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Calculus IB Final Review
Name _________________________
6
11
14 16
17
1) Find each limit algebraically.
⎛ sin(x) ⎞
⎛ sin(x) ⎞⎛ 1 ⎞
1 1
a. lim ⎜ 2
⎟ = lim ⎜
⎟⎜
⎟ = 1⋅ =
3 3
x →0⎝ x + 3x ⎠ x →0⎝ x ⎠⎝ x + 3 ⎠
a.
€
b.
1
3
⎛ 9x ⎞
⎛ 9x ⎞ 9
⎟⎟ = lim ⎜⎜
⎟⎟ = = 9
lim ⎜⎜ €
x → ∞⎝ x 2 + x ⎠ x →∞⎝ x 2 ⎠ 1
€ b. ____9_____
c.
€
⎛ 9x ⎞
⎛ 9x ⎞ −9
⎟⎟ = lim ⎜⎜
⎟⎟ =
lim ⎜⎜
= −9
x →−∞⎝ €x 2 + x ⎠ x →−∞⎝ x 2 ⎠ 1
c. ____-9_____
2) Find the first and second derivatives of the functions.
€
€
8
a. f(x) = 4 x + 7x 3 − 9x + 2
x
2
16
+ 21x 2 − 9 − 3
x
x
f ’(x) =
€
f “(x) =
€
b. g(x) = tan(x) + sec(x)
−1
48
3 / 2 + 42x + 4
x
x
g’(x) = sec2(x) + sec(x)tan(x)
g“(x) = 2sec2(x)tan(x)€+ sec(x)tan2 (x) + sec3(x)
€
3) Find any relative extrema using the first derivative test and the second derivative test.
x2 + 3
f(x) =
x
rel. max @ x = − 3
rel. min. @ x = 3
x(2x) − (x 2 + 3)⋅ 1
2x 2 − x 2 − 3 x 2 − 3
=
=
, critical values: x = 0, ± 3
€
x2
x2
x2
1st Derivative Test
2nd Derivative Test
€
x 2 (2x) − (x 2 + 3)⋅ 2x 2x(x 2 − x 2 − 3) 6
f “ (x) = €
=
= 3
x4
x4
x
€
€
€
f ’ (x) =
f ’ (-2)
1
+
4
f ’ (-1)
f ’ (1)
-2
-2
Rel. max. @ x = − 3 ,
€
rel. min. @ x =
€
€
f “ ( − 3 ) = –, concave up, rel. max. @ x = − 3
f ’ (2)
1
+
4
€
€
€
€
f “ ( 3 ) = +, concave down, rel. min. @ x =
€
€
3
€
€
3
4) Find the indefinite integrals.
⎛
1⎞
a. ∫ ⎜ sec(x)tan(x) − 3x 4 + x + 3 ⎟ dx = ∫ sec(x)tan(x) − 3x 4 + x 1/ 2 + x −3 dx
⎝
x ⎠
⎛
⎞ ⎛
⎞
3x 5 2 3 / 2 1 −2
3x 5 2 3 / 2
1
= ⎜sec(x) −
+ x
− x + C ⎟ = ⎜sec(x) −
+ x
− 2 + C⎟
5
3
2
5
3
2x
⎝
⎠ ⎝
⎠
€
€
(
b.
(x +1) 2
dx =
x
∫
€
∫
(x 2 + 2x +1)
dx =
€
x
∫ (x
3
2
)
+ 2x
1
+x
2
€
€
−1
2
)dx€ = ⎛⎜⎝ 25 x
5
2
+2•
1 ⎞
2 32
x + 2 • 1x 2 ⎟ + C
⎠
3
⎛2 5 4 3
1 ⎞
= ⎜ x 2 + x 2 + 2x 2 ⎟ + C
⎝5
⎠
3
€
5) Find the area of the region bounded by the curves. Write as a definite integral.
€
a. y = 0, y = 3x 2 − 4 x +1, x = -3, and x = 2.
2
2
∫ −3 (3x 2 − 4 x +1)dx = [ x 3 − 2x 2 + x ]−3 = [(2) − (−48)]
= 50
€
€ π
€
b. y = 0, y = 2sin(x) , x = 0, and x = .
2
€
π/2
∫ 2sin( x)dx = 2[sin(u)]
π/2
= 2[ −cos(x)] 0
0
€
€
€
€
€
6) Given
a.
∫
b
∫
x 2 dx = 10,
a
a
∫ a (9πesin(
€
€
4 x3
€
b.
= 2[ −cos(π /2) − ( −cos(0))] = 2[0+1] = 2
a
€
b
b
∫ dx = -4, find
xdx = 7,
a
a
)dx = 0
€
a
_____0______
b
a
∫ b x 2dx + ∫ b xdx + ∫ b dx = - ∫ x dx –
2
a
∫
b
xdx –
a
b
∫ dx = -10 – 7 – (-4) = -13
a
____-13_____
€
c.
€b
∫a
€
€
(3x 2 − 6x + 2)dx = 3
∫
€
x 2 dx – 6
b
a
∫
€
xdx + 2
b
a
b
∫ dx = 3(10) – 6(7) + 2(-4) = -20
a
____-20_____
€
€
€
€