Calculus 2 - Examination
Concepts that you need to know:
Two methods for showing that a function is 1 − 1 :
a) Showing the function is monotonic.
b) Assuming that f (x1 ) = f (x2 ) and showing x1 = x2 .
Horizontal Line Test: A function is 1 − 1 (and has an inverse) iff no
horizontal line intersects the graph of f more than once.
Finding inverses of functions:
1) Switch x ’s and y ’s.
2) Solve for y .
Properties of f and f −1 :
1)
2)
3)
4)
5)
Domain of f = range of f −1 .
Range of f = domain of f −1 .
f (f −1 (x)) = x for all x in domain of f −1 .
f −1 (f (x)) = x for all x in domain of f .
The graphs of f and f −1 are symmetric about y = x .
6) If g = f −1 with f and g differentiable, then g 0 (x) =
1
f 0 (g(x))
.
The graphs of y = ex and y = ln x .
Properties of log x :
1) log(xy)
= log x + log y
x
2) log
= log x − log y
y
3) log xp = p log x
NOTE: The previous three properties work for any base including ln x .
Change of Base Formula:
loga x =
ln x
ln a
2
Properties of ln x and ex :
1) eln Allison = Allison .
2) ln(eAllison ) = Allison .
Logarithmic Differentiation:
a) Take natural logs of both sides.
b) Differentiate implicitly with respect to x .
c) Solve for y entirely in terms of x .
Exponential Growth and Decay:
y(t) = y(0)ekt
y(t) - Population at time t .
y(0) - Population at time 0 .
k - Growth or decay constant.
e - Rumored to be just a number. ≈ 2.71828
Inverse Trig Functions:
The graphs of y = sin−1 x , y = cos−1 x , and y = tan−1 x .
How to compute things similar to sin(arccosx) .
Indeterminate Forms and l’Hôpital’s Rule:
The two forms where l’Hôpital’s Rule is applicable - ∞
1
∞
0
and
.
0
∞
How to manipulate the other forms 0 · ∞ , ∞ − ∞ , 00 , ∞0 , and
into the two forms where you can use l’Hôpital’s Rule.
Rates of growth:
How to determine if a function grows faster or slower than another
function.
Integration by Parts:
R
R
u dv = uv − v du
Goal: To choose u and dv so that the product of v and du is
as simple as possible to antidifferentiate.
How to integrate by parts multiple times in one problem.
3
Trig Integrals:
Knowing the three pythagorean identities:
sin2 x + cos2 x = 1
tan2 x + 1 = sec2 x
1 + cot2 x = csc2 x
Other:
sin2 x = 12 (1 − cos 2x)
cos2 x = 12 (1 + cos 2x)
sin 2x = 2 sin x cos x
Goal 1: To find du . Once you identify du , you can use the trig
identities to substitute for whatever remains.
NOTE: Usually you want to keep the trig functions raised to even
powers and pull off one sin x or one cos x to become du . Sometimes
when presented with even powers, you can try pulling off a sec2 x or
csc2 x .
Trig Substitution:
NOTE: Try a non-trig substitution first...
If you see something like 4 − 9x2 you should think about sin θ .
If you see something like 9x2 − 4 you should think about sec θ .
If there is only addition like 9x2 + 4 you should think about tan θ .
NOTE: Remember that if you are working with an indefinite integral that is in terms of x , you have to label/create a TRIANGLE to
get everything back in terms of x at the end.
Integration by Partial Fractions:
Be able to rewrite a fraction using partial fractions. (Non-repeating
linear factors only.)
Be able to identify what A , B , C , etc. are by picking convenient
values for x .
Be able to integrate each piece once you have decomposed the
original fraction. With linear factors, each piece should integrate to
ln |something| .
4
DERIVATIVES THAT YOU NEED TO KNOW!!!
y = xn
y 0 = nxn−1
y = ex
y 0 = ex
y = eu
y 0 = eu ·
y = ax
y 0 = ax ln a where a > 0 and a 6= 1
y = au
y 0 = au ln a ·
y = ln x
y0 =
y = ln u
y0 =
du
dx
du
where a > 0 and a 6= 1
dx
1
x
y = loga x
1 du
·
u dx
1
y0 =
x ln a
y = loga u
y0 =
y = sin x
y 0 = cos x
y = cos x
y 0 = − sin x
y = tan x
y 0 = sec2 x
y = cot x
y 0 = − csc2 x
y = sec x
y 0 = sec x tan x
y = csc x
y 0 = − csc x cot x
y = sin−1 x
y0 = √
y = sec−1 x
y0 =
1
√
|x| x2 − 1
y = tan−1 x
y0 =
1
1 + x2
y = sin−1 u
y0 = √
y = sec−1 u
y0 =
1
du
|u| u2 − 1 dx
y = tan−1 u
y0 =
1 du
1 + u2 dx
1
du
·
u ln a dx
1
1 − x2
1
du
1 − u2 dx
√
5
INTEGRALS THAT YOU NEED TO KNOW!!!
R
xn dx
R dx
x
xn+1
+ C provided n 6= −1
n+1
ln |x| + C
R
ex dx
ex + C
R
ax dx
ax
+ C provided a > 0 and a 6= 1
ln a
R
cos x dx
sin x + C
R
sin x dx
− cos x + C
R
sec2 x dx
tan x + C
R
√
1
dx
1 − x2
sin−1 x + C
R
1
√
dx
|x| x2 − 1
sec−1 x + C
R
x2
1
dx
+1
tan−1 x + C
Direct substitution is the easiest method and should be tried first!!!
You should know the following forms:
If you see something involving:
Transform it by introducing:
# − #x2
sin θ
#x2 − #
sec θ
# + #x2
tan θ
6
Sample Exam 1 (1-12)
1. Given f (x) = x2 − 1 , x > 0 . Let g(x) = f −1 (x) and H(x) = g 3 (x) .
Then H 0 (8) =
R
x
2. Determine the integral
dx .
4
x +1
3. Find lim (1 + ax)1/x , where a is a fixed non-zero number.
x→0
4. The derivative of y = ln(sin−1 x) , where 0 < x ≤ 1 , is
5. Solve ln x = ln(10) − ln(x + 3) for x .
6. (P-C) Apply logarithmic differentiation to find the derivative of
f (x) = x(x cos x) .
R 2 x3
7. (P-C) Find
x e dx .
R1 √
1 − x2 dx .
8. (P-C) Use trignometric substitution to evaluate 0
R
9. (P-C) Find
sin−1 x dx .
R
10. (P-C) Find
sec4 x dx .
1
11. (P-C) After 11 days, a sample of radon-222 decayed to
of its
8
original amount. What is the half-life of radon-222 in days? Simplify as
much as possible.
12. (P-C) Find
R
x2
10
dx .
− 3x − 4
Sample Exam 2 (13-25)
13. log7 x written in terms of natural logarithms is
−1 1
14. Find the exact value of sin tan
.
x
15. If f is an increasing function, then what is true about f −1 ?
The choices were: A) increasing B) decreasing C) differentiable D)
continuous E) not a function
ex + e−x − 2
.
x→0
x sin(2x)
16. Evaluate lim
17. Find the derivative of f (x) = x sin−1 (ex ) .
R
x
18. After trig substitution, the integral
dx becomes
(4 − x2 )5/2
7
19. Evaluate
20. Evaluate
lim (e2x − 2)1/x .
x→∞
R
π
2
0
x cos x dx .
21. (P-C) Differentiate f (x) =
xx
.
cos2 x
22. (P-C) Evaluate
R sin3 θ
dθ .
cos2 θ
23. (P-C) Evaluate
R
x2
dx .
x2 + 4
24. (P-C) Same as # 9. :)
25. (P-C) The half-life of radioactive cobalt is 5.27 years.
a) If a sample has a mass of 400 mg, find a formula for the function
y(t) which represents the mass of the sample in mg that remains after
t years.
b) When will the mass of the sample be reduced to 20 mg?
Additional problems from other exams/books (26-29):
26. (P-C) Evaluate
27. Evaluate
x2 sin x dx.
lim (1 + sin x)1/x .
x→0+
28. (P-C) Evaluate
29. Evaluate
R
R
π
2
0
R
sin3 x cos4 x dx .
sin x
dx .
cos x − 2
8
Answers:
1.
H 0 (x) = 3[g(x)]2 · g 0 (x)
H 0 (8) = 3[g(8)]2 · g 0 (8)
1
H 0 (8) = 3[g(8)]2 · 0
f (g(8))
To go any further, we need to know g(8) . Since g = f −1 , I know that to
find g(8) I just have to find something that I can put into the function
f that produces 8 . Since f (3) = 8 , I know that g(8) = 3 .
So
H 0 (8) = 3[3]2 ·
1
f 0 (3)
Since f 0 (x) = 2x , we have f 0 (3) = 6 and
H 0 (8) = 27 ·
1
27
9
=
= .
6
6
2
2. The first thing that I noticed about this integral is that the deR
1
du is
nominator was something squared +1 . I know that
u2 + 1
tan−1 u + C , so I want to manipulate this into that form.
Z
x
dx
x4 + 1
Let u = x2 . Then du = 2x . Introduce a 2 and a
1
2
to get
1 R 2x
1 R du
1
1
−1
dx
=
=
tan
u
+
C
=
tan−1 (x2 ) + C
2 x4 + 1
2 u2 + 1
2
2
3. Anytime that you have a variable in the exponent remember that
you can use the properties of logarithms to pull it in front of the logarithm. With this problem, you have to identify the form first.
The form is 1∞ which is indeterminate. I have to set it equal to y
and then take the natural log of both sides to try to express it in a form
where l’Hôpital’s Rule is applicable.
y = lim (1 + ax)1/x
x→0
ln y = lim ln(1 + ax)1/x
x→0
1
ln(1 + ax)
x→0 x
ln(1 + ax)
ln y = lim
x→0
x
ln y = lim
9
The form is now
0
so I can use l’Hôpital’s Rule.
0
a
ln y = lim 1 + ax
x→0
1
ln y = a
y = ea
4. y 0 =
−1
sin
1
√
x 1 − x2
5.
ln x = ln 10 − ln(x + 3)
10
ln x = ln
x+3
10
x=
x+3
2
x + 3x = 10
x2 + 3x − 10 = 0
(x + 5)(x − 2) = 0
2 is the only solution as −5 is not in the domain of ln x .
6.
y = xx cos x
ln y = ln xx cos x
ln y = x cos x ln x
x cos x
1 0
y = (cos x − x sin x) ln x +
y
x
0
y = y [(cos x − x sin x) ln x + cos x]
y 0 = xx cos x [(cos x − x sin x) ln x + cos x]
7.
Z
2 x3
x e
1
dx =
3
Z
3
3x2 ex dx
Let u = x3 . Then du = 3x2 dx and we have
1
3
Z
eu du =
1 u
1 3
e +C = ex +C
3
3
10
8. Since I have an integral of something of the form a2 − u2 , I want
to try a trig substitution involving sin θ . It would be p
nice if x2 = sin2 θ
√
because√then I would be able to replace
1 − x2 with
1 − sin2 θ which
is just
cos2 θ or cos θ . That is how I choose what to substitute.
2
Let x = sin2 θ . Then x = sin θ and dx = cos θ dθ .
Z
1
p
1−
x2
0
Z
dx =
π/2
cos θ cos θ dθ
0
Since this is a deinite integral, I decided to change the limits of integration. If you plug 0 in for x in the equation x = sin θ , we find that
θ = 0 . Plugging in 1 for x yields that θ = π/2 .
Z
π/2
2
Z
π/2
cos θ dθ =
0
0
π/2
1
1
1
(1 + cos(2θ)) dθ = θ + sin(2θ)
2
2
4
0
If this were an indefinite integral, I would have to switch everything
back into x by using a right triangle where sin θ = x . I would also need
the identity sin(2θ) = 2 sin θ cos θ . Since the integral I have is a definite
integral and I have already changed the limits of integration, I do not
switch back. I just plug in π/2 and 0 and subtract.
π
π
+ 0 − (0) =
4
4
9.
Z
sin−1 x dx
There are five main methods of integration that you have learned.
When I look at a problem, I try the following order:
1.
2.
3.
4.
5.
Integration by
Integration by
Trig integral.
Integration by
Integration by
substitution.
partial fractions.
trig substitution.
parts.
The middle three techniques should be fairly obvious. Also, the order
of the middle three is not important. The important part is that you
should try Integraion by substitution FIRST and Integration by parts
LAST.
Attacking this problem:
1. The only option for integration by substitution is to let
1
, this won’t
u = sin−1 x . As there is no du which would be √
1 − x2
work.
2. There is no fraction. If there was, I would see if I could factor the denominator so that I could use integration by partial fractions.
11
3. Trig integrals consist of the six trig functions sin x, cos x, . . . ,
NOT their inverses.
4. There is nothing of the form a2 − u2 , u2 − a2 , or u2 + a2 ,
so integration by trig substitution is OUT.
5. I have to integrate this by parts...
u = sin−1 x
v=x
1
du = √
dx
v = dx
1 − x2
Z
Z
−1
−1
sin x dx = x sin x −
√
x
dx
1 − x2
Now, I have another integral to evaluate, so I start back at integration
by substitution. Here it works. If I choose u = 1 − x2 , then du = −2x .
−1
By introducting a
and a −2 , we have
2
Z
x
−1
√
dx =
2
1 − x2
Z
−2x
−1
√
dx =
2
1 − x2
Z
u−1/2 du =
−1 u1/2
=
2 1/2
p
= − 1 − x2 + C
Replacing this into the earlier computation gives us
Z
p
sin−1 x dx = x sin−1 x + 1 − x2 + C
10.
Z
Z
4
sec x dx =
2
sec2 x sec
| {zx dx}
du
so u = tan x
and
Z
=
(u2 + 1) du
=
11.
sec2 x = tan2 x + 1
u3
tan3 x
+u+C =
+ tan x + C
3
3
y(t) = y(0)ekt
1
y(0) = y(0)e11k
8
1/8 = e11k
ln(1/8) = 11k
(ln(1/8))/11 = k
12
So the original equation becomes
1
y(t) = y(0)e(t ln 8 )/11
t/11
y(t) = y(0)eln( 8 )
t/11
1
y(t) = y(0)
8
1
We want the half-life, so
t/11
1
1
y(0) = y(0)
2
8
t/11
1
1
=
2
8
1
1
t
ln =
ln
2
11
8
11 ln 12
=t
ln 81
12.
Z
→
−11 ln 2
=t
−3 ln 2
10
=
2
x − 3x − 4
Z
11
=t
3
→
10
(x − 4)(x + 1)
10
A
B
can be written in the form
+
,
(x − 4)(x + 1)
x−4 x+1
with A and B real numbers.
A
B
10
=
+
. By multiplying both sides by
We have
(x − 4)(x + 1)
x−4 x+1
(x − 4)(x + 1) , we obtain
I know that
10 = A(x + 1) + B(x − 4).
This is true for any value of x . So I pick two convenient values for x .
Let x = 4 . Then we have
10 = A(5) + 0
so
A=2
10 = 0 + B(−5)
so
B = −2
10
dx =
(x − 4)(x + 1)
Z Let x = −1 . Then we have
Then
Z
2
−2
+
x−4 x+1
which is
2 ln |x − 4| − 2 ln |x + 1| + C
dx
13
13. ln7 x =
ln x
ln 7
−1
14. sin tan
{z
|
J
1
x
tan−1
= sin(J)
1
=J
x
→
tan J =
}
So sin(J) = √
15.
1
f 0 (g(x))
g 0 (x) =
1
x
1
f 0 (g(x))
Since
f
is always increasing,
1
.
1 + x2
f 0 > 0 , so
> 0 and g must be increasing.
16.
lim
x→0
ex + e1x − 2 H
ex − e−x
lim
x sin 2x = x→0 sin 2x + 2x cos 2x
ex + e−x
H
1+1
1
lim
=
=
= x→0 2 cos 2x + 2 cos 2x − 4x sin 2x
2+2−0
2
17. f 0 (x) = sin−1 (ex ) + x p
ex
1 − (ex )2
18. Although trig substitution is not the best way to do this problem,
since i have 4 − x2 , I would very much like the x2 to be 4 sin θ because
then it would just be 4(1 − sin2 θ) = 4 cos2 θ . So I set them equal to each
other.
x2 = 4 sin2 θ
x = 2 sin θ
dx = 2 cos θ dθ
Z
19.
x
2 sin θ(2 cos θ) dθ
dx =
2
5/2
(4 − x )
(4 cos2 θ)5/2
Z
Z
4 sin θ cos θ dθ
1
=
=
32 cos5 θ
8
y = lim (e2x − 2)1/x
x→∞
ln(e2x − 2)
x→∞
x
2x
2e
2x
ln y = lim e − 2
x→∞
1
ln y = 2
ln y = lim
y = e2
sin θ
dθ
cos4 θ
14
20.
Z
x cos x dx
u=x
du = dx
v = sin x
v = cos x dx
Z
Z
x cos x dx = x sin x −
21.
sin x dx = x sin x + cos x + C
xx
cos2 x
xx
ln y = ln
cos2 x
y=
ln y = ln xx − ln(cos2 x)
ln y = x ln x − 2 ln(cos x)
1 0
2
y = ln x + 1 −
(− sin x)
y
cos x
2 sin x
0
y = y ln x + 1 +
cos x
x
x
y0 =
[ln x + 1 + 2 tan x]
cos2 x
22.
Z
sin3 θ
dθ = −
cos2 θ
Z
sin2 θ
(− sin θ dθ)
cos2 θ | {z }
du
2
1−u
du
u2
Z 1
=−
− 1 du
u2
1
=+ +u+C
u
= sec θ + cos θ + C
Z
=−
23.
Z
x2
dx =
x2 + 4
x2 + 4
4
− 2
2
x +4 x +4
Z 4
=
1− 2
dx
x +4
Z
4
=x−
dx
2
x +4
Z dx
15
To compute this second integral, we need to recognize the need for
a trig substitution. I see this because of the x2 + 4 . I would like this
to become 4 tan2 θ + 4 = 4(tan2 θ + 1) = 4 sec2 θ , so I make the following
substitutions:
x2 = 4 tan2 θ
x = 2 tan θ
Z
dx = 2 sec2 θ dθ
Z
4
4
dx
=
2 sec2 θ dθ
x2 + 4
4 sec2 θ
Z
=2
dθ
= 2θ + C
x
+C
2
Replacing this into what was above, we obtain
Z
x2
x
dx = x − 2 tan−1 + C
2
x +4
2
= 2 tan−1
24. Same as #9.
25.
y(t) = y(0)ekt
1/2 = e5.27k
ln(1/2) = 5.27k
ln(1/2)
=k
5.27
y(t) = 400et(ln(1/2))/5.27
t/5.27
y(t) = 400eln(1/2)
y(t) = 400(1/2)t/5.27
20 = (1/2)t/5.27
t
ln 20 =
ln(1/2)
5.27
5.27 ln(1/20)
=t
ln(1/2)
16
26.
Z
x2 sin x dx
u = x2
v = − cos x
du = 2xdx
v R= sin xdx
= −x2 cos x − R −2x cos x dx
= −x2 cos x + 2 x cos x dx
u=x
du = dx
R
= −x2 cos x + 2 x sin x − sin x dx
= −x2 cos x + 2x sin x + 2 cos x + C
27.
v = sin x
v = cos xdx
y = lim+ (1 + sin x)1/x
x→0
ln y = lim
x→0+
ln y = lim
x→0+
ln(1 + sin x)
x
cos x
1 + sin x
1
ln y = 1
y=e
28.
Z
3
Z
4
sin x cos x dx = −
sin2 x cos4 x(− sin x dx)
| {z }
du
Z
=−
Z
=−
(1 − u2 )u4 du
(u4 − u6 ) du
u7
u5
=− +
+C
5
7
1
1
= − cos5 x + cos7 x + C
5
7
29.
Z
0
π/2
sin x
dx
cos x − 2
The first thing that you should try when integrating is direct substitution. It is the easiest of the methods that we have. For this problem,
I choose u = cos x − 2 . Then du is − sin x dx
By introducing two negatives, we have
Z
−
0
π/2
− sin x
dx = −
cos x − 2
Z
−2
−1
−2
du
= − ln |u| = − ln | − 2| − (− ln | − 1|) = − ln 2
u
−1