Sample Midterm Exam 1 Answers
(1) Solve for
Z
x
1
3
1
dt = 2
t
Z
x
1
1
dt
t
We first simplify the left hand side of the equation.
Since
Z
Z
b
f (x)dx =
x
1
3
Since
Z
1
dt =
t
c
Z
1
1
3
1
dt +
t
Z
b
a
Z
x
1
3
Z
x
1
1
dt
t
a
f (x)dx = −
Hence
f (x)dx
a
Z
b
f (x)dx +
a
We have
Z
c
f (x)dx
b
1
dt = −
t
Z
1
1
3
1
dt +
t
Z
1
x
1
dt
t
By the definition of natural log function, we have
Z x
1
1
dt = − ln ( ) + ln x
1
t
3
3
Z x
1
dt = ln 3 + ln x
1
t
3
The right hand side of the original equation is 2 ln x, by the definition
of the nature log function.
So that finally we have
ln 3 + ln x = 2 ln x
ln x = ln 3
x=3
(2) (a) Since its derivative is positive.
(b)
1
15
r
(3) (a) y =
3
4−x
x−2
1
(b) y =
(4) (a) xx
2 − x5
1
x5 − 1
x +x
(ln x + (ln x)2 +
1
)
x
(b) covered in class
(c) (ln x2 )2x+3 (2 ln (2 ln x) +
2x + 3
)
x ln x
(5) (a) e2
1
(b)
e
(c) e3
(6) (a) y0 is the initial population, whereas k is the growth rate.
5
(b) 10000 · 2 2
(c) 10 days
(7) See lecture notes or refer to book page 368
(8) See lecture notes or refer to book page 369
(9) (a)
x2 sin x − 2 sin x + 2x cos x + C
(b)
−
(c)
1 ln x
1
−
+C
2013
2013 x
(2013)2 x2013
1 2
1
1
y arctan y − y + arctan y + C
2
2
2
(d)
x sinh x − cosh x
(e)
(801)2 (x ln2 x − 2x ln x + 2x) + C
(f)
1
1
((t − 1)927 t −
(t − 1)928 ) + C
927
928
(g)
t arctan
1 1
+ ln (t2 + 1) + C
t
2
(10) (a)
3 ln (x2 + x − 5) + C
(b)
ln (ex+3 + 1)
e
(c)
− arctan (ln x) + C
(d)
1
2014x
arctan (
)+C
5 ln (2014)
5