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Universidad Carlos III de Madrid
Escuela Polit´ecnica Superior
Department of Mathematics
CALCULUS I
First Year Degree in Biomedical Engineering Final Exam January 12, 2016
Time: 3 hours
- No electronic devices —including calculators—, books, or notes can be used in the exam.
- All answers must be properly justified —otherwise they will not be considered.
- Respond exclusively to what you are being asked. Anything else that you add may be detrimental to you.
Problem 1. (2 points)
Determine for which values x > 0 the series
∑^ ∞
n=
xn (1 + x)(1 + x^2 ) · · · (1 + xn) converges.
Solution: Let us apply the quotient test. Since
an = x
n (1 + x)(1 + x^2 ) · · · (1 + xn) >^0 ,
the quotien test requires computing the limit
nlim→∞^ an+ an
= x
n+ (1 + x)(1 + x^2 ) · · · (1 + xn)(1 + xn+1)
(1 + x)(1 + x^2 ) · · · (1 + xn) xn
= lim n→∞
x (1 + xn+1) =
x, if x < 1, 1 2 ,^ if^ x^ = 1, 0 , if x > 1.
Whichever the cases, this limit is always smaller than 1, hence the series coverges for any x > 0.
Problem 2. (2 points)
Given the function
f (x) =
ex^ − 1 − x x^2 ,^ x <^0 ,
a + b
∫ (^) x
0
e−t^4 dt, x > 0 ,
calculate a and b so that it is continuous and differentiable.
Solution: If the function must be continuous at 0 then
lim x→ 0 −^
f (x) = lim x→ 0 +^
f (x).
But
lim x→ 0 −^
f (x) = lim x→ 0 e
x (^) − 1 − x x^2 = lim x→ 0
1 + x + x^2 /2 + o(x^2 ) − 1 − x x^2 = lim x→ 0
x^2 /2 + o(x^2 ) x^2 = lim x→ 0
[
2 +^ o(1)
]
lim x→ 0 +^
f (x) = lim x→ 0
a + b
∫ (^) x
0
e−t^4 dt
= a + b
0
e−t^4 dt = a.
Hence a = 1/2. Now, for the function to be differentiable at x = 0 it must hold
lim x→ 0 −
f (x) − f (0) x =^ xlim→ 0 +
f (x) − f (0) x.
Since f (0) = 1/2,
lim x→ 0 −
f (x) − f (0) x
= lim x→ 0
ex^ − 1 − x x^2
x
= lim x→ 0 e
x (^) − 1 − x − x (^2) / 2 x^3 = lim x→ 0
1 + x + x^2 /2 + x^3 /6 + o(x^3 ) − 1 − x − x^2 / 2 x^3 = lim x→ 0
x^3 /6 + o(x^3 ) x^3 = lim x→ 0
[
]
=^1
lim x→ 0 +
f (x) − f (0) x
= lim x→ 0
2 +^ b
∫ (^) x
0
e−t^4 dt − (^12) x
= lim x→ 0 b x
∫ (^) x
0
e−t^4 dt = b d dx
(∫ (^) x
0
e−t^4 dt
x= = be−x^4
∣x=0 =^ b.
Therefore b = 1/6.
Here is a shorter alternative. We can Taylor expand both functions to first order. On the one hand
ex^ = 1 + x + x
2 2
3 6
therefore ex^ − 1 − x x^2 =
2 +^
x 6 +^ o(x). On the other hand if g(x) =
∫ (^) x
0
e−t^4 dt
then g(0) = 0, g′(x) = e−x^4 and g′(0) = 1, so
g(x) = x + o(x),
therefore a + b
∫ (^) x
0
e−t
4 dt = a + bx + o(x).
If f (x) has to be continuous and differentiable at x = 0 both expansions must coincide up to first order, hence we obtain the same values for a and b.
Problem 4. (2 points)
Calculate the area of the figure delimited by the curves y = log(x + 7), y = 2 log(x + 1), and the Y axis.
Solution: This is the plot of the two curves and the region whose area we are asked to compute:
The curve y = log(x + 7) cuts the Y axis (x = 0) at y = log 7 > 0. The curve y = 2 log(x + 1) cuts the Y axis at y = 0, so the former is above the latter. They meet at the solution of
log(x + 7) = 2 log(x + 1) ⇔ x + 7 = (x + 1)^2 ⇔ x^2 + x − 6 = 0.
The two roots are x = 2 and x = −3, but the latter is outside the domain of the second curve, so x = 2 is the meeting point. The area is therefore
A =
0
[
log(x + 7) − 2 log(x + 1)
]
dx.
Now, with the change of variable t = x + a, ∫ log(x+a) dx =
log t dt = t log t−t+c′^ = (x+a) log(x+a)−x−a+c′^ = (x+a) log(x+a)−x+c,
where c′^ and c are arbitrary constants (we can choose, e.g., c = 0 for simplicity). Therefore
A =
[
(x + 7) log(x + 7) − x − 2(x + 1) log(x + 1) + 2x
]∣∣
2 0 =
[
(x + 7) log(x + 7) − 2(x + 1) log(x + 1) + x
]∣∣
2 0 = 9 log 9 − 6 log 3 + 2 − 7 log 7 = 9 log 9 − 3 log 9 + 2 − 7 log 7 = 6 log 9 + 2 − 7 log 7.
Problem 5. (2 points)
Plot the function f (x) =
1 + e^3 x 1 − e^2 x^.
hint: 2 + 3t − t^3 = (1 + t)^2 (2 − t).
Solution: The domain of this function excludes the point 1 − e^2 x^ = 0, i.e., x = 0 —which is a vertical asymptote. In R − { 0 } the function is continuous and differentiable infinitely often. Growth is determined by the sign of
f ′(x) =^3 e
3 x(1 − e 2 x) + 2e 2 x(1 + e 3 x) (1 − e^2 x)^2
=^3 e
3 x (^) − 3 e 5 x (^) + 2e 2 x (^) + 2e 5 x (1 − e^2 x)^2
= e
2 x(2 + 3ex (^) − e 3 x) (1 − e^2 x)^2
and following the hint
f ′(x) = e
2 x(1 + ex) (^2) (2 − ex) (1 − e^2 x)^2
All factors are positive except 2 − ex. If 2 > ex, i.e., x < log 2, the function f (x) will increase; if 2 < ex, i.e., x > log 2, the function will decrease. Therefore it has got a maximum at x = log 2.
Finally,
x→−∞lim f^ (x) =^ x→−∞lim^ 1 +^ e
3 x 1 − e^2 x^ = 1,
so y = 1 is a horizontal asymptote when x → −∞. But on the other hand,
x→lim+∞ f^ (x) =^ x→lim+∞^ 1 +^ e
3 x 1 − e^2 x^
= (^) x→lim+∞ ex^ e
− 3 x (^) + 1 e−^2 x^ − 1
and as a matter of fact, f (x) = −ex^ + o(ex) (x → ∞).
From all these features the plot of the function turns out to be
