MTH4100 Exercise sheet 5 SOLUTIONS
Q1(a) Use l’Höpital’s rule to calculate the following limits:
√
1 − cos(6x)
x2 + 12 − 4
(ii) lim
(i) lim
x→0
x→2
x−2
36x2
(b) Estimate the area bounded by the graph of f (x) = x3 , the x-axis and the line
x = 1 by partitioning the interval [0, 1] into
Pn n equal subintervals of width ∆ = 1/n,
and calculating the upper sum Un = k=1 f (ck )∆. Determine limn→∞ Un and
R1
compare it with 0 x3 dx.
Solution (a)(i) Applying l’Hopital’s rule gives
√
x(x2 + 12)−1/2
x2 + 12 − 4
lim
= lim
= 1/2 .
x→2
x→2
x−2
1
(ii) Applying l’Hopital’s rule twice gives
6 sin(6x)
36 cos(6x)
1 − cos(6x)
= lim
= lim
= 1/2 .
2
x→0
x→0
x→0
36x
72x
72
lim
[6]
(b) Since f (x) is increasing on [0, 1] we take ck to be the right hand end-point of
the k’th subinterval. Since the intervals have equal width 1/n, the k’th subinterval
, nk ] and we have ck = nk . Hence
is [ k−1
n
n
X
n
X
n
1
1 X 3 n2 (n + 1)2
(n + 1)2
Un =
f (ck )∆ =
f (k/n) = 4
k =
=
.
n
n k=1
4n4
4n2
k=1
k=1
We have
(n + 1)2
(1 + 1/n)2
=
lim
= 1/4
n→∞
n→∞
4n2
4
lim Un = lim
n→∞
and
Z
1
x3 dx = [x4 /4]10 = 1/4 .
0
[4]
Q2(a) Suppose that f (x) is differentiable for all values of x, and that f (1) = 0
and f 0 (1) < 0. Which of the following statements must be true for the function
Z x
F (x) =
f (t)dt ?
0
Give reasons for your answers.
(i) F is a twice-differentiable function of x.
(ii) F and dF/dx are both continuous.
(iii) F has a local maximum at x = 1.
Z π/6
(b) Find
d
cos(t3 ) dt .
3x
dx √
Solution (a) (i) We have F 0 (x) = f (x) so F is differentiable. Since f is differentiable, we have F 00 (x) = f 0 (x) so F is twice differentiable.
[1]
(ii) F and F 0 are both continous since they are both differentiable.
[1]
0
00
0
(iii) F (1) = f (1) = 0 so x = 1 is a critical point of F . Since F (1) = f (1) < 0,
x = 1 is a local maximum of F .
[2]
√
du
1 −2/3
3
and
(b) Let u = x. Then dx = 3 x
!
Z π/6
Z π/6
d
du
d
3
3
cos(t
)
dt
=
cos(t
)
dt
3x
dx √
du u
dx
Z u
1 −2/3 d
3
cos(t ) dt
= − x
3
du π/6
1
= − x−2/3 cos(u3 )
3
1
= − x−2/3 cos(x) .
3
[6]
Q3(a) Find
Z q
1 + sin2 (x − 1) sin(x − 1) cos(x − 1)dx .
(b) Find the area enclosed by the two curves y = x2 − 2 and y = 2.
Solution (a) Put u = 1 + sin2 (x − 1). Then du = 2 sin(x − 1) cos(x − 1)dx. Hence
Z q
Z
1 √
1
1
2
1 + sin (x − 1) sin(x−1) cos(x−1)dx =
udu = u3/2 +C = (1+sin2 (x−1))3/2 +C .
2
3
3
[5]
(b)
2
-2
Curves intersect when x2 − 2 = 2 so x = ±2. We have
Z 2
Area =
2 − (x2 − 2) dx = [4x − x3 /3]2−2 = 2(8 − 8/3) = 32/3 .
−2
[5]
Question for discussion in exercise class
(a)Let f be a function defined on an interval [a, b].
(i) Explain what is meant by a Riemann sum for f over [a, b] with respect to the
partition of [a, b] into n equal subintervals.
(ii) Explain what it means to say that f is integrable on [a, b].
(b) Calculate the upper sum Un for f (x) = x2 on [0, 1] with respect to the partition
of [a, b] into n equal subintervals. Determine the limit of Un as n → ∞.
P
(a)(i) A Riemann sum is an expression of the form ni=1 f (ci ) n1 where ci is some
point in the i’th subinterval.
(ii) f is integrable if all Riemann sums tend to the same limit as n → ∞.
(b) The upper sum Un is defined by taking ci to be a point in the i’th subinterval
such that f (ci ) is as large as possible. Since f (x) = x2 is increasing on [0, 1], we
will take ci to be the right hand endpoint of the i’th subinterval i.e. ci = i/n.
Hence
n
n 2
n
X
1 X i
1
1 X 2 n(n + 1)(2n + 1)
Un =
f (ci ) =
= 3
i =
n
n n
n i=1
6n3
i=1
i=1
and
2 + n3 +
n(n + 1)(2n + 1)
lim
= lim
n→∞
n→∞
6n3
6
1
n2
=
1
3