Question 5.3 #66. Evaluate the limit by first recognizing the sum
as a Riemann sum for a function defined on [0,1].
1
n→∞ n
lim
q
1
n
+
q
2
n
n q
P
i
lim n1
n
n→∞
i=1
n q
P
i 1
lim
n·n
n→∞ i=1
n
P
=
=
=
lim
f (xi )·∆x
n→∞ i=1
R1√
xdx
R01 1/2
x
dx
0
i1
x3/2
(3/2)
0
2
3/2
− 03/2
3 1
=
=
=
+
q
3
n
+ ... +
where f (x) =
pn
√
n
x
=
= 2/3
Question 5.3 #73. Find a function f and a number a such that
Z x
√
f (t)
dt = 2 x, for, all x > 0.
6+
2
t
a
Rx
Hints. The a should look familiar from the statement of the fundamental
theorem of calculus, and so you might think to differentiate both sides. What
do you obtain? Then plug the result back into the original equation... The
solution is below, but try first to do it with the hint.
Solution.
d
dx
6+
Rx
f (t)
dt
a t2
=
d
dx
√
(2 x)
0 + f (x)/x2 = 2 · 21 · x−1/2 = x−1/2
f (x) = x−1/2 x2
f (x) = x3/2
Now plug f (x) = x3/2 back into the left hand side of the given equation:
Rx
LHS = 6 + a ft(t)
2 dt
R x t3/2
= 6 + Ra t2 dt
x
= 6 + a t−1/2
i dt
1/2 x
t
= 6 + 1/2
√ a
= 6 + 2√t]xa √
= 6 + 2 x − 2 a.
Since
√
RHS = 2 x,
1
we have
√
√
6 + 2 x − 2√ a
6−2 a
√6
a
a
2
=
=
=
=
=
√
2 x
0√
2 a
3
9.