper bounds are [ a ] and [ x ], respectively.*
F'(x) = f(x) for all x in the interval
∫ [ f(x) ]dx.
157*Lower and upper bounds are [ a ] and [ b ], respectively.*
F(b) - F(a), where F' = f
∫ [ f(x) ]dx.
158*Lower and upper bounds are [ b ] and [ a ], respectively.*
-∫ [ f(x) ]dx.
*Lower and upper bounds are [ a ] and [ b ], respectively.*
∫ [ f(x) ]dx + ...
*Lower and upper bounds are [ a ] and [ b ], respectively.*
... ∫ [ f(x) ]dx
159*Lower and upper bounds are [ b ] and [ c ], respectively.*
∫ [ f(x) ]dx.
*Lower and upper bounds are [ a ] and [ c ], respectively.*
d/dx ∫ [ f(t) ]dt
160*Lower and upper bounds are [ a ] and [ x ], respectively.*
f(x)
d/dx ∫ [ f(t) ]dt
161*Lower and upper bounds are [ a ] and [ g(x) ], respectively.*
f( g(x) ) × g'(x)
∫ [ f(x) ]dx
162*Lower and upper bounds are [ a ] and [ a ], respectively.*
0
163∫[ k × f(x) ]dx
k ∫[ f(x) ]dx
164∫[ f(x) ± g(x) ]du =
∫[ f(x) ]du ± ∫[ g(x) ]du =
Trapezoid Rule
∫ [ f(x) ]dx ≈
165*Lower and upper bounds are [ a ] and [ b ], respectively.*
ʰ⁄₂ (y₀ + 2y₁ + 2y₂ + ... + 2yₙ₋₁ + yₙ)
If f is increasing on the interval [a, b], then a Left Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
166*Lower and upper bounds are [ a ] and [ b ], respectively.*
underestimates
If f is decreasing on the interval [a, b], then a Left Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
167*Lower and upper bounds are [ a ] and [ b ], respectively.*
overestimates
If f is increasing on the interval [a, b], then a Right Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
168*Lower and upper bounds are [ a ] and [ b ], respectively.*
overestimates
If f is decreasing on the interval [a, b], then a Right Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
169*Lower and upper bounds are [ a ] and [ b ], respectively.*
underestimates
If f is concave upward on the interval [a, b], then a Midpoint Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
170*Lower and upper bounds are [ a ] and [ b ], respectively.*
underestimates
If f is concave downward on the interval [a, b], then a Midpoint Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
171*Lower and upper bounds are [ a ] and [ b ], respectively.*
overestimates
If f is concave upward on the interval [a, b], then a Trapezoid Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
172*Lower and upper bounds are [ a ] and [ b ], respectively.*
overestimates
If f is concave downward on the interval [a, b], then a Trapezoid Riemann Sum ________ the true value of ∫ [ f(x) ]dx.
173*Lower and upper bounds are [ a ] and [ b ], respectively.*
underestimates
174∫[ sin(u) ]du =
-cos(u) + C
175∫[ uⁿ ]du =
(uⁿ⁺¹) / (n + 1) + C, n ≠ -1
176∫[ cos(u) ]du =
sin(u) + C
177∫[ sec²(u) ]du =
tan(u) + C
178∫[ csc²(u) ]du =
-cot(u) + C
179∫[ sec(u) × tan(u) ]du =
sec(u) + C
180∫[ csc(u) × cot(u) ]du =
-csc(u) + C
181∫[ tan(u) ]du =
ln ∣sec(u)∣ + C
182∫[ cot(u) ]du =
-ln ∣csc(u)∣ + C
183∫[ sec(u) ]du =
ln ∣sec(u) + tan(u)∣ + C
184∫[ csc(u) ]du =
-ln ∣csc(u) + cot(u)∣ + C
185∫[ 1 / √(a² - u²) ]du =
arcsin(ᵘ⁄ₐ) + C
186∫[ 1 / u√(u² - a²) ]du =
¹⁄ₐ arcsec ∣ᵘ∣⁄ₐ + C
187∫[ 1 / (a² + u²) ]du =
¹⁄ₐ arctan(ᵘ⁄ₐ) + C
189∫[ aᵘ ]du =
aᵘ / ln(a) + C
190∫[ 1 ]du =
ln ∣u∣ + C
Partial fractions
1911 / (x - r)(x - s) =
A/(x-r) + B/(x-s)
Integration by parts
192∫[ u ]dv =
uv - ∫[ v ]du
193Displacement
∫ [ v(t) ]dt
*Lower and upper bounds are [ a ] and [ b ], respectively.*
194Distance
∫ [ |v(t)| ]dt
*Lower and upper bounds are [ a ] and [ b ], respectively.*
195Area under f on [a, b], were f(x) > 0
∫ [ f(x) ]dx
*Lower and upper bounds are [ a ] and [ b ], respectively.*
196Area under f on [a, b], were f(x) < 0
-∫ [ f(x) ]dx
*Lower and upper bounds are [ a ] and [ b ], respectively.*
197Area between f and g, where f(x) ≥ g(x)
∫ [ f(x) - g(x) ]dx
*Lower and upper bounds are [ a ] and [ b ], respectively.*
If R(t) represents a rate of change, then ∫ [ R(t) ]dt represents...
198*Lower and upper bounds are [ a ] and [ b ], respectively.*
the accumulated amount of change during the time interval a ≤ t ≤ b
199Volume (by disks)
ΔV = π r² Δx
200Volume (by washers)
ΔV = π (R² - r²) Δx
Volume (by shells*)
201*Not a required topic
ΔV = 2π rh Δx
202Arc length for y = f(x) between points (a,c) and (b,d)
∫ [ √(1 + (dy/dx)² ]dx
*Lower and upper bounds are [ a ] and [ b ], respectively.*
203Arc length for x = f(y) between points (a, c) and (b, d)
∫ [ √(1 + (dx/dy)² ]dx
*Lower and upper bounds are [ c ] and [ d ], respectively.*
204Find the limit of [ (x² + 5x) / 5x²) ] as x → 0.
Limit of [ (x(x + 5) / 5x²) ] as x → 0 =
Limit of [ ((x + 5) / 5x) ] as x → 0 : *does not exist*
205Find the limit of [ (x² + 5x) / 5x²) ] as x → ∞.
Limit of [ (1 + 5/x) / 5) ] as x → ∞ = *⅕*
List all values of x where f is discontinuous.
⠀⠀ { (x + 3) / (2x² - 18), x < 0
f(x) { x², 0 ≤ x ≤ 2
206⠀ ⠀ { (8) / (4-x), x > 2
*x = -3*, because f(-3) does not exist.
*x = 0* because f(0) = 0 ≠ limit of [ (x + 3) / (2x² -18) ] as x → 0⁻ = -⅙
*x = 4* because the limit of [ (8) / (4-x) ] as x → 4 does not exist.
207If y = (x³)tan(x), find dy/dx.
dy/dx =* (x³)sec²(x) + (3x²)tan(x)*
208f(x) cos(x) / (e³ˣ); f'(x)
f'(x) = ( e³ˣ( -sin(x) ) - 3e³ˣ( cos(x) ) ) / (e³ˣ)²
= *-( sin(x) + 3cos(x) ) / e³ˣ*
209Classify the extreme points on the graph of y = 12x - x³ for the interval -3 ≤ x ≤ 5.
y' = 0 ⇒ 12 - 3x² = 3(2 − x)(2 + x) = 0
*(-2, -16)* is a local min.
*(2, 16)* is the absolute max.
*(5, -65)* is the absolute min.
210For what values of x is the graph of y = eˣ - x² concave downward?
y" = eˣ - 2 < 0 for *x < ln(2)*
If f'(x) = (x)In(x), at what point(s) does
211the graph of f have a point of inflection?
f"(x) = x(¹⁄ₓ) + In(x) changes sign from negative
to positive at x = ¹⁄ₑ; the point of inflection
is *( ¹⁄ₑ, - ¹⁄ₑ)*
The position of an object in motion along the x-axis for 0 ≤ t ≤ 9 is given by x(t) = 12t - t².
(a) When was the object at rest?
212(b) How far did it travel?
(a) v(t) = 12 - 2t = 0 at t = 6.
(b) ∫ [ v(t) ]dt - ...
*Lower and upper bounds are [ 0 ] and [ 6 ], respectively.*
... ∫ [ v(t) ] dt = ( x(6) - x(0) ) - ( x(9) - x(6) )
*Lower and upper bounds are [ 6 ] and [ 9 ], respectively.*
= (36 - 0) - (27 - 36) = *45 units*
If we estimate ∫ [ x² ]dx using a Midpoint Rule approximation with 6 subintervals, how large is the error?
213*Lower and upper bounds are [ 0 ] and [ 6 ], respectively.*
∫ [ x² ]dx = 3 = 72
M(3) = 2 × 1² + 2 × 3² + 2 × 5² = 70
The error is *2 square units*.
*Lower and upper bounds are [ 0 ] and [ 6 ], respectively.*
214Average value of f on [a, b]
1/(b-a) ∫ [ f(x) ]dx
*Lower and upper bounds are [ a ] and [ b ], respectively.*
Mean Value Theorem for integrals
215If f is continuous on [a, b], then there exists a c in [a, b] such that...
(b - a) f(c) = ∫ [ f(x) ]dx
*Lower and upper bounds are [ a ] and [ b ], respectively.*
216Separable differential equation
∫[ f(y) ]dy = ∫[ g(x) ]dx
217General solution of dy/dx = kx
x = ceᵏᵗ
218The general solution of a differential equation is...
a family of functions that satisfy the
differential equation
219The domain of a particular solution of a differential equation is...
an interval that contains the initial point for which the particular solution is continuous
220A particular solution of a differential equation is...
a continuous function that satisfies the
differential equation and the initial conditions
221Differential equation modeling exponential growth or decay
dy/dx = kx
222Differential equation modeling limited growth
dx/dt = k(L - x)
If we estimate ∫ [ x² ]dx using a Trapezoid Rule approximation with 3 subintervals, how large is the error?
223*Lower and upper bounds are [ 0 ] and [ 6 ], respectively.*
∫ [ x² ]dx = [ x³ / 3 ] = 72
*Lower and upper bounds are [ 0 ] and [ 6 ], respectively.*
T(3) = 2( (0² + 2²) / 2 ) + 2( (2² + 4²) / 2 ) + 2( (4² + 6²) / 2 ) = 76
The error is *4 square units*.
The table shows some values for a continuous function f.
⠀_________________________
︱⠀x ︱1︱4︱5︱7︱
︱f(x)︱8︱2︱4︱5︱
⠀--------------------
224Estimate f'(6).
f'(6) ≈ ( 5 - 4 ) / (7 - 5) = ½
The table shows some values for a continuous function f.
⠀_________________________
︱⠀x ︱1︱4︱5︱7︱
︱f(x)︱8︱2︱4︱5︱
⠀--------------------
225Using a Left Riemann Sum, estimate the area under f on this interval.
(4 - 1)8 + (5 - 4)2 + (7 - 5)4 = *34*
A tank contains 100 gallons of diesel fuel at time t = 1. The table shows R(t), the rate (in gallons per hour) at
which fuel is being drained out at selected times.
⠀_____________________________________
︱⠀⠀ t(hr)⠀⠀⠀︱1︱4︱5︱7︱
︱R(t) (gal/hr)︱8︱2︱4︱5︱
⠀-----------------------------
226Using a Trapezoidal Sum, estimate the amount of fuel that remains at t = 7.
Approximately
( (8 + 2) / 2 ) × 3 + ( (2 + 4) / 2 ) × 1 + ( (4 + 5) / 2 ) × 2 =
27 gallons have drained out so there are about 100 - 27 = *73 gallons* left.
227Let R be the region bounded by the curve y = e²ˣ, the line y = 10, and the y-axis. Find the area of R.
e²ˣ = 10 ⇒ x = ½ln(10)
∫ [ 10 - e²ˣ ]dx = (10x - ½e²ˣ)^(½ln(10))...
*Lower and upper bounds are [ 0 ] and [ ½ln(10) ], respectively.*
...= 10(½ln(10)) - ½e^(2½ln(10)) - (10 × 0 - ½e²⁽⁰⁾)
...= *5ln(10) - ⁹⁄₂*
Let R be the region bounded by the curve y = e²ˣ, the line y = 10, and the y-axis.
228Set up an integral for the volume of the solid formed when R is rotated around the x-axis.
e²ˣ = 10 = ½ln(10)
V = *π∫ [ 10² - (e²ˣ)² ]dx*
*Lower and upper bounds are [ 0 ] and [ ½ln(10) ], respectively.*
229Let R be the region bounded by the curve y = e²ˣ, the line y = 10, and the y-axis. Setup an integral for the volume of the solid formed when R is rotated around y = 10.
e²ˣ = 10 = ½ln(10)
V = *π∫ [ (10 - e²ˣ)² ]dx*
*Lower and upper bounds are [ 0 ] and [ ½ln(10) ], respectively.*
Let R be the region bounded by the curve y = e²ˣ, the line y = 10, and the y-axis.
230Set up an integral for the volume of the solid formed when R is rotated around the y-axis.
e²ˣ = 10 = ½ln(y)
V = *π∫ [ ½ln(y) ]dy*
*Lower and upper bounds are [ 0 ] and [ 10 ], respectively.*
231Find the area of the region bounded by y = x + 8 and y = 10 - x².
10 − x² = x + 8 ⇒ x = −2, 1
∫ [ (10 - x²) - (x + 8) ]dx = ...
*Lower and upper bounds are [ 1 ] and [ -2 ], respectively.*
... ∫ [ (2 - x - x²) ]dx
*Lower and upper bounds are [ 1 ] and [ -2 ], respectively.*
[ 2x - (x² / 2) - (x³ / 3) ] = *⁹⁄₂*
*Lower and upper bounds are [ 1 ] and [ -2 ], respectively.*
232Find the volume of the solid formed when the region bounded by y = x + 8 and y = 10 - x² is rotated about the x-axis.
10 - x² = x + 8 ⇒ x = -2, 1
∫ [ (10 - x²)² - (x + 8)² ]dx = 237.504
*Lower and upper bounds are [ 1 ] and [ -2 ], respectively.*
233Find the volume of the solid whose base is the region bounded by y = x + 8 and y = 10 - x², if cross-sections parallel to the y-axis are squares with their bases in xy-plane.
10 - x² = x + 8 ⇒ x = -2, 1
∫ [ (10 - x²) - (x + 8) ]² =* 8.1*
*Lower and upper bounds are [ 2 ] and [ 1 ], respectively.*
For what value(s) of k is f differentiable?
f(x) { (k)sin(4x), x < 0
234 ⠀⠀ { k + e²ˣ, x ≥ 0
*None*; f would be continuous at x = 0 only if k = -1. However, if k = -1, the one-sided derivatives -4 cos(4×0) and 2e²⁽⁰⁾ are not equal.
What integral is equivalent to ∫ [ 1 / √(1 + x²) ]dx
using the substitution x = tan(θ)?
235*Lower and upper bounds are [ 0 ] and [ 1 ], respectively.*
tan(θ) = 0 ⇒ θ = 0; tan(θ) = 1 ⇒ θ = π/4
∫ [ 1 / √(1 + x²) ]dx = ...
*Lower and upper bounds are [ 0 ] and [ 1 ], respectively.*
... ∫ [ sec²(θ) / √(1 + tan²(θ) ]dθ = ...
*Lower and upper bounds are [ 0 ] and [ π/4 ], respectively.*
... ∫ [ sec²(θ) / sec(θ) ]dθ = ...
*Lower and upper bounds are [ 0 ] and [ π/4 ], respectively.*
... *∫ [ sec(θ) ]dθ*
*Lower and upper bounds are [ 0 ] and [ π/4 ], respectively.*
The table shows some values of a differentiable function f and its derivative.
If g(x) = √(2x + 1) and h(x) = f(g(x)), find h'(4).
⠀_____________________
︱*x*︱*f(x)*︱*f'(x)*︱
︱1 ︱ 2⠀︱ -1⠀︱
︱2︱ -3 ︱ 8⠀ ︱
︱3︱ 1⠀ ︱ 2⠀ ︱
︱4︱ 7⠀︱ 5⠀ ︱
236⠀⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻
g′(x) = ½(2x + 1)^(-½) × 2; g'(4) = ⅓
h'(4) = f'(g(4)) × g'(4) = f'(3) × g'(4) = 2 × ⅓ = *⅔*
At which of the points shown on the graph of function h is h(x) × h'(x) × h"(x) < 0?
The points are *not* given, but graph is *similar* to ½[ x(x - 2)(x + 1) ].
237 the points are around A: (-1.25, -0.5), B: (-0.75, 0.25), C:(-0.25, 0.21), D: (0.75, -0.8), E: (1.7, -0.68), and F: (2.15, 0.5).
At *B*, h > 0, h' > 0, h" < 0.
At *E*, h < 0, h' > 0, h" > 0.
A particle moves along the x-axis for t ≥ 0 with position given by x(t) = e^(-t²).
238When is its speed increasing?
Velocity x'(t) = -2te^(-t²) and acceleration x(t) = 2(2t² - 1)( e^(-t²) ) are both negative for *0 < t < 1/√2*
The graph of function f consists of two segments and a semicircle.
Find ∫ [ f(x) ]dx.
*Lower and upper bounds are [ -2 ] and [ 7 ], respectively.*
For f(x), go to Desmos and input 3 functions
- polygon((-1,2)(0,2))
- (x - 2)² + (y - 2)² = 4 {y ≤ 2}
239- y = -2x + 10 {-4 ≤ y ≤ 2}
A = rectangle + (rectangle - semicircle) + triangle - triangle
A = 1 × 2 + (4 × 2 - ½π × 2²) + ½(1) × 2 - ½(2) × 4 = *7 - 2π*
240Without using absolute value, set up an integral expression for the area of the region bounded by the curve y = x³ - 3x² - 4x and the x-axis.
x(x - 4)(x + 1) = 0 at x = -1, 0, and 4
y > 0 for -1 < x < 0 and y < 0 for 0 < x < 4
A = ∫ [ x³ - 3x² - 4x ]dx - ...
*Lower and upper bounds are [ -1 ] and [ 0 ], respectively.*
... *∫ [ x³ - 3x² - 4x ]dx*
*Lower and upper bounds are [ 0 ] and [ 4 ], respectively.*
241Find limit of [ (1/h)( cos(π/3 + h) ) - ½ ] as h → 0.
For f(x) = cos(x), f'(π/3) = -sin(π/3) = *-√(3)/2*
242Line l is tangent to the curve y = 2x / (x - 4) at its y-intercept. At what other point on this curve is the tangent line parallel to l?
y' = -8 / (x - 4)²; y' = -1/2 at (0, 0) and again at *(8, 4)*.
The table shows some values of a differentiable function f and its first derivative f'. If g(x) = f ⁻¹(x), find g'(3).
⠀___________________________
︱⠀x⠀︱0︱1︱2 ︱3︱
︱f(x) ︱0︱3︱5︱6︱
︱f'(x)︱5︱4︱3︱2︱
243⠀---------------------
g'(x=3) = 1 / f'(y=3) =* ¼*
244Find the equations of all asymptotes to the graph of y = 2x² - 10x / 25 - x².
*x = -5* and *y = -2*
245Classify the extreme points of the function y = x² / eˣ
y = ( eˣ(2x) - x²eˣ ) / (eˣ)² = 0 ⇒ x = 0, 2
*(0, 0) *is the absolute minimum; *(2, 4 / e²)* is a relative maximum.
Shown is the graph of f'. Let g(x) = ∫ [ f(t) ]dt.
*Lower and upper bounds are [ b ] and [ x ], respectively.*
At what value(s) of x do the graphs of f and g have points of inflection?
246***theres a graph j
f' changes from decreasing to increasing at x = a and from increasing to decreasing at x = c, so f has point of inflection at x = a, c.
g' = f ⇒ g" = f', which changes from positive to negative at x = d; hence, g has a point of inflection at *x = d*.
247If y is a differentiable function of x and (x)In(y) + 5 = y³, find dy/dx
x(1/y) (dy/dx) + ln(y) = 3y² (dy/dx)
dy/dx = *(y)ln(y) / (3y³ - x)*
248Find the particular solution of dx = x(dy) - y²(dx) that passes through the point (1, 1).
[ 1 / (1+y²) ]dy = [ 1/x ]dx
arctan(y) = ln(x) + C
*y = tan(ln(x) + π/4)*
249The rate at which a particle approaches a charged barrier is inversely proportional to its distance from the barrier. If the particle is 13 mm from the barrier at t = 2 sec, and only 12 mm from the barrier at t = 3 sec, how far from the barrier was it at t = 0 sec?
ds/dt = k/s ⇒ ∫[ s ]ds = k∫[ 1 ]dt
s²/2 = kt + c ⇒ s = √(kt + C)
Use (2, 13) and (3, 12) to solve simultaneous equations for k and c.
s = √(219 - 50t) ⇒ s(0) = √219 ≈* 14.799 mm*
If g(x) = ∫ [ ∛(t² - 17) ]dt, find g'(5).
250*Lower and upper bounds are [ 10 ] and [ 2x-1 ], respectively.*
g'(x) = ∛( (2x - 1)² - 17) × 2
g'(5) = *8*
Describe the behavior (increasing/decreasing and concavity) of the graph of f for x > 1 if f'(x) = ∫ [ (t)cos²(t) ]dt.
251*Lower and upper bounds are [ 0 ] and [ x ], respectively.*
For x > 1, f" (x) = (x)cos²(x) > 0 and f'(x) > 0; hence, the graph of f is *increasing* and *concave upward*.
The curve y = x³ - 5x² + 4x + 3 passes through the points (1, 3) and (4, 3).
252Find the value c in the interval 1 < c < 4 where the function satisfies Rolle's Theorem.
f'(c) = 3c² - 10c + 4 = 0 ⇒ c = *2.869* (in this interval)
The graph of the function f(x) = x³ - 5x² + 4x + 3 passes through the points (2, -1) and (4, 3). The Intermediate Value Theorem predicts there is a value c in the interval 2 < c < 4 where f(c) = 0.
253Find c.
x³ - 5x² + 4x + 3 = 0 ⇒ x = *3.700* (in this interval)
The curve y = x³ - 5x² + 4x + 3 passes through the points (2, -1) and (4, 3).
254Find the value c in the interval 2 < c < 4 where the function satisfies the Mean Value Theorem.
f'(c) = 3c² - 10c + 4 = [ (3 - (-1)) / (4 - 2) ] ⇒ *c = 3.120* (in this interval)
An object in motion has velocity v(t) = cos(e²ᵗ + 1) for 0 ≤ t ≤ 3 sec.
255Find the object's average acceleration for this time interval.
1/(3-0) ∫ [ a(t) ]dt = ⅓(v(3) - v(0)) = *-0.085 units/sec²*
*Lower and upper bounds are [ 0 ] and [ 3 ], respectively.*
256Find the average value of y = 2ˣ on the interval [1, 4].
1/(4-1) ∫ [ 2ˣ ]dx = *6.733*
257Find the average rate of change of y = 2ˣ on the interval [1, 4].
(2⁴ - 2¹)/(4-1) = ¹⁴⁄₃
258Find the slope of the curve y = x + 4 cos(x) at its largest x-intercept.
x + 4 cos(x) = 0 ⇒ x = 3.595 (largest root)
y'(3.595) = *2.753*
259∫[ x / (9-x²) ]dx =
-½∫ [ -2x(dx) / (9-x²) ] = *-½ ln∣9-x²∣ + C*
260∫[ x / √(9-x²) ]dx =
-½∫ [(9 - x²)^(-½) × (-2x) ]dx =* -√(9-x²) + C*
261∫[ 1 / √(9-x²) ]dx =
∫ [ ⅓ / √(1 - (x/3)²) ]dx = *arcsin(x/3) + C*
262∫[ 1 / x√(9-x²) ]dx =
⅓∫ [ 1 / x√(1 - x²/9) ]dx = ⅓∫[ ⅓ / (x/3)√(1 - (x/3)²) ]dx = *⅓ arcsec(x/3) + C*
263∫[ 1 / (9-x²) ]dx =
⅑∫ [ 1 / (1 + (x/3)² ]dx = 3 × ⅑∫[ ⅓ / (1 + (x/3)² ] = *⅓ arctan(x/3) + C*
264∫[ tan(5x) ]dx =
*⅕ ln∣sec(5x)∣ + C*
265∫[ 1 / (x)sin²(ln(x)) ]dx=
∫ [ csc²(ln(x)) × (1/x) ]dx = *-cot(lnx) + C*
266∫[ sin(2X) / (1 + cos²(x) ]dx =
-∫ [ (-2cos²(x) × sin(x)) / (1 + cos²(x)) ]dx = *-ln(1+cos²(x)) + C*
267∫[ csc(√x) × cot(√x) / (√x) ]dx = ?
2∫ [ (csc(√x) × cot(√x) × (½x^(-½)) ]dx = *-2csc√x + C*
268If f is a continuous function with f(2) = 10 and f'(2) = -7/4, estimate f(2.008).
f(x) = f(2) + f'(2)(x-2)
f(2.008)10 + 0.008(-7/4) = *9.986*
269What point on the curve y = √(1+x²) is closest to the point (2,0)?
D = (x-2)² + (√(1+x²) - 0)²
D' = 2(x-2)+ 2x
D'= 0 and D" > at x = 1
The point is *(1, √2)*.
270An object moves around the ellipse "4x² + y² = 40". When it passes through the point (3,2) its x-coordinate is decreasing at the rate of 5 units per second. How is its y coordinate changing there?
8x(dx/dt) + 2y(dy/dt) = 0
8(3)(-5) + 2(2) (dy/dt) = 0
dy/dt = 30
Its y-coordinate is increasing at a rate of *30 units/sec*.
The length and width of a rectangle are changing, but the area remains constant.
When the length is 20 inches and the width is 15 inches, the width is increasing at the rate of 3 inches per minute.
271How is the perimeter changing then?
l×w = k {k is constant} P = 2l + 2w
l(dw/dt) + w(dl/dt) = 0 dP/dt + 2(dl/dt) + 2(dw/dt)
20(3) + 15(dl/dt) = 0 ⇒ dl/dt = -4
dP/dt = 2(3) + 2(-4) = *-2 inches/min*
For differentiable functions f and g, f(2) = 6, f'(2) = -2, g(2) = 3, and g'(2) = ½.
272If h(x) = (f)(x), find h'(2).
h'(2) = ( g(2) × f'(2) × g'(2) ) / ( g(2) )² = ( 3(-2) - 6(½) ) / 3² =* -1*
273A particle moves along the x-axis for t ≥ 0 with position given x(t) = e^(-t^2). When is its speed increasing
Velocity x'(t) = -2te^(-t^2) and acceleration x"(t) = 2(2t^2 - 1)e^(-t^2) are both negative for 0<t<(1/√2).