Calculus II Diagnostic Test 1220-004
Fall 2014
Your Name:
Your UID :
1. Answer the following questions
(1’) Your grades will show up on
(a)Canvas
(b)My Webpage
(c) Both (a) and (b)
(2’) Where do midterm review exams, homework assignments and class
notes appear?
(a)Canvas
(b)My Webpage
(c) Both (a) and (b)
2. Calculate the limit below
sin 5x
x→0 3x
lim
Solutions:
sin 5x
x→0 3x
sin 5x 5
= lim
∗
x→0 5x
3
5
=1∗
3
5
=
3
lim
1
3. Suppose that f (2) = 3, f 0 (2) = 4, f ”(2) = −1, g(2) = 2, g 0 (2) = 1
Find
¯
¯
d
3
[f (x)g(x) + sin (g (x))]¯¯
dx
x=2
Solutions:
¯
¯
d
3
[f (x)g(x) + sin (g (x))]¯¯
dx
x=2
¯
= (f 0 (x)g(x) + f (x)g 0 (x) + cos (g 3 (x)) ∗ 3g 2 (x) ∗ g 0 (x))¯x=2
= (f 0 (2)g(2) + f (2)g 0 (2) + cos (g 3 (2)) ∗ 3g 2 (2) ∗ g 0 (2))
= 4 ∗ 2 + 3 ∗ 1 + cos 8 ∗ 12
= 11 + 12 cos 8
4. Calculate the following integral
Z 2
√
0
t3
dt
t4 + 9
Solutions:
Z
2
t3
√
dt
t4 + 9
0
Z 2
d( 41 t4 )
=
1
4
0 (t + 9) 2
Z
1 2 d(t4 + 9)
=
4 0 (t4 + 9) 21
Z
1
1 2 4
(t + 9)− 2 d(t4 + 9)
=
4 0
¯t=2
1¯
1
4
= ∗ 2 ∗ (t + 9) 2 ¯¯
4
t=0
=1
2
5. If G(x) =
R −x
0
f (−t)dt, Find G’(x).
Solutions:
Z
−x
G(x) =
f (−t)dt
0
Z
−x
=−
f (−t)d(−t)
0
Z
=−
−x
f (u)du
(let u = −t)
0
It follows that:
d(−x)
f (−x)
dx
= −(−1)f (−x)
= f (−x)
G0 (x) = −
Keywords:
Problem 2: Trigonometry, Limits
Problem 3: Product Rule, Chain Rule
Problem 4: Antiderivative, Substitution Rule for Definite Integrals,
Change of Variables
Problem 5: Chain Rule, Change of Variables,
Fundamental Theorem of Calculus
3