SOLUTIONS TO EXAM 1, MATH 10560
1. The function
f (x) = x3 + ln(x) + 1
is a one-to-one function (there is no need to check this). What is (f −1 )0 (2)?
Solution:
We would like to use the formula
1
(f −1 )0 (y) = 0 −1
.
f (f (y))
This means we need to compute f 0 and f −1 (2). By inspection we see f −1 (2) = 1. Now
f 0 (x) = 3x2 + x1 . So
1
1
1
(f −1 )0 (2) = 0 −1
= 0
= .
f (f (y))
f (1)
4
1
4
(f −1 )0 (2) =
2. Solve for x in the following equation:
log2 (2x + 1) − log2 (2x − 1) = 2.
Solution:
Combining logs we have
log2
2x + 1
2x − 1
=2.
Now exponentiate
2x + 1
= 22 = 4 .
2x − 1
Cross multiply
2x + 1 = 4(2x − 1) = 4 · 2x − 4 .
Combine like terms
3 · 2x = 5 .
Divide by 3
2x =
5
.
3
Take log2
5
x = log2
.
3
5
x = log2
3
1
2
SOLUTIONS TO EXAM 1
d
3. Find the derivative of dx
(x2 + 1)cos(x) .
Solution:
Let y = x2 + 1)cos(x) . If we use logorithmic differentiation we have
dy
d
=y
ln(x2 + 1)cos(x) ) .
dx
dx
d
Let us concentrate on dx
ln(x2 + 1)cos(x) ). Now
(0.1)
ln(x2 + 1)cos(x) ) = cos(x) ln(x2 + 1) .
Therefore using the product rule and chain rule we have
d
d
2x cos(x)
ln(x2 + 1)cos(x) ) =
cos(x) ln(x2 + 1) = − sin(x) ln(x2 + 1) +
.
dx
dx
x2 + 1
Combining this with (0.1) we have
dy
2x cos(x)
2
= (x2 + 1)cos(x)
−
sin(x)
ln(x
+
1)
.
dx
x2 + 1
d 2
cos(x)
2
cos(x) 2x cos(x)
2
(x + 1)
= (x + 1)
− sin(x) ln(x + 1)
dx
x2 + 1
Z
4. Compute
0
1/2
earcsin x
√
dx.
1 − x2
Solution:
Let u = arcsin(x) then du =
substitution
Z 1/2
0
√ 1
,
1+x2
u(0) = 0, and u(1/2) = π/6. Under this
earcsin x
√
dx =
1 − x2
Z
π/6
eu du
0
π/6
=e u
0
= eπ/6 − 1 .
Z
0
1/2
earcsin x
√
dx = eπ/6 − 1
2
1−x
5. Evaluate the derivative of :
ln(arctan(2x )).
Solution:
(arctan(2x ))0
arctan(2x )
1
1
=
(2x )0
arctan(2x ) 1 + (2x )2
2x ln 2
=
(1 + 22x ) arctan(2x )
ln(arctan(2x ))0 =
ln(arctan(2x ))0 =
2x ln 2
(1 + 22x ) arctan(2x )
SOLUTIONS TO EXAM 1
3
6. Determine the following limit.
lim x e1/x − 1 .
x→∞
Solution:
lim x e
1/x
x→∞
− 1 = lim
e1/x − 1
1
x
− 12 e1/x
lim x 1
x→∞
− x2
1/x
lim e
=1
x→∞
x→∞
(L0 Hospital)
=
=
lim x e1/x − 1 = 1
x→∞
7. Evaluate the integral
π/3
Z
x cos(3x)dx.
0
Solution: Set u = x and v =
parts.
Z
0
π/3
1
3
sin(3x) , then dv = cos(3x)dx. Use integration by
π/3
1
x cos(3x) dx = x sin(3x)
−
3
0
π/3
1
= x sin(3x)
+
3
0
2
=−
9
Z
Z
1 π/3
sin(3x) dx
3 0
π/3
1
cos(3x)
9
0
π/3
x cos(3x) dx = −
0
π
4
Z
8. Find
2
9
tan3 x sec4 x dx.
0
Solution:
Recall the identities (tan x)0 = sec2 x and sec2 x = 1 + tan2 x. We can rewrite the
integral as follows:
Z
3
4
tan x sec xdx =
Z
3
2
2
tan x sec x(sec xdx) =
Z
tan3 x(1 + tan2 x)(sec2 xdx).
4
SOLUTIONS TO EXAM 1
Let us use the substitution u = tan x with du = sec2 xdx. The limits of integration
are u = tan π4 = 1 and u = tan 0 = 0. Our integral becomes
Z π
Z 1
4
tan3 x(1 + tan2 x) sec2 xdx =
u3 (1 + u2 )du
0
0
Z 1
=
(u3 + u5 )du
0
u4 u6
=
+
4
6
1 1
= +
4 6
5
= .
12
1
0
To sum up,
π
4
Z
tan3 x sec4 x dx =
0
9. Evaluate the integral
√
Z
5
.
12
2p
4 − x2 dx.
0
Solution:
Let us use the trigonometric substitution x = 2 sin θ, where |θ| ≤ π2 . Now dx =
√
2 cos θdθ and θ = arcsin( x2 ). The limits of integration are θ = π4 (for x = 2) and θ = 0
(corresponds to x = 0). We are ready to compute the integral
Z √2 p
Z πp
4
2
4 − x dx =
4 − 4 sin2 θ 2 cos θdθ
0
0
π
4
Z
=
4 cos2 θdθ
0
Z
π
4
1 + cos 2θ
dθ
2
0
π
4
1
= 2 · θ + sin 2θ
2
0
π π 1
=2
+ sin 2
4 2
4
π 1
=2
+
4 2
π
= + 1.
2
=4
Hence the answer is
√
Z
0
2p
4 − x2 dx =
π
+ 1.
2
SOLUTIONS TO EXAM 1
5
10. Which of the following expressions gives the partial fraction decomposition of the
function
x2 + 3x + 1
f (x) =
?
(x − 1)(x2 − 1)(x2 + 1)
Solution:
Note that x2 − 1 = (x − 1)(x + 1). Then
x2 + 3x + 1
x2 + 3x + 1
=
.
(x − 1)(x2 − 1)(x2 + 1)
(x + 1)(x − 1)2 (x2 + 1)
These denominators contribute to the partial fraction decomposition as follows:
1
A
1
B
C
1
Dx + E
as
,
as
+
and 2
as 2
.
2
2
x+1
x + 1 (x − 1)
(x − 1) (x − 1)
x +1
x +1
Let’s get it together
x2 + 3x + 1
A
B
C
Dx + E
=
+
+
+ 2
.
(x + 1)(x − 1)2 (x2 + 1)
x + 1 (x − 1) (x − 1)2
x +1
11. A sample of a Cobalt-57 has an initial mass of 6 grams. Let M (t) denote the mass
of the sample after t days, M (t) decreases at a rate that is proportional to the amount
of the substance present at time t. That is
M 0 (t) = kM (t).
Cobalt-57 has a half-life of 272 days.
(a) Give a formula for M (t). (Solve for all unknown constants).
(b) How long (how many days) will it take for the sample to decrease from 6 grams
to 2 grams?
Solution:
For part (a), we recognize from the condition M 0 (t) = kM (t) that M (t) should be an
exponential (decay) function (think about the derivatives of exponentials and what they
look like), so it will have the form
M (t) = M0 e−kt
where M0 = 6 is the initial mass (in grams), t is time (in years), and k is a constant to
be determined. Since we are told the half life is 272 days, we have
0.5M0 = M0 e−k(272) ⇐⇒ ln(0.5) = −k · 272
0.5
which gives k = − ln272
. Hence our function is
M (t) = 6e
ln(0.5)
·t
272
For part (b), we need to solve for the value of t in
2 = 6e
ln(0.5)
·t
272
⇐⇒ ln(1/3) =
ln(1/2)
ln(1/3)
· t ⇐⇒ t = 272
272
ln(1/2)
or alternately
t = 272
ln(3)
ln(2)
6
12. Evaluate
SOLUTIONS TO EXAM 1
Z
2
dx .
(x + 1)(x2 + 1)
Solution: We approach this problem using the method of partial fractions. Notice
that the denominator of our integrand is already factored in irreducible factors, as x2 + 1
has no real roots and cannot be factored further. First we write the partial fraction
decomposition of our integrand:
2
A
Bx + C
=
+ 2
(Bx + C is one degree less than x2 + 1)
2
(x + 1)(x + 1)
x+1
x +1
A(x2 + 1) + (Bx + C)(x + 1)
=
(recombine into one fraction)
(x + 1)(x2 + 1)
and equating numerators gives us
2 = A(x2 + 1) + (Bx + C)(x + 1)
Plugging in x = 1 immediately gives A = 1 . To find B and C we must expand the RHS
and equate powers of x:
0x2 + 0x + 2 = Ax2 + A + Bx2 + Bx + Cx + C
= (A + B)x2 + (B + C)x + (A + C)
On the RHS the coefficient of x2 is 0 = (A + B), so B = −1 . Looking at the constant
term, A + C = 2, so C = 1 . Hence our original integral can be rewritten as
Z
Z 2
1
−x + 1
dx =
+ 2
dx
(x + 1)(x2 + 1)
x+1
x +1
Z
Z
Z
1
x
1
=
dx −
dx +
dx
x+1
x2 + 1
x2 + 1
After making the mental substitution u = x2 + 1 in the second integral, we get
Z
2
1
dx = ln |x + 1| − ln(x2 + 1) + arctan(x) + C
2
(x + 1)(x + 1)
2
13. Compute
Z
√
x2
1
dx
− 6x + 10
Solution:
Regular substitution doesn’t work here: if we let u = x2 − 6x + 10, there is no factor
on top that would be a scalar multiple of the differential du = (2x − 6)dx. Instead we
try trigonometric substitution, but to figure out the appropriate substitution, we must
first complete the square in the denominator:
Z
Z
Z
dx
dx
dx
√
√
p
=
=
2
2
x − 6x + 10
x − 2 · 3x + 9 + 1
(x − 3)2 + 1
and (x−3)2 +1 resembles the trig identity tan2 (x)+1 = sec2 (x), so we let x−3 = tan(θ),
which gives dx = sec2 (θ)dθ, and the entire denominator becomes
q
p
p
(x − 3)2 + 1 = tan2 (x) + 1 = sec2 (x) = sec(θ)
SOLUTIONS TO EXAM 1
7
After the substitution, our original integral becomes
Z
Z
dx
sec2 (θ)dθ
√
=
sec(θ)
x2 − 6x + 10
Z
= sec(θ)dθ
= ln | sec(θ) + tan(θ)| + C
Now, since x − 3 = tan θ, the sec(θ) term inside the log can be expressed as a function of
x by using an appropriate right triangle whose legs have lengths x−3 and 1, respectively.
Hence our integral equals:
Z
p
dx
2
√
= ln x − 6x + 10 + x − 3 + C
2
x − 6x + 10