Download Solutions to the first midterm

Math 252 Midterm 1
March 5, 2015
Help sheets are at the end.
(15)
1. A bacteria culture starts with 500 bacteria and after 4 hours there are
8000 bacteria.
a. Find the expression for the number of bacteria after t hours.
Solution: The general equation is A(t) = A0 ek t = A0 (ek )t .
Clearly A0 = 500 so A(4) = 8000 give 8000 = 500(ek )4 . Solving for ek (rather than k) gives ek = 161/4 = 2. So the answer
is
A(t) = 500 · 2t
b. Find the number of bacteria after 6 hours.
32, 000
c. At what time will the population be 256000?
t=9
(15)
2. Find each of the following limits, if it exists. If the limit does not exist,
write “does not exist”.
x2 − x − 2
x→2 x4 − 8x
a. lim
cos(x) − 1 + x2 /2
x→0
x4
b. lim
1
c. lim(1 + t) 2t
t→0
Solution:
1
1
(1 + t) 2t = eln((1+t) 2t ) = e
ln(1+t)
2t
By l’Hospital’s rule the limit of the exponent is 1/2, so the answer
is
e1/2
(20)
3. Differentiate
x3
a. ln √
x2 + 3
b. sec(etan(x) )
c. (arcsin(x3 ))4
d. xx
Solution:
x
xx = eln(x ) = ex ln(x)
So
(xx )0 = ex ln(x) (1 + ln x) = xx (1 + ln x)
(60)
4. Evaluate the following integrals. Hint: Sometimes integrals involving
tan’s and sec’s can be made easier by converting to sin’s and cos’s.
Z 4
ln t
√ dt
a.
t
1
Solution: Use integration by parts with u = ln t and dv =
The answer is
4 ln 4 − 4
Z
b.
dt
√
.
t
sin2 x cos5 x dx
Solution: Substituting u = sin x (so du = cos x dx) and using
cos2 x = 1 − sin2 x transforms this to the integral
Z
u2 (1 − u2 )2 du
Z
c.
0
∞
x2 dx
(1 + x2 )7/2
People had trouble with the limits of integration. You use trig
substation with θ = arctan(x). But note that when x goes to ∞,
θ goes to π/2. We convert to sin’s and cos’s and the substitute
u = sin(θ) as in the last part. The u-limits are 0 to 1. So we get
Z
0
∞
x2 dx
=
(1 + x2 )7/2
Z
π/2
0
Z
tan2 (θ) sec2 (θ dθ
sec7 (θ)
π/2
=
sin2 (θ) cos3 (θ) dθ
Z0 1
u2 (1 − u2 ) du
0
1
3
5
= u /3 − u /5 =
0
2
1 1
= − =
3 5
15
Z
d.
x2 ex dx
Z
e.
√
dx
Hint: Write (16 + x2 )2 as ( 16 + x2 )4 and use trig substitution.
2
2
(16 + x )
Solution: The indicated trig substitution gives
Z
Z
1
4 sec2 (θ) dθ
1
=
(θ + sin(θ) cos(θ)) + C
cos2 (θ) =
4
4
4 sec (θ)
64
128
Using the triangle from the sheet, we change back to x to get the
answer
1
1
x
arctan(x/4) +
·
+C
128
32 16 + x2
Z
f.
1
∞
6x2
dx
+ 7x + 2
Solution: 6x2 + 7x + 2 = (3x + 2)(2x + 1). The partial fractions
decomposition is
6x2
2
3
1
=
−
+ 7x + 2
2x + 1 3x + 2
Integrating this gives two logs. We need to put these together to
take the limit:
ln(2x + 1) − ln(3x + 2) = ln(
So the answer is
2x + 1
ln
3x + 2
2x + 1
)
3x + 2
∞
= ln(2/3) − ln(3/5) = ln(10/9)
1
Integrals
Z
sin2 (x) dx =
x 1
− sin(x) cos(x) + C
2 2
Z
cos2 (x) dx =
x 1
+ sin(x) cos(x) + C
2 2
Z
1
n−1
n−1
sin (x) dx = − cos(x) sin (x) +
sinn−2 (x) dx
n
n
Z
Z
n−1
1
n−1
n
cosn−2 (x) dx
cos (x) dx = cos (x) sin(x) +
n
n
Z
sec(x) dx = ln | sec(x) + tan(x)| + C
Z
n
Z
sec2 (x) dx = tan(x) + C
Z
sec3 (x) dx =
Z
Z
Z
√
1
sec(x) tan(x) + ln | sec(x) + tan(x)| + C
2
dx
= arcsin(x/a) + C,
− x2
a2
a>0
dx
= (1/a) arctan(x/a) + C,
a2 + x 2
a>0
dx
= (1/a) arcsec(|x/a|) + C,
x x 2 − a2
√
|x| > a > 0
Trigonometric Substitutions
Math 121 Calculus II
D Joyce, Spring 2013
Now that we have trig functions and their inverses, we can use trig subs. They’re special
kinds of substitution that involves these functions. For these, you start out with an integral
that doesn’t have any trig functions in them, but you introduce trig functions to evaluate the
integrals. These depend on knowing
(1) the Pythagorean identities
sin2 θ + cos2 θ = 1
sec2 θ = 1 + tan2 θ
(2) the definitions
sin θ
cos θ
tan θ =
(3) the derivatives
sec θ =
(sin θ)′ = cos θ
(tan θ)′ = sec2 θ
1
cos θ
(cos θ)′ = − sin θ
(sec θ)′ = sec θ tan θ
There are three kinds
You use√them when you see as part of the integrand
√
√ of trig subs.
one of the expressions a2 − x2 , a2 + x2 , or x2 − a2 , where a is some constant. In each
kind you substitute for x a certain trig function of a new variable θ. The substitution will
simplify the integrand since it will eliminate the square root. Here’s a table summarizing the
substitution to make in each of the three kinds.
If use see use the sub
√
and
a2 − x2
x = a sin θ
dx = a cos θ dθ
√
a2 + x2
x = a tan θ
dx = a sec2 θ dθ
√
x2 − a2
x = a sec θ
dx = a sec θ tan θ dθ
√
√
√
so that
a2 − x2 = a cos θ
a2 + x2 = a sec θ
x2 − a2 = a tan θ
In each line, the last entry follows from the second entry by one of the Pythagorean identities.
There are also right triangles you can draw to make the connections between x, a, and θ.
The three triangles below refer to the three trig subs, respectively.
a ✚✚
✚
✚
✚θ
✚
√
✚
✚
✚
✚
x
√
✚
a2 + x2✚
✚
✚
✚
✚
✚
✚ θ
a2 − x2
✚
✚
x ✚✚
x
✚
✚
✚
a
✚
✚θ
a
1
✚
✚
✚
√
x 2 − a2